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An electrician, a bank worker, a plumber and so on all have tools of their trade. Without these tools,and a good working knowledge of how to use them, it would be impossible for them to perform theirjobs successfully.In the same way, mathematics practitioners have various tools of their trade. The most vital of thesetools is algebra, since it underpins so much of mathematics and its study.In this chapter, techniques of algebra (both old and new) are developed, so that you will be equippedwith this vital tool for your further study of mathematics.

AlgebraictechniquesAlgebraic

techniques

11

MMCAS34SB_2ed_01_3pp.fm Tuesday, June 23, 2009 3:44 PM

MathsWorld

Mathematical Methods CAS Units 3 & 4

2

Algebra of polynomials

Polynomials

A

polynomial

is an algebraic expression of the form

where

n

is a positive integer or zero and the coefficients

a

n

,

a

n

−

1

, …,

a

0

are real numbers. The term

a

n

x

n

,

a

n

≠

0, is called the

leading term

of the polynomial. The

degree

of the polynomial is

n

.

The remainder and factor theorems

The

remainder theorem

states that if a polynomial

P

(

x

) is divided by the linear expression(

x

−

a

) then the remainder is given by

P

(

a

).

If the remainder is 0 then (

x

−

a

) is a factor of

P

(

x

). The

factor theorem

states that if (

x

−

a

) is a factor of

P

(

x

) then

P

(

a

)

=

0 and, conversely, if

P

(

a

)

=

0 then (

x

−

a

) is a factor of

P

(

x

).

Dividing by (

bx

−

c

) is the same as dividing by . In this case the remainder is

given by .

Example

1

If (

x

−

1) is a factor of

P

(

x

)

=

3

x

3

+

kx

2

+

bx

−

5 and

P

(

−

2)

=

−

45, find the values of

k

and

b

. Hence find the remainder when

P

(

x

) is divided by (

x

+

1).

Solution

(

x

−

1) is a factor of

P

(

x

) so

P

(1)

=

0.

3

+

k

+

b

−

5

=

0

k

+

b

=

2 (1)

P

(

−

2)

=

−

45

−

24

+

4

k

− 2b − 5 = −454k − 2b = −162k − b = −8 (2)

3k = −6 (1) + (2)k = −2

Substitute k = −2 into equation (1).−2 + b = 2

b = 4Hence P(x) = 3x3 − 2x2 + 4x − 5.The required remainder is given by P(−1).

P(−1) = −3 − 2 − 4 − 5= −14

P x( ) anxn an 1– xn 1– … a1x a0+ + + +=

b x cb---–⎝ ⎠

⎛ ⎞

Pcb---⎝ ⎠

⎛ ⎞

TipThe Solve command of a CAS can be used to solve for b and k.

Define P(x) then solve P(1) = 0 and P(−2) = −45 for b and k.

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1 Algebraic techniques

3

Long division of polynomials and factorisationLong division is a useful technique, particularly for factorising polynomials by hand.

Example 2Factorise:

a x3 + 4x2 − 7x − 10 b x3 − 7x + 6

Solutiona Let P(x) = x3 + 4x2 − 7x − 10.

Use the factor theorem to find a factor.

P(−1) = (−1)3 + 4(−1)2 − 7(−1) − 10= 0

So (x + 1) is a factor.

Now use long division to find the other factor.

x2 + 3x − 10

x + 1 ) x3 + 4x2 − 7x − 10

x3 + x2

3x2 − 7x3x2 + 3x

−10x − 10−10x − 10

0

So, P(x) = x3 + 3x2 − 4x − 10= (x + 1)(x2 + 3x − 10)= (x + 1)(x + 5)(x − 2)

b Let P(x) = x3 − 7x + 6.

Use the factor theorem to find a factor.

P(1) = (1)3 − 7(1) + 6= 0

So (x − 1) is a factor. Now use long division to find the other factor.

x2 + x − 6

x − 1 ) x3 + 0x2 − 7x + 6

x3 − x2

x2 − 7xx2 − x

−6x + 6−6x + 6

0

So, P(x) = x3 − 7x + 6= (x − 1)(x2 + x − 6)= (x − 1)(x + 3)(x − 2)

TipIf the coefficient of the leading term of P(x) is 1, when using the factor theorem to find a factor of P(x) (i.e. finding a value of a for which P(a) = 0), the value of a must be a factor of the constant term. In this case you would only test the values 1, −1, 2, −2, 5, −5 and 10, −10.

TipWhen performing long division it is recommended to keep the same powers of x in vertical columns. In some cases it may be necessary to include zero terms to make this easier.

1.1


MathsWorld Mathematical Methods CAS Units 3 & 4

4

Other techniques that do not involve the use of long division can also be used to factorise cubic polynomials.

Equality of polynomialsTwo polynomials P1(x) and P2(x) such that

P1(x) = anxn + an − 1xn − 1 + an − 2xn − 2 + … + a1x + a0

P2(x) = bnxn + bn − 1xn − 1 + bn − 2xn − 2 + … + b1x + b0

are equal for all values of x if and only if

an = bn, an − 1 = bn − 1, …, a1 = b1, a0 = b0.

Example 3Factorise:

a x3 + 5x2 + 6x b x3 + 4x2 − 4x − 16

Solutiona x3 + 5x2 + 6x

= x(x2 + 5x + 6) (taking out a common factor first)

= x(x + 3)(x + 2)

b x3 + 4x2 − 4x − 16

= x2(x + 4) − 4(x + 4) (grouping terms)

= (x + 4)(x2 − 4)

= (x + 4)(x + 2)(x − 2)

Example 4Find the values of a, b and k for which 8x3 + (a − 6)x2 + (b − a)x = kx3 for all values of x.

Solution8x3 + (a − 6)x2 + (b − a)x = kx3

8x3 + (a − 6)x2 + (b − a)x = kx3 + 0x2 + 0x

Equating coefficients gives

k = 8 (x3)a − 6 = 0 (x2)

a = 6b − a = 0 (x)

b = a= 6

TipThe Factor command of a CAS can be used to factorise polynomials such as those in example 3.

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5

exercise 1.11 Find the remainder when 2x3 + 3x2 − 4x + 1 is divided by (x + 2).

2 Use the remainder theorem to find the remainder when 2x4 − x2 + 3x − 2 is divided by (2x − 1).

3 When x4 + ax2 + bx − 2 is divided by (x − 1) the remainder is −2, and when it is divided by (x + 1) the remainder is −6. Find the values of a and b.

4 x3 + ax2 + 4x − 5 and x2 + 4x − 8 both have the same remainder when divided by (x − 1). Find the value of a.

5 The polynomial x3 + px2 + qx + 6 is exactly divisible by (x + 2) and (x − 3). Find the values of p and q.

6 Factorise:

a x3 − 2x2 − 11x + 12 b x3 + x2 − x − 1 c 3x3 + x2 − 10x

d x3 − 2x2 − 9x + 18 e 6x3 + x2 − 5x − 2 f 4x3 − 4x2 − 15x + 18

g 2x3 − 18x2 + 54x − 54 h 3x3 + 9x2 − 18x − 24

7 If x3 + 4x2 − 3x + 2 = (x + 1)(ax2 + bx + c) + d for all values of x, find the values of a, b, c and d.

8 If x3 + 4x2 − 12x + 14 = x3 + (mx + n)2 + 5 for all values of x, find the values of m and n.

9 a If x2 = a(x + 2)2 + b(x + 2) + c for all values of x, find the values of a, b and c.

b If 3x2 − 12x + 11 = A(x − h)2 + k, find the values of A, h, and k.

Example 5Factorise 2x3 − 5x2 − 4x + 3 without using long division.

SolutionUse the factor theorem to find one factor.

P(−1) = 2(−1)3 − 5(−1)2 − 4(−1) + 3= 0

So (x + 1) is a factor. The other factor must be quadratic (i.e. of the form ax2 + bx + c).

(x + 1)(ax2 + bx + c) = 2x3 − 5x2 − 4x + 3

Expanding gives

ax3 + bx2 + cx + ax2 + bx + c = 2x3 − 5x2 − 4x + 3ax3 + (a + b)x2 + (b + c)x + c = 2x3 − 5x2 − 4x + 3

Equating coefficients of like powers of x:

a = 2, c = 3a + b = −52 + b = −5

b = −7So, 2x3 − 5x2 − 4x + 3 = (x + 1)(2x2 − 7x + 3)

= (x + 1)(2x − 1)(x − 3)

Tipa and c can be found by inspection before expanding. Then it is only necessary to determine either thex2 or x term in the expansion ofthe left hand side of the equationin order to find b.

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6

Solution of polynomial equationsSolving an equation means finding the value(s) of the variable that makes the equation true.

Linear equationsA linear equation in one variable is an equation involving polynomials whose degree is 1. Thus 2x + 5 = 3 − 2x is a linear equation in the variable x.

Some equations reduce to linear equations when suitably transformed. For example, the

equation can be expressed as x = 7(2x − 3).

Linear equations are usually solved by isolating the variable on one side of the equation.

Quadratic equationsThe standard form of a quadratic equation is ax2 + bx + c = 0, where x is the variable and a, b and c are constants with a ≠ 0. Methods for solving quadratic equations include factorisation, completing the square, using the quadratic formula and numerical or graphical approximation. When quadratic equations are expressed as the product of linear factors, the null factor law can be used to find solutions.

Example 6Solve for x the equation 2(2x − 3) − 5 = 3(x + 2) + 5.

SolutionExpand the bracketed terms and simplify.

4x − 11 = 3x + 11

Isolate the variable by subtracting 3x from both sides and adding 11 to both sides.

x = 22Check that x = 22 is the correct solution.LHS = 2(44 − 3) − 5 = 77; RHS = 3(22 + 2) + 5 = 77 = LHS

Example 7

Solve for x.

SolutionClear the fractions by multiplying both sides by (2x − 1)(x + 2). Thus:

3(x + 2) = 2(2x − 1)3x + 6 = 4x − 2

8 = xHence x = 8.

Null factor lawIf ab = 0, then either a = 0 or b = 0 or both a and b = 0.

x2x 3–--------------- 7=

TipThe Solve command of a CAS can be used as shown.

32x 1–--------------- 2

x 2+------------=

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7

Example 8Solve each of the following for x.

a 3x2 − 48 = 0 b 2x2 = 15 − xc x2 + 6x − 3 = 0 d 3x2 + 2x − 6 = 0

Solutiona 3x2 − 48 = 0

3(x2 − 16) = 0 (common factor)

3(x − 4)(x + 4) = 0 (difference of two squares)

x = 4, −4 (using the null factor law)

b 2x2 = 15 − x

2x2 + x − 15 = 0 (taking all terms to one side)

(2x − 5)(x + 3) = 0 (factorising by inspection)

x = (using the null factor law)

c x2 + 6x − 3 cannot be factorised by inspection so complete the square.

x2 + 6x − 3 = 0

x2 + 6x + 9 − 9 − 3 = 0

(x + 3)2 − 12 = 0

= 0 (since )

So x = .

d 3x2 + 2x − 6 cannot be factorised by inspection so use the quadratic formula.

If ax2 + bx + c = 0 then .

Here, a = 3, b = 2 and c = −6.

x =

=

=

=

52--- 3–,

x 3 2 3–+( ) x 3 2 3+ +( ) 12 2 3=3– 2 3 3– 2 3–,+

TipTo avoid errors when cancelling, taking out a common factor in the numerator may be advantageous.

=

=

=

2– ± 2 196

------------------------------

2 1– ± 19( )6

-----------------------------------

1– ± 193

--------------------------

xb– ± b2 4ac–

2a-----------------------------------------=

2– ± 4 4 3( ) 6–( )–6

----------------------------------------------------

2– ± 766

--------------------------

2– ± 2 196

------------------------------

1– ± 193

--------------------------

TipThe Solve command of a CAS can be used to find exact or approximate solutions to a quadratic equation such as that in example 8 part d. Press the appropriate key combination to obtain the numerical approximations.

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8

Equations reducible to quadraticsSome equations are reducible to quadratics, as shown in the following examples.

Example 9

Solve for x: .

Solution

Let so the equation becomes a2 − 5a + 6 = 0.

Now factorise and solve:

(a − 3)(a − 2) = 0a − 3 = 0 or a − 2 = 0

So, or .

x2 − 3x − 1 = 0 or x2 − 2x − 1 = 0

Using the quadratic formula:

x = or

x = or

Example 10Solve for x: .

SolutionIsolate the square root term to one side and then square both sides.

(subtract 2 from both sides)x2 − 6x + 9 = 3x − 5 (square both sides; note (−1)2 = 1)

x2 − 9x + 14 = 0(x − 2)(x − 7) = 0

x = 2, 7Check:

x = 2: LHS = 2 − 1 = 1; RHS = 2 − = 1 = LHS

x = 7: LHS = 7 − 1 = 6; RHS = 2 − = −2 ≠ LHS

So the only solution is x = 2.

x 1x---–⎝ ⎠

⎛ ⎞ 2

5 x 1x---–⎝ ⎠

⎛ ⎞– 6+ 0=

x 1x---–⎝ ⎠

⎛ ⎞ 2

5 x 1x---–⎝ ⎠

⎛ ⎞– 6+ 0=

a x 1x---–=

x 1x---– 3– 0= x 1

x---– 2– 0=

3 ± 3–( )2 4 1( ) 1–( )–2

----------------------------------------------------------- x2 ± 2–( )2 4 1( ) 1–( )–

2-----------------------------------------------------------=

3 ± 132

----------------------- x 1 ± 2=

x 1– 2 3x 5––=

TipWhen solving equations involving surds, it is necessary to always check solutions in the original equation.

x 3– 3x 5––=

1

16



9

Equations involving higher degree polynomialsA polynomial equation has the form anxn + an − 1xn − 1 + … + a1x + a0 = 0. Techniques for solving linear (degree 1) and quadratic (degree 2) equations have already been considered. With cubic (degree 3) and higher degree polynomial equations it is necessary to use either factorisation or technology to find solutions.

Example 11Solve each of the following for x, giving exact solutions where possible.

a x3 + 3x2 − 5x − 15 = 0 b x5 − 3x3 + 4x = 0 c x3 = x − 1

Solution a x3 + 3x2 − 5x − 15 = 0

Factorising the left-hand side by grouping gives:

x2(x + 3) − 5(x + 3) = 0(x + 3)(x2 − 5) = 0

= 0x =

b x5 − 3x3 + 4x = 0

Factorising gives:

x(x4 − 3x2 + 4) = 0x(x2 − 4)(x2 + 1) = 0

x(x − 2)(x + 2)(x2 + 1) = 0x = 0, 2, −2

c x3 = x − 1

x3 − x + 1 = 0

However, x3 − x + 1 does not have a rational linear factor, so the equation is solved using technology. A graphical approach can be used.

Plot the graph of y = x3 − x + 1 and find any zeros or x-intercepts using the appropriate ‘jump-to’ feature of a CAS. From the screenshot, there is exactly one solution, x = −1.325, correct to 3 decimal places. (An alternative graphical method is to plot the graphs of y = x3 and y = x − 1 and find any points of intersection.)

Alternatively, the Solve command of a CAS can be used as shown.

TipThe factor theorem could also be used to identify the factor (x + 3).

x 3+( ) x 5–( ) x 5+( )3– 5 5–, ,

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10

exercise 1.110 Solve each of the following for x.

a 2(x + 1) = 3(x + 2) − 14 b 9(3x + 4) − 55x = 64

c d

e (x + 2)2 = (x − 1)2 + 6 f (x + 1)(x − 2) − 2(x − 2) = (x + 1)2

11 Find all the exact solution(s) to each of the following equations. Check the solution(s) with your CAS.

a (x + 3)(x − 3) + 5 = 0 b 4x2 + 6x − 2 = 0 c x2 + 6x + 7 = 0

d 5x2 − 16x − 12 = 0 e (x + 5)2 = 15 − 10x f 2x4 = 32x2

g x4 + 5x2 − 36 = 0 h i x3 − 2x2 − 7x − 4 = 0

j 2x3 − 3x2 − 11x + 6 = 0 k x4 − x3 + x − 1 = 0 l x6 = 8 − 7x3

m x + 1 = n 1 + = −x o

(Hint: In part o, square both sides, taking care with the left-hand side, then isolate the remaining square root term and square both sides a second time.)

12 Use your CAS to solve each of the following equations, giving answers correct to 3 decimal places.

a x2 − 2x − 5 = 0 b 2x2 = 5 − 2x c x3 − 3x2 + x − 1 = 0

d x4 − 4x2 − 2 = 0 e x3 = −3 + x f x6 = 5 + 2x3

Binomial expansions Pascal’s trianglePascal’s triangle is named after the French mathematician Blaise Pascal. Although Pascal was not responsible for its discovery (it was discovered by the Chinese about 500 years before his birth), he was the first mathematician to perform any extensive study of it.

Constructing Pascal’s triangle. At the top we place the number 1. This is referred to as row zero.

. At each end of every subsequent row, place the number 1.

. Each other element is found by adding the two numbers diagonally above it in the previous row.

continued

5x 6+4

--------------- 100 4x–3

--------------------- 6–= 6x--- 4+ 15

x------ 3–=

16 x3x---+⎝ ⎠

⎛ ⎞2

– 0=

4x 1+ x 3+ 5 x– x 2x 1–=–



11

Rows 0 to 6 of Pascal’s triangle are shown below.

For example, in row 4, the second element 4 is obtained by adding the numbers 1 and 3 diagonally above it. Similarly, the third element 6 is found by adding the numbers 3 and 3.

Pascal’s triangle and binomial expansionsConsider the expansion of (a + b)n for n = 0, 1, 2, 3, 4 and 5. Since the bracketed expression consists of two terms, it is said to be binomial and hence its expansion is known as a binomial expansion.

(a + b)0 = 1(a + b)1 = a + b(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

It is important to note the following features of each expansion.

. There are n + 1 terms in each expansion.

. In each term the sum of the powers of a and b is n.

. In successive terms, the powers of a decrease from n to 0 and the powers of b increase from 0 to n.

. The coefficients of the terms in each expansion relate to the rows of Pascal’s triangle as shown below.

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

(a + b)0 1

(a + b)1 1 1

(a + b)2 1 2 1

(a + b)3 1 3 3 1

(a + b)4 1 4 6 4 1

(a + b)5 1 5 10 10 5 1

1.1



12

Example 12Use Pascal’s triangle to expand:

a (x − 3)4 b (2x − 3)6 c

Solutiona To expand (x − 3)4 we obtain the coefficients from the 4th row of Pascal’s triangle.

So the expansion of (a + b)4 is (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4.

Here a = x, b = −3.

(x − 3)4 = x4 + 4x3(−3) + 6x2(−3)2 + 4x(−3)3 + (−3)4

= x4 − 12x3 + 54x2 − 108x + 81b To expand (2x − 3)6 we obtain the coefficients from the 6th row of Pascal’s triangle.

So the expansion of (a + b)6 is

(a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6

Here a = 2x, b = −3.

(2x − 3)6 = (2x)6 + 6(2x)5(−3) + 15(2x)4(−3)2 + 20(2x)3(−3)3 + 15(2x)2(−3)4 + 6(2x)(−3)5 + (−3)6

= 64x6 − 576x5 + 2160x4 − 4320x3 + 4860x2 − 2916x + 729

c To expand we obtain the coefficients from the 3rd row of Pascal’s triangle.

So the expansion of (a + b)3 is

(a + b)3 = a3 + 3a2b + 3ab2 + b3

Here a = 2x, b = .

=

=

2x 1x---–⎝ ⎠

⎛ ⎞ 3

1 4 6 4 1

1 6 15 20 15 6 1

2x 1x---–⎝ ⎠

⎛ ⎞ 3

1 3 3 1

1x---

2x 1x---–⎝ ⎠

⎛ ⎞ 32x( )3 3 2x( )2 1

x---–⎝ ⎠

⎛ ⎞ 3 2x( ) 1x---–⎝ ⎠

⎛ ⎞ 2 1x---–⎝ ⎠

⎛ ⎞ 3+ + +

8x3 12x–6x--- 1

x3-----–+

TipThe Expand command of a CAS can be used to obtain binomial expansions. For example the screenshot on the right shows the results for example 12 parts a and b.

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13

Expansion of (a + b)nRecall from Year 11 Combinatorics that . The elements in Pascal’s

triangle can be expressed as combinations, as shown below.

For example, and .

These agree with the values in row 5 of Pascal’s triangle (see page 11).

Using combinations, the expansion for a high power of (a + b)n can be found without having to write out all the previous rows of Pascal’s triangle.

Binomial expansionIn general:

(a + b)n = nC0anb0 + nC1an − 1b1 + … + nCran − rbr + … + nCn − 1a1bn − 1 + nCna0bn

= an + nC1an − 1b + … + nCran − rbr + … + nCn − 1abn − 1 + bn

where n ∈ N.

The (r + 1)th term, or general term, is given by nCr an − rbr, r = 0, 1, 2, …, n.

n

r⎝ ⎠⎛ ⎞ Cn

rn!

r! n r–( )!-----------------------= =

0

0⎝ ⎠⎛ ⎞

1

0⎝ ⎠⎛ ⎞ 1

1⎝ ⎠⎛ ⎞

2

0⎝ ⎠⎛ ⎞ 2

1⎝ ⎠⎛ ⎞ 2

2⎝ ⎠⎛ ⎞

3

0⎝ ⎠⎛ ⎞ 3

1⎝ ⎠⎛ ⎞ 3

2⎝ ⎠⎛ ⎞ 3

3⎝ ⎠⎛ ⎞

4

0⎝ ⎠⎛ ⎞ 4

1⎝ ⎠⎛ ⎞ 4

2⎝ ⎠⎛ ⎞ 4

3⎝ ⎠⎛ ⎞ 4

4⎝ ⎠⎛ ⎞

5

0⎝ ⎠⎛ ⎞ 5

1⎝ ⎠⎛ ⎞ 5

2⎝ ⎠⎛ ⎞ 5

3⎝ ⎠⎛ ⎞ 5

4⎝ ⎠⎛ ⎞ 5

5⎝ ⎠⎛ ⎞

5

1⎝ ⎠⎛ ⎞ C5

15!

1!4!---------- 5 4!×

4!-------------- 5= = = =

5

2⎝ ⎠⎛ ⎞ C5

25!

2!3!---------- 5 4× 3!×

2 3!×----------------------- 10= = = =

TipA CAS can be used to calculate a combination as shown in the following screenshot.TI-N

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14

exercise 1.113 State the number of terms in the expanded form of:

a (x + 4)7 b (2x − 5)10 c (1 − 4z)14

14 Use Pascal’s triangle to expand:

a (x + 3)5 b (x − 2)4 c (x − 1)5 d (2x + 1)6

e (3x − 2)5 f (1 − 3x)4 g (2t + 3y)4 h

15 Use (a + b)n = an + nC1an − 1b + nC2an − 2b2 + … + bn to expand:

a (2y − x)6 b (x2 − y)4 c (x2 + x)3 d (x2 − y2)5

e f g h

16 Find the 4th term in the expansion of each of the following.

a (x − 2)8 b (2x − 3t)4

Example 13Expand (3x + y)5 using (a + b)n = an + nC1an − 1b + nC2an − 2b2 + … + bn

Solution(3x + y)5 = (3x)5 + 5C1(3x)4y + 5C2(3x)3y2 + 5C3(3x)2y3 + 5C4(3x)y4 + y5

= 243x5 + 5(81x4)y + 10(27x3)y2 + 10(9x2)y3 + 5(3x)y4 + y5

= 243x5 + 405x4y + 270x3y2 + 90x2y3 + 15xy4 + y5

continued

2x 1x---–⎝ ⎠

⎛ ⎞ 6

x2--- 2+⎝ ⎠

⎛ ⎞4

x1x---+⎝ ⎠

⎛ ⎞5 2

x2----- x–⎝ ⎠

⎛ ⎞ 7

2x 1–( )6

In this analysis task, you will be asked to determine a general expression for the greatest coefficient in the expansion of (1 + x)n. Note that the binomial coefficients are all positive.

Part 1 Gathering the tools for analysisLet Tr + 1 be the (r + 1)th term and Tr be the rth term in the expansion of (1 + x)n for 1 ≤ r ≤ n.

a Write down general expressions for Tr + 1 and Tr.

b Find a general simplified expression for the ratio of the coefficients of Tr + 1 and Tr.

c Under what condition relating n and r must the coefficient of Tr + 1 always be greater than or equal to the coefficient of Tr?

Part 2d By using Pascal’s triangle and the results from the previous part, determine a general

expression for the greatest coefficient in the expansion of (1 + x)n. Explain your reasoning carefully and illustrate your findings with some specific examples.

Analysis task 1—the greatest coefficient

SAC

The greatest coefficientSAC analysis task

1.1



15

Expansions of binomial expressions are not limited to positive integer powers. The expansion of (1 + x)−1 can be used to approximate reciprocals of numbers close to 1; the expansion of can be used to approximate square roots of numbers close to 1.

Part 1(1 + x)–1 = 1 − x + x2 − x3 + …, where the nth term is (−x)n − 1

a Write down the next four terms in the series.

b i What is the value of x in this expansion to approximate ?

ii Use technology to evaluate and write down the result correct to 4 decimal places.

iii Use the binomial expansion above with 4 terms to approximate . To how many decimal places is this accurate?

iv How many terms are required to approximate correct to 4 decimal places?

Part 2

where the nth term is .

c Simplify this series expansion.

d What are the next three terms in this series in simplest form?

e Write down the value of x if this expansion is used to approximate:

i ii iii iv

v vi vii viii

f Complete the following table.

i In each case, how many terms are required for 4 decimal place accuracy?

ii As each extra term is added the approximation approaches the actual value. This is referred to as convergence. How does the value of x affect the rate of convergence?

x Value from technology

Approximation using

1 term 2 terms 3 terms 4 terms 5 terms 6 terms 7 terms

1 x+( )1 2⁄

11.3--------

11.3--------

11.3--------

11.3--------

1 x+( )1 2⁄ 112---x

12---⎝ ⎠

⎛ ⎞ 12---–⎝ ⎠

⎛ ⎞ x2

2----- 1

2---⎝ ⎠

⎛ ⎞ 12---–⎝ ⎠

⎛ ⎞ 32---–⎝ ⎠

⎛ ⎞ x3

6----- …,+ + + +=

12---⎝ ⎠

⎛ ⎞ 12---–⎝ ⎠

⎛ ⎞ 32---–⎝ ⎠

⎛ ⎞ 52---–⎝ ⎠

⎛ ⎞ … xn 1–

n 1–( )!------------------- n 1>,

(n − 1) terms

1.4 0.6 1.3 0.7

1.2 0.8 1.1 0.9

1 x+

1.4

0.6

1.3

0.7

1.2

0.8

1.1

0.9

Analysis task 2—approximations using binomial expansions

SAC

Approxim

ations using binomial expansions

SAC analysis task

1.2

1.1



16

Systems of simultaneous linear equations

A set of two or more equations in the same variables is referred to as a system of equations. In this section we will consider systems of linear equations.

Systems of linear equations in two unknownsRecall that solving simultaneous linear equations in two unknowns algebraically involves obtaining a single linear equation in one variable by eliminating the other variable.

Example 1Solve these simultaneous linear equations using an algebraic method:

y = 2x − 3 and 3x + 2y = 8

Solutiony = 2x − 3 (1)

3x + 2y = 8 (2)Substitute 2x − 3 for y in equation (2).

3x + 2(2x − 3) = 83x + 4x − 6 = 8

7x = 14x = 2

Substitute x = 2 in equation (1).

y = 2(2) − 3y = 1

So the solution is x = 2 and y = 1.

Example 2Solve these simultaneous linear equations:

x + 5y = −13 and 2x − y = 7

SolutionThese equations can be solved algebraically.

x + 5y = −13 (1)2x − y = 7 (2)

x + 5y = −13 (1)10x − 5y = 35 (3) = (2) × 5

11x = 22 (1) + (3)x = 2

Substitute x = 2 in equation (2).4 − y = 7

y = −3So the solution is x = 2 and y = −3.

TipIt is a good idea to check your answer. Substitute x = 2 and y = 1 into the left-hand side of equation (2). LHS = 3(2) + 2(1) = 8 = RHS.

1.21.2



17

Simultaneous linear equations can also be solved graphically by finding the point of intersection of the straight lines corresponding to each equation.

They can also be solved using matrix methods on a CAS. The command rref uses elimination methods to solve the equations. To use this method, first write each equation in the form ax + by = c. Then create a matrix whose elements are the coefficients of x and y and the constant terms in the equations.

x + 5y = −13

2x − y = 7

To solve the equations, apply the command rref to the matrix above. The result when rref is applied is shown in the screenshot at right.

The rref command has transformed the original pair of equations into:

1x + 0y = 20x + 1y = −3

So the solution is x = 2, y = −3.

(Further details and examples of the use of matrices in the solution of simultaneous linear equations are found on pages 21–23.)

TI-Nspire 2.4

ClassPad 2.4

[−10, 10] by [−6.7, 6.7]

1 5 13–

2 1– 7⇒

TI-Nspire 8.1, 8.2

ClassPad 8.1, 8.2

TipThe matrix used with the rref command can be stored and then used as shown. Alternatively, the Solve command can be used to solve simultaneous linear equations as shown.

TI-Nspire 8.1, 8.2, 10.2

ClassPad 8.1, 8.2, 10.2

1.2



18

Parallel and coincident linesThe graphs of the two equations are straight lines. If the graphs are of parallel lines, which do not intersect, there will be no solutions to the simultaneous equations. If the graphs are coincident, that is, the same straight line, there will be an infinite number of solutions to the simultaneous equations.

Example 3Solve the simultaneous equations 3x − 2y = 3 and 6x − 4y = k if:

a k = 6 b k = 0

Solutiona If k = 6, then:

3x − 2y = 3 (1)6x − 4y = 6 (2)6x − 4y = 6 (3) = (1) × 26x − 4y = 6 (2)

The graphs of these two equations are the same straight line, that is, the two lines are coincident. Any point on the first line will also lie on the second. There is an infinite number of solutions to the simultaneous equations consisting of the points lying on the line.

Let y = t. Then:

6x − 4t = 66x = 6 + 4t

x =

So, for any real value of t, and y = t satisfy the equations.

b If k = 0, then:

3x − 2y = 3 (1)

6x − 4y = 0 (2)

6x − 4y = 6 (3) = (1) × 26x − 4y = 0 (2)

Subtracting these equations gives 0 = 6.This is impossible, so these simultaneous equations have no solution. This is becausethe graphs of two original equations areparallel lines that never intersect.

123---t+

x 123---t+=

y

x

2

0

–2

–4

4

2–2–4 4

6x – 4y = 0

3x – 2y = 3



19

exercise 1.21 Solve each of the following pairs of simultaneous equations by hand then check your

answers using technology.

a 3x + 2y = 4 b = 4 c y = 14 − 4x

x − 3y = 5 = 10

3x + 2y = 8

d = 3 e = 11 f x = 3 − y

5x + 3y = 17 = 22 = 8

2 Find the value(s) of k for which the simultaneous equations have:

i no solution ii an infinite number of solutionsa x + y = 3 b x − y = k c 4x + 9y = k

2x + 2y = k −3x + 3y = 6 0.8x + 1.8y = 2

3 A straight line has equation y = mx + c. Use simultaneous equations to find the equation of the straight line that passes through the points (−2, 11) and (3, −4).

4 A parabola that passes through the points (−1, 11) and (2, 5) has equation y = x2 + bx + c. Use simultaneous equations to find the values of b and c.

Simultaneous linear equations in more than two unknowns Situations often arise in which there are three or more linear equations with corresponding numbers of unknowns. Such systems of equations can be solved by hand, eliminating one variable at a time until a single variable remains and solving the resulting linear equation. The values of the other variables can then be found by substitution. Systems of equations are more easily solved using technology.

Example 4Solve the following system of equations, using:

a algebra (by hand) b technology.

x − 2y + 3z = 3x + 4y + 2z = 42x − 2y + z = 4

Solutiona x − 2y + 3z = 3 (1)

x + 4y + 2z = 4 (2)2x − 2y + z = 4 (3)

2x − 4y + 6z = 6 (3) = (1) × 2x + 4y + 2z = 4 (2)

4x − 4y + 2z = 8 (4) = (3) × 2

32---x y–

12---x

34---y+

xy2---+ x

3--- y

2---+

x y3---–

13---x

23---y–

continued �

1.2



20

Now y can be eliminated.3x + 8z = 10 (5) = (3) + (2)5x + 4z = 12 (6) = (4) + (2)3x + 8z = 10 (5)

10x + 8z = 24 (7) = (6) × 27x = 14x = 2

Substitute x = 2 in equation (6) to find z.10 + 4z = 12

4z = 2

z =

Substitute x = 2 and in equation (2) to find y.

2 + 4y + 1 = 44y = 1

y =

The solution is x = 2, and .

b First create a matrix whose elements are the coefficients of the variables and the constant terms in the equations.

Now use the rref command.

This has transformed the equations into

1x + 0y + 0z = 2

0x + 1y + 0z =

0x + 0y + 1z =

Read off the solution: x = 2, and .

12---

z12---=

14---

y14---= z

12---=

1 2– 3 31 4 2 42 2– 1 4

TI-Nspire 8.1, 8.2

ClassPad 8.1, 8.2

TipThe matrix can be storedfirst, then rref applied tothe stored matrix.

14---

12---

y14---= z

12---=



21

Matrix representation of systems of linear equationsThe system of equations in example 4 can be written in the form of a matrix equation as follows.

x − 2y + 3z = 3

x + 4y + 2z = 4 ⇒

2x − 2y + z = 4

A X B

The matrix form on the right is written as AX = B, where A is the matrix formed by the coefficients of the variables, X is a column matrix containing the variables and B is a column matrix formed by the constants on the right hand side of the equations.

To recover the original equations from the matrix equation, start by multiplying the first row of A by the column matrix X and equate the result to the first entry in B:

(1)(x) + (−2)(y) + (3)(z) = 3, i.e. x − 2y + 3z = 3.

The remaining equations can be recovered in the same way.

When written in this form the equations can be solved on a TI-Nspire using the Simult command, accessible from the Matrix menu or the CATALOG. The screenshot at right shows how, with the matrices A and B above stored as a and b.

Of course, the system of equations can also be solved using the rref or Solve commands.

The role of parametersA system of linear equations in which there are fewer equations than unknowns can have infinitely many solutions. The solutions are then given in terms of a parameter, another variable. Then substitution of a particular value for the parameter gives one particular solution for the other variables.

In some cases, a parametric solution occurs for an obvious geometrical reason, as in example 3 part a, where the two equations represent coincident lines.

Example 5Solve the system of equations:

x + 4z = 132x − 2y − 7z = −194x − 2y + z = 7

SolutionFirst create a matrix whose elements are the coefficients of x, y and z and the constant terms in the equations.

1 2– 31 4 22 2– 1

xyz

344

=

1 0 4 132 2– 7– 19–

4 2– 1 7

1.2

continued �



22

Inconsistent systemsSometimes there are no solutions to a system of linear equations. In such cases, the system is said to be inconsistent.

Now use the rref command, as shown on the right.

The resulting matrix shows x + 4z = 13 and ,

so there is an infinite set of solutions.

This set of solutions can be written in terms of a parameter.

Let z = t.

x + 4t = 13 and = x = 13 − 4t

y =

The solutions are x = 13 − 4t, , z = t, where t ∈ R.

If t = 0, the solution is x = 13, , z = 0. If t = 1, then x = 9, y = 15, z = 1.

Similarly, further specific solutions can be found by assigning t any real number value.

Example 6Solve the system of equations:

3x − y − 4z = 11x + 4y + 3z = 102x + y − z = −1

SolutionFirst create a matrix whose elements are the coefficientsof x, y and z and the constant terms in the equations.

TI-Nspire 8.1, 8.2

ClassPad 8.1, 8.2

y152

------z+ 452

------=

y152

------t+ 452

------

152

------ 3 t–( )

y152

------ 3 t–( )=

y452

------=

TipUsing the Solve command, the same set of solutions as found above is identified. Note that the TI-Nspire expresses these in terms of a parameter c1; the ClassPad is interpreted as ‘z can take any real value’, and then x and y are determined in terms of that value (so z is taken to be a parameter).

TI-Nspire 10.2, 10.3

ClassPad 10.2, 10.3

3 1– 4– 111 4 3 102 1 1– 1–

continued �



23

exercise 1.25 Solve the following systems of linear simultaneous equations by hand. Check your

answers using technology.

a x + y + z = 3 b x + 2y − z = 2 c 2x + y + z = 104x + 2y + z = 6 2x − y + 2z = 5 x + 5y − 3z = −79x + 3y + z = 13 x + y − z = 1 x + 2y + 3z = 23

d 5x + z = 2 e x − 2y + 2z = 8 f x + 3y + 2z = 1

y + z = 5 3x − 2y + z = 8 2x − y − z = 6−x + y = 1 2x + y − z = 1 3x + y − 4z = −1

6 A parabola with equation y = ax2 + bx + c passes through the points (−1, −10), (2, −1) and (3, −6). Find the values of a, b and c.

Now use the rref command as shown in the screenshot.

The last row of this matrix reads 0x + 0y + 0z = 1. This cannot be true, so the system of equations has no solution. Alternatively, using the Solve command returns the value false on a TI-Nspire or No Solution on a ClassPad, which again means that the system is inconsistent.

Warning Interpreting error messagesThe alternative matrix method of solving systems of equations using the TI-Nspire command Simult only works if there is a single (unique) solution to the simultaneous system. The screenshot below shows what happens when this method is applied in examples 5 and 6.

The error message means that there is a problem with the matrix of coefficients. So this method is unable to deal with examples in which there are infinitely many solutions, or there are inconsistent equations. In such cases the rref or Solve commands are more informative.

continued

1.2



24

7 For each of the following:

i determine whether the system has a unique solution or an infinite number of solutions or no solution

ii find the solutions to those systems of simultaneous equations for which solutions exist.

a x + 3y + 4z = 2 b x + 4y + 3z = 10 c x − y + z = 62x + y + z = 5 2x + y − z = −1 x − 4y + 5z = 7x − y − 2z = 4 3x − y − 4z = 11 x + 2y − 3z = 5

d 3x + 2y − 5z = 4 e 2x − 3y + z = −1 f 3x − 3y − z = 15x + 3y − 8z = 6 x + 2y − z = 0 2x − 3y + z = 14x + 2y − 6z = 5 3x − y + 2z = 3 x − 2z = 0

8 Express the following systems of equations as matrix equations and use technology to solve them.

a x − 2y − 5z − w = 0 b x − 3y + 6w = 13x − 5y − z − 3w = 1 x − 5y + z − w = 1

2x − 8y − 2z − 10w = −4 y + z + w = 12x − 5y + 2z − 5w = −1 x + y + z = 0

9 A curve passes through the points (−1, −9), (1, −5) (2, −3) and (3, 7). If the curve has equation y = ax3 + bx2 + cx + d, find the values of a, b, c and d.

10 Five numbers add to zero. The first is equal to the sum of the second and the fourth. The third is equal to the sum of the fourth and the fifth. The sum of the first two numbers is 2 more than the fifth number. The fifth number is three times the sum of the third and fourth numbers. Find the five numbers.

11 Consider the pair of linear equations below, where k is a real constant.

5x + ky = kkx + 20y = 2k

a Find the solution if k = 5.

b Show that the equations are inconsistent if k = −10.

c Solve the pair of equations if k = 10.

d Solve the pair of equations for x and y in terms of k for all .

12 Consider the following system of linear equations, where n is a real constant.

2x + (n + 3)y − (n + 4)z = 12y + (n + 4)z = 2

x − 2y + z = 1

a Find the solution if n = −3.

b Are there any values of n for which the equations are inconsistent?

c Are there any values of n for which the equations have an infinite set of solutions? State the solutions in terms of a parameter t in this case.

k R\ 10– 10,{ }∈



25

Algebra of exponential and logarithmic functions

Definition of a logarithmRecall that if a > 0 and ax = y then loga y = x. This means that x is the power to which a is raised in order to obtain y. For example, log3 9 = 2, since 32 = 9.

Laws of logarithmsThe laws of logarithms can be deduced from the index laws.

Logarithm of a product (first logarithm law)loga (xy) = loga x + loga y, where x, y > 0.

(In words, the logarithm of a product is equal to the sum of the logarithms.)

Let m = loga x and n = loga y.

x = am and y = an (definition of a logarithm)xy = am × an

= am + n (1st index law)loga (xy) = m + n (logarithm form)

= loga x + loga y

Logarithm of a quotient (second logarithm law)loga = loga x − loga y, where x, y > 0.

(In words, the logarithm of a quotient is equal to the difference of the logarithms.)

Let m = loga x and n = loga y.

x = am and y = an (definition of a logarithm)

=

= am − n (2nd index law)loga = m − n (logarithm form)

= loga x − loga y

Logarithm of a power (third logarithm law)loga (xn) = n loga x, where x > 0.

(In words, the logarithm of a power is equal to the product of the power and the logarithm.)

Let m = loga x.

x = am (definition of a logarithm)xn = (am)n

= amn (3rd index law)loga (xn) = mn (logarithm form)

= n loga x

xy---⎝ ⎠

⎛ ⎞

xy--- am

an------

xy---⎝ ⎠

⎛ ⎞

1.31.3



26

Two special cases of the third law are important.

If n = 0, loga (x0) = 0, i.e. loga 1 = 0. Of course, this also follows from the fact that a0 = 1.

If n = −1, loga (x−1) = −loga x, i.e. loga = −loga x.

Changing the base of logarithmsWhen a decimal approximation to a logarithm is required, the change of base rule is useful, together with a calculator that will evaluate logarithms with base 10.

Suppose we need an approximation for loga x.

Let y = loga x, so that x = ay.

Take logarithms (base b) of each side:logb x = logb ay

= y logb a= (loga x)(logb a)

Therefore loga x = .

In the special case where b = 10, we have:

.

Comparison of index and logarithm lawsIndex form Logarithm form

. First law: am × an = am + n loga (xy) = loga x + loga y

. Second law: = am − n loga = loga x − loga y

. Third law: (am)n = amn loga (xn) = n loga x

. Special value: a0 = 1 loga 1 = 0

Example 1Correct to 3 decimal places, log3 5 = 1.465 and log3 2 = 0.631. Use these to find an approximate value for the following.a log3 10 b log3 40 c d log3 45

Solutiona log3 10 = log3 (5 × 2) b log3 40 = log3 (8 × 5)

= log3 5 + log3 2 = log3 8 + log3 5= 1.465 + 0.631 = log3 23 + log3 5= 2.096 = 3 log3 2 + log3 5

= 3 × 0.631 + 1.465= 3.358

c = log3 4 − log3 5 d log3 45 = log3 (9 × 5)= log3 22 − log3 5 = log3 9 + log3 5= 2 log3 2 − log3 5 = 2 + 1.465= 2 × 0.631 − 1.465 = 3.465= −0.203

1x---

am

an------ x

y---⎝ ⎠

⎛ ⎞

log b xlog b a---------------

log a xlog 10 xlog 10 a-----------------=

log 345---

log 345---



27

Example 2

Evaluate

Solution

=

=

= 3

Example 3Express a in terms of b if:

a log10 a = 2 log10 b + 1 b log4 a = 0.5b + 3

Solutiona log10 a = 2 log10 b + 1 b log4 a = 0.5b + 3

= log10 b2 + log10 10 a = 40.5b + 3

= log10 10b2 = 40.5b × 43

a = 10b2 = (40.5)b × 64

= 64 × 2b

Example 4Find the value of log11 13, correct to 3 decimal places, using common (base 10) logarithms.

SolutionUse the change of base rule to change the logarithm base 11 to base 10.

loga x =

log11 13 =

= 1.070 (correct to 3 decimal places)

WarningIt is of course completely incorrect to ‘cancel the logs’ as in:

.log 5 8log5 2--------------- 8

2--- 4= =

log 5 8log5 2---------------

log 5 8log5 2---------------

log 5 23

log 5 2-----------------

3 log5 2log5 2

-------------------

log 10 xlog 10 a-----------------

log 10 13log10 11--------------------

TipThe calculation using base 10 is shown in the screenshot opposite. However, note that a CAS can find the value directly as shown. (Remember to press the appropriate key combination to obtain the numerical approximation.)

TI-Nspire 1.1

ClassPad 1.1

1.3



28

exercise 1.31 Evaluate each of the following without the use of a calculator.

a log4 64 b log9 3 c

d e f log3 18 − log3 2

g log6 9 + log6 24 h 2 log2 6 − log2 9 i

2 Express each of the following as a single logarithm.

a log5 x − 3 log5 (x + 1) b 2 log10 3 + 5 log10 x

c 2 log3 (x − 1) − log3 4 − log3 x d

e f

3 If loga x = 0.3 and loga y = 0.2, find:

a loga (xy) b c d 2 loga xy3

4 Given that log5 2 = 0.43 and log5 3 = 0.68 correct to 2 decimal places, find an approximate value for each of the following.

a log5 4 b log5 15 c log5 60

d log5 24 e log5 40 f

5 Find an expression for a in terms of b, if:

a log10 a = 2 − b b log10 a = 2b − 1

c log10 a = 2 log10 b − 3 d

6 If x = log2 a and y = log2 b, express in terms of x and y.

7 Use the change of base rule and common logarithms (base 10) to evaluate the following, correct to 3 decimal places.

a log4 7 b log5 9 c log2 10

Solving exponential equations and inequationsConsider the exponential equation 3x = 81. The solution can be obtained as follows:

3x = 81

= 34

x = 4

It is not always possible to express the term on the right-hand side in a simple way involving indices as in the example above. In such a case, logarithms can be used.

log 3127------

log 5 81log5 3

------------------log 3 64log3 16------------------

log 10 14 log1075---–

log 2 x2y 2 log2 xy2 log 2xy2-----+–

log 3 x2 13--- log 3 x6y3( ) 4 log3 x–+ 1

2--- log 10 x log 10 x+

log axy2----- log a

xy

--------

log 5103

------

log 10 a 212---log 10 b–=

log 24a3b2

b--------------⎝ ⎠

⎛ ⎞



29

Example 5Solve for x, giving answers correct to 3 decimal places.

a 2x − 3 = 15 b 5−0.4x = 1.7 c 3x + 1 = 5x d 2x + 3 = 2x

Solutiona One solution method is to use common logarithms as follows.

log10 2x − 3 = log10 15 (taking the common logarithm of both sides)

(x − 3) log10 2 = log10 15

x − 3 =

x = = 6.907 (correct to 3 decimal places)

Alternatively, 2x − 3 = 15 is equivalent to x − 3 = log215, i.e. x = log215 + 3 (note how the change of base rule can be used here to reveal the answer as expressed in the solution above). Hence, correct to 3 decimal places, the answer is x = 6.907 (refer to the screenshot).

b 5−0.4x = 1.7 is equivalent to −0.4x = log51.7, i.e. x = .

Hence, correct to 3 decimal places, the answer is x = −0.824 (refer to the screenshot).

c 3x + 1 = 5x

log10 3x + 1 = log10 5x

(x + 1) log10 3 = x log10 5log10 3 = x log10 5 − x log10 3

x = = 2.151 (correct to 3 decimal places)

d 2x + 3 = 2x

Technology is required in this case. One method is to plot the graphs of y = 2x + 3 and y = 2x with a CAS and findthe x-coordinates of the points of intersection as shown.Alternatively, the Solve command can be used.Hence, correct to 3 decimal places, the solution isx = −1.296 or x = 3.247.

log 10 15log10 2

--------------------

log 10 15log10 2

-------------------- 3+

TI-Nspire 1.1

ClassPad 1.1 log 51.70.4–

-------------------

log 10 3log10 5 log10 3–-----------------------------------------

TI-Nspire 2.4, 10.2

ClassPad 2.4.10.2

1.3



30

Example 6Find, correct to 3 decimal places where necessary:

a {x : 3x + 1 < 4} b {x : 4(0.5)x ≥ 1}

Solutiona {x : 3x + 1 < 4}

3x + 1 < 4log10 3x + 1 < log10 4

(x + 1) log10 3 < log10 4x log10 3 < log10 4 − log10 3

x <

x < 0.262The solution set is {x : x < 0.262}.

TipThe Solve command can be used to find the exact or approximate solutions to exponential equations. For example, consider example 5 part a. The screenshot

opposite gives the exact answer in the form ,

where the symbol ln stands for log natural, which uses a new base, e, as the default base for logarithms (ln can also be written as loge). This new base e is explored on page 33 of this chapter. The numerical approximation is 6.907 as before.The change of base rule can be used to convert answers to different bases. Thus:

x =

= log2120

= (and so on)

Note that the TI-Nspire has a command Convert to logbase (found in the Algebra menu) that can be used as shown in the screenshot.

ln120ln2

---------------

120ln2ln

---------------

log 12010

log 210---------------------

TI-Nspire 10.2

ClassPad 10.2

b {x : 4(0.5)x ≥ 1}

4(0.5)x ≥ 1

(0.5)x ≥

We could take the logarithm of both sides and solve as in part a, taking care to note that log10 0.5 is negative. However, in this case it is easier to use indices. Since

, the inequality becomes

2x ≤ 4 (taking the reciprocal of both sides)2x ≤ 22

x ≤ 2

The solution set is {x : x ≤ 2}.

14---

0.512---= 1

2---⎝ ⎠

⎛ ⎞x 1

4---≥

log 10 4 log10 3–log 10 3

------------------------------------------



31

Solving logarithmic equationsThe definition of a logarithm and the laws of logarithms are used in solving logarithmic equations.

Example 7Solve for x, giving exact answers.

a 52x − 4 × 5x − 5 = 0 b 2x = 10 × 2−x + 9

Solutiona 52x − 4 × 5x − 5 = 0

(5x)2 − 4 × 5x − 5 = 0

Factorising gives:(5x + 1)(5x − 5) = 0

Using the null factor law:5x = −1 (which is impossible as 5x is positive for all x), or5x = 5x = 1

b 2x = 10 × 2−x + 9

Multiply both sides by 2x:(2x)

2 = 10 + 9 × 2x

(2x)2 − 9 × 2x − 10 = 0

Factorising gives:(2x + 1)(2x − 10) = 0

2x = −1 (which is impossible), or2x = 10x = log2 10

The screenshot at right shows that the answer using the

Solve command is expressed as x = , which by

the change of base rule is equivalent to x = log2 10. If required, the numerical approximation can be obtained by pressing the appropriate key combination.

Example 8Solve for x.

a log2 (x + 3) = 5 b log2 (x + 3) − log2 (x − 2) = 3 c log3 x + log3 (x − 8) = 2

Solutiona log2 (x + 3) = 5

x + 3 = 25

x + 3 = 32x = 29

10ln2ln

------------

Check: LHS = log2 (29 + 3)= log2 32= 5= RHS

1.3

continued �



32

exercise 1.38 Solve for x.

a 5x = 25 b c 16x = 4

d 22x − 5 = 1 e f

9 Solve for x.

a 42x + 1 = 22x − 3 b c

d e f

10 Solve for x, giving answers correct to 3 decimal places.

a 2x = 10 b c 5x = 0.5

d 100.2x = 35 e 2 × 3−0.01x = 0.32 f 3 × 10x − 1 = 125

11 Solve for x, giving answers correct to 3 decimal places.

a 2x = x + 1 b 3 − 2x = 5−0.2x c x2 − x = 3−x

12 Working correct to 3 decimal places where necessary, find

a {x : 2x > 5} b {x : 3−x < 0.4} c {x : 52x ≥ 22}

d e {x : 2x + 4 > 0.2} f

b log2 (x + 3) − log2 (x − 2) = 3

= 3

= 23

x + 3 = 8(x − 2)x + 3 = 8x − 16

19 = 7x

x =

Check:LHS =

=

=

= log2 8= 3= RHS

c log3 x + log3 (x − 8) = 2log3 (x(x − 8)) = 2

x(x − 8) = 32

x2 − 8x − 9 = 0(x − 9)(x + 1) = 0

x = 9, −1

Check:x = 9:LHS = log3 9 + log3 (9 − 8)

= 2 + log3 1= 2= RHS

x = −1:LHS = log3 (−1) + log3 (−9)

It is not possible to evaluate the logarithm of a negative number.

So, x = 9 is the only solution.

log 2x 3+x 2–------------⎝ ⎠

⎛ ⎞

x 3+x 2–------------

197

------

log 2197

------ 3+⎝ ⎠⎛ ⎞ log 2

197

------ 2–⎝ ⎠⎛ ⎞–

log 2407

------ log 257---–

log 2

407

------

57---

TipWhen solving logarithmic equations it is necessary to always check solutions. Be sure to check the original equation.

continued

3x 19---=

49x 7= 32x 1

2--------=

5x 1

25x 2+--------------= 3x2 5x– 6– 1=

6( )x 3– 365 x–= 7x2 7x( )2= 27x 1+ 3x2 1–=

13---⎝ ⎠

⎛ ⎞ x

18=

x :12---⎝ ⎠

⎛ ⎞x

0.25≤⎩ ⎭⎨ ⎬⎧ ⎫

x : 32 x– 13---≤

⎩ ⎭⎨ ⎬⎧ ⎫



33

13 Solve for x.

a 22x − 6 × 2x + 8 = 0 b 32x − 12 × 3x + 27 = 0

c 52x − 30 × 5x + 125 = 0 d 22x − 15 × 2x − 16 = 0

e 2x + 4 × 2−x − 5 = 0 f 3x − 8 = 9 × 3−x

g 22x + 1 − 17 × 2x + 8 = 0 h 21 + x = 4 − 21 − x

14 Solve for x.

a log2 (x + 1) = 3 b logx 125 = 3

c 2 log3 x = 4 d log3 x2 = 4

e log3 (x + 2) − log3 (x + 1) = 4 f log2 (x + 1) − log2 x = log2 (x + 2)

g log10 5x = 2 log10 x h logx (x2 + 6x) = 3

i log2 x2 = (log2 x)2 j log2 (x + 2) + log2 (x − 2) = 1

k (log3 x)2 − 18 log3 x + 81 = 0 l log10 x − 3 log10 4 = log10 (x − 1)

The base eConsider the binomial expansion of , where n is a natural number.

=

=

=

=

=

As

So,

The irrational number e is defined as:

e =

=

= 2.71828 …In many applications of exponential and logarithmic functions, the base number e is used.

It is referred to as the natural base. The function y = ex, x ∈ R, is referred to as the natural exponential function and the function y = loge x, x > 0 is referred to as the natural logarithmic function. Note that loge x is also denoted by ln x. Natural logarithms can be evaluated directly using the ln function of a CAS.

11n---+⎝ ⎠

⎛ ⎞n

11n---+⎝ ⎠

⎛ ⎞n

1 Cn1

1n---⎝ ⎠

⎛ ⎞ Cn2

1n---⎝ ⎠

⎛ ⎞2

Cn3

1n---⎝ ⎠

⎛ ⎞3

…+ + + +

1n!

n 1–( )!------------------- 1

n---⎝ ⎠

⎛ ⎞ n!2! n 2–( )!------------------------ 1

n---⎝ ⎠

⎛ ⎞2 n!

3! n 3–( )!------------------------ 1

n---⎝ ⎠

⎛ ⎞3

…+ + + +

1n n 1–( )!

n 1–( )!----------------------- 1

n---⎝ ⎠

⎛ ⎞ n n 1–( ) n 2–( )!2! n 2–( )!

----------------------------------------- 1n---⎝ ⎠

⎛ ⎞2 n n 1–( ) n 2–( ) n 3–( )!

3! n 3–( )!----------------------------------------------------------- 1

n---⎝ ⎠

⎛ ⎞3

…+ + + +

1 1n n 1–( )

2!--------------------- 1

n---⎝ ⎠

⎛ ⎞2 n n 1–( ) n 2–( )

3!--------------------------------------- 1

n---⎝ ⎠

⎛ ⎞3

…+ + + +

1 112!----- 1 1

n---–⎝ ⎠

⎛ ⎞ 13!----- 1 1

n---–⎝ ⎠

⎛ ⎞ 1 2n---–⎝ ⎠

⎛ ⎞ …+ + + +

n ∞→ 1n--- 0→ 2

n--- 0→ …, , ,

1 1n---+⎝ ⎠

⎛ ⎞ n

n ∞→lim 1 1

12!----- 1

3!----- …+ + + +=

11n---+⎝ ⎠

⎛ ⎞n

n ∞→lim

1 112!----- 1

3!----- …+ + + +

1.3



34

The techniques for solving equations, already covered in this chapter, are also valid for equations involving the natural base.

Example 9Evaluate each of the following correct to 4 decimal places.

a 2e3 b c 2 loge 5 d

SolutionThe screenshot at right shows the relevant calculations, obtained by pressing the appropriate key combination to give numerical approximations. Hence the answers are:

a 40.1711 b 0.3679

c 3.2189 d 0.3466

Example 10Solve for x, giving exact answers in terms of the natural logarithm function.

a 3e2x = 7 b e2x − 5ex + 4 = 0

Solutiona 3e2x = 7

e2x =

2x =

x =

A decimal approximation can also be obtained, if necessary. Thus, correct to 3 decimal places, the answers are 0.424 in part a, and 0 or 1.386 in part b.

1e--- ln 2

b e2x − 5ex + 4 = 0(ex)

2 − 5ex + 4 = 0

(ex − 1)(ex − 4) = 0ex = 1 or ex = 4x = ln 1 or x = ln 4x = 0 or x = ln 4

73---

loge73---

12---loge

73---



35

exercise 1.315 Solve for x in each of the following giving

i exact answers. ii answers correct to 4 decimal places.

a ex = 5 b e2x = 3 c 2e−x = 8

d ex + 1 = 2 e f 30e−0.2x + 5 = 65

g h e2x − 9ex − 10 = 0 i ex = 6e−x + 5

16 Solve for x in each of the following, giving exact answers.

a ln x = 2 b 3 loge 4x = 9

c ln(2x − 3) = 0 d ln(x2) = 4

e ln(3x − 2) = ln(x + 1) f loge (15 − x2) = loge (x2 − x)

g (ln x)(ln x − 3) = 0 h

i ln(x + 4) − ln(x + 1) = ln x

17 Let a and b be positive real numbers.

a If ax + 1 = bx − 1, express x in terms of a and b using natural logarithms.

b Hence show that the solution to 2x + 1 = 10x − 1 can be written in the form , where p, q and r are positive integers with p > 1.

18 Let a and b be positive real numbers.

a Prove that (loga b) (logb a) = 1

b Prove that = loga b + logb a

c Simplify:

Example 11Solve for x giving exact answers.

a b 3 ln (2x) = 6

Solution

a loge x =

x =

=

log e x12---=

TipAlways give an exact answer unless a decimal approximation is required or necessary.

b 3 ln (2x) = 6ln (2x) = 2

2x = e2

x = 12---e2

12---

e1 2⁄

e

continued

2ex 1–------------- 5=

3e2 x–---------- 9=

log e1x--- 3=

1p log e qlog e r

-------------------+

1loga b-------------- 1

logb a---------------+

1

loge1e---⎝ ⎠

⎛ ⎞------------------- 1

log1e---

e---------------+

1.3



36

Algebra of circular functions

The unit circle, symmetry properties and exact valuesThe unit circle is a circle of radius 1 unit with centre at the origin, O. Let P(x, y) be a point on the unit circle such that OP makes an angle θ with the positive direction of the x-axis. A tangent is drawn to this circle through the point A(1, 0). When the line OP is produced it meets this tangent at Q.

In general, for any point P on the unit circle, we define the functions cosine and sine by:

cosine θ = cos θ = x and sine θ = sin θ = y

where x is the x-coordinate of P and y is the y-coordinate of P.

Using Pythagoras’ theorem on triangle OMP:

OM2 + MP2 = OP2

cos2θ + sin2θ = 1

This is known as an identity, since it is true for all values of θ.

We define the tangent function by:

tangent θ = tan θ = AQ, the y-coordinate of Q

tan θ = = = , cos θ ≠ 0

The angle θ is positive if the angle of rotation from the positive direction of the x-axis is in an anticlockwise direction and negative if the angle of rotation from the positive direction of the x-axis is in a clockwise direction.

The four quadrants in the Cartesian plane are numbered according to the system shown on the right.

Recall the following properties of symmetry of the trigonometric functions.

Radians Degrees

Quadrant 2 sin (π − θ ) = sin θcos (π − θ ) = −cos θtan (π − θ ) = −tan θ

sin (180 − θ )° = sin θ °cos (180 − θ )° = −cos θ °tan (180 − θ )° = −tan θ °

Quadrant 3 sin (π + θ ) = −sin θcos (π + θ ) = −cos θtan (π + θ ) = tan θ

sin (180 + θ )° = −sin θ °cos (180 + θ )° = −cos θ °tan (180 + θ )° = tan θ °

Quadrant 4 sin (2π − θ ) = −sin θcos (2π − θ ) = cos θtan (2π − θ ) = −tan θ

sin (360 − θ )° = −sin θ °cos (360 − θ )° = cos θ °tan (360 − θ )° = −tan θ °

y

xA (1, 0)(–1, 0)

(0, –1)

(0, 1)

P

M

Q

y

xθ

O

AQ1

-------- yx--- sin θ

cos θ-------------

Quadrant 1Quadrant 2

Quadrant 4Quadrant 3

x

y

1.41.4



37

For negative angles: sin (−θ) = −sin θ cos (−θ) = cos θ tan (−θ) = −tan θThe signs of the circular functions can be remembered using the following diagrams.

This is sometimes known as the CAST rule.

The exact values of sin, cos and tan for come from the triangles drawn below.

Other exact values are obtained directly from the unit circle.

θ 0 π 2π

sin θ 0 1 0 −1 0

cos θ 1 0 −1 0 1

tan θ 0 1 undefined 0 undefined 0

Example 1Find the exact value of:

a sin 120° b c

Solutiona sin 120° = sin (180° − 60°) b = c =

= sin 60° = =

= = = = −1

SIN ispositive

ALL functionsare positive

TAN ispositive

COS ispositive

x

y

x

y

S A

T C

π6--- π

4--- π

3---, ,

π4

π4

21

1

π6

π3

2

1

3

(1, 0)0, 2

(0, 1)

(–1, 0)

(0, –1)

O x

y

23

2

π6--- π

4--- π

3--- π

2--- 3π

2------

12---

1

2-------- 3

2--------

32

--------1

2-------- 1

2---

1

3-------- 3

cos11π6

--------- tan 5π4

------–⎝ ⎠⎛ ⎞

cos11π6

--------- cos 2π π6---–⎝ ⎠

⎛ ⎞ tan 5π4

------–⎝ ⎠⎛ ⎞ tan–

5π4

------

cosπ6--- tan π π

4---+⎝ ⎠

⎛ ⎞–

32

-------- 32

-------- tanπ4---–

1.4



38

Solving equations involving circular functionsDue to the periodic nature of circular functions, a trigonometric equation will have an infinite number of solutions. However, in most situations, only particular solutions are of interest and therefore we usually solve these equations over a finite interval.

In some instances, the equations we wish to solve have recognizable exact solutions and we can solve these by hand, i.e. without the use of technology. Other equations require the use of technology to find solutions.

Example 2If cos θ = −0.6 and , find the exact values ofa sin θ b tan θ

Solutiona cos2 θ + sin2 θ = 1

(−0.6)2 + sin2 θ = 10.36 + sin2 θ = 1

sin2 θ = 0.64sin θ = −0.8 (θ in 3rd quadrant)

Note that tan θ is positive as expected (θ in 3rd quadrant)

Example 3Solve for x: ,

SolutionIdentify the quadrants in which the solutions lie.

The value of cos x is positive so there is a solution in the first quadrant and a solution in the fourth quadrant (CAST rule).

Find the reference angle in the first quadrant: cos 30° = .

So the solution in the first quadrant is x = 30.

By symmetry, the solution in the fourth quadrant is x = 360 − 30 = 330.

So x = 30 or x = 330.

TipA CAS can be used to find exact values of sin θ, cos θ and tan θ . With the mode set to Radian, you can still find the values for angles in degrees by using the ° symbol. However, it is important that you learn the exact values for the special angles, so that you can answer questions about these without the use of technology.

TI-Nspire 1.6

ClassPad 1.6

π θ 3π2

------< <

b tan θ =

=

=

sinθcosθ------------

0.8–0.6–

-----------

43---

cos x° 32

--------= 0 x 360≤ ≤

S

T

A

C

32

--------



39

Example 4Solve for x: sin x = 0.32, 0 ≤ x ≤ 2π for x. Give your answers correct to 4 decimal places.

SolutionMethod 1: using symmetry propertiesIdentify the quadrants in which the solutions lie.

The value of sin x is positive so there is a solution in the first quadrant and a solution in the second quadrant.

Find the reference angle in the first quadrant. With your CAS in Radian mode, the reference angle is 0.3257. So the solution in the first quadrant is x = 0.3257.

By symmetry the solution in the second quadrant is x = π − 0.3257 = 2.8159.

Hence x = 0.3257 or x = 2.8159.

Method 2: using the Solve command of a CASSolve the equation as shown in the screenshot opposite. (Note that it is important to include the restriction on x.)

Hence x = 0.3257 or x = 2.8159.

Example 5Solve for θ : , 0 ≤ θ ≤ 2π

Solutionsin θ =

=

tan θ = , since

Identify the quadrants in which the solutions lie.

The value of tan θ is negative, so there is a solution in the second quadrant and a solution in the fourth quadrant (CAST rule).

Find the reference angle in the first quadrant: .

By symmetry, the solution in the second quadrant is ,

and the solution in the fourth quadrant is .

Hence or .

S

T

A

C

TI-Nspire 1.5, 1.6, 10.2

ClassPad 1.5, 1.6, 10.2

Warning Check the mode!When solving an equation where the variable is in degrees, make sure your CAS is set to Degree mode. When solving an equation where the variable is in radians, make sure your CAS is set to Radian mode.

sin θ 3– cos θ=

3– cos θsin θcos θ------------- 3–

3–sin θcos θ------------- tan θ=

S

T

A

Ctan

π3--- 3=

x π π3---–

2π3

------= =

x 2π π3---–

5π3

------= =

x2π3

------= x5π3

------=

1.4



40

Example 6Solve each of the following equations for x if 0 ≤ x ≤ 2π.

a 2 sin2x − sin x = 0 b tan2x − 3 = 0 c 2 sin2x + 3 cos x −3 = 0

Solutiona 2 sin2x − sin x = 0

Take out the common factor sin x.

sin x (2 sin x − 1) = 0

sin x = 0 or sin x = (null factor law)

If sin x = 0 then x = 0 or x = π or x = 2π.

If then .

Hence the solutions are .

b tan2x − 3 = 0

= 0 (difference of two squares)

or

If then . If then .


TipThe Solve command can be used to find the exact solutions to trigonometric equations where possible. For example, consider the equation in example 3. With the mode set to Degree, the solution is x = 30 or x = 330, agreeing with the solution to example 3.

By changing the mode to Radian, the solution to the

equation in example 5 is shown as or ,

agreeing with the solution to example 5.

However, it is important that you are able to solve equations like these without the aid of technology.

x2π3

------= x5π3

------=

TI-Nspire 10.2

ClassPad 10.2

S

T

A

C

12---

sin x12---= x

π6--- or x

5π6

------==

0π6--- 5π

6------ π 2π, , , ,

tan x 3–( ) tan x 3+( )tan x 3= tan x 3–=

tan x 3= xπ3--- or x

4π3

------== tan x 3–= x2π3

------ or x5π3

------= =

S

T

A

C

S

T

A

C

π3--- 2π

3------ 4π

3------ 5π

3------, , ,

continued �



41

exercise 1.41 Find the exact value of:

a sin 240° b cos 300° c tan 135° d sin (−150°)

e cos (−180°) f tan (−210°) g h

i j k l

m n o

2 a If and θ is in the second quadrant, find the exact values of

i cosθ ii tanθb If cosθ = k, −1 < k < 0, and θ is in the third quadrant, express in terms of k:

i sinθ ii tanθ iii cos2θ − sin2θ3 Solve each of the following for x if 0 ≤ x ≤ 360. Give exact values for x.

a b c

d e f

4 Solve each of the following for x if 0 ≤ x ≤ 360, correct to the nearest tenth of a degree.a sin x° = 0.6 b 3 cos x° = −1 c 3 tan x° − 1 = 3

5 Solve each of the following for x if 0 ≤ x ≤ 2π. Give exact values for x.

a b c

d 2 cos x − 1 = 0 e sin x + cos x = 0 f 4 cos2 x = 1

g cos x = 1 h sin x = −1 i tan x = 0

6 Solve each of the following for x if 0 ≤ x ≤ 2π, correct to 4 decimal places.a cos x = 0.64 b sin x = −0.25 c 2 tan x = 3

7 Solve each of the following for x if 0 ≤ x ≤ 2π. Give exact values for x.

a 4 sin2x − 1 = 0 b tan2x − tan x = 0 c 2 cos2x − 1 = 0

d 2 cos2x + 3 cos x + 1 = 0 e 3 tan3x = tan x

c 2 sin2x + 3 cos x − 3 = 0

2(1 − cos2x) + 3 cos x − 3 = 0

2 − 2 cos2x + 3 cos x − 3 = 0

2 cos2x − 3 cos x + 1 = 0

(2 cos x − 1)(cos x − 1) = 0

or cos x = 1

If .

If cos x = 1 then x = 0 or x = .


TipThe identity sin2x + cos2x = 1 can be useful when solving equations. If sin2x is made the subject, then sin2x = 1 − cos2x and if cos2x is made the subject, then cos2x = 1 − sin2x.

S

T

A

C

cos x12---=

cos x12--- then x

π3--- or x

5π3

------== =

2π

0π3--- 5π

3------ 2π, , ,

sin3π4

------ cos5π6

------

tan4π3

------ sin11π6

--------- cos5π3

------ tan5π4

------

sin 5π3

------–⎝ ⎠⎛ ⎞ cos 3π

4------–⎝ ⎠

⎛ ⎞ tan 11π6

---------–⎝ ⎠⎛ ⎞

sinθ 513------=

cos x°1

2--------= sin x° 1

2---–= tan x° 3–=

2 sin x° 3– 0= 3 tan x° 3= sin x° 3 cos x°– 0=

tan x 3= cos x 32

--------–= sin x22

--------=

1.4



42

Equations on other intervalsIn all of the previous examples, solutions were found on the interval [0°, 360°] or [0, 2π]. It is sometimes necessary to obtain solutions of equations on other intervals.

Equations of the form Af(bx) = d, where f is one of sine, cosine or tangent

Example 7Find the exact values of x for which .

SolutionFind the solutions between 0 and 2π first.


The value of cos x is negative so there is a solution in the second quadrant and a solution in the third quadrant.


By symmetry, the solution in the second quadrant is

.

Similarly, the solution in the third quadrant is .

The solutions between 2π and 4π can be found by ‘going round’ the unit circle again, that is, by adding 2π to those already found.

x = x =

= =


Example 8If 0 ≤ x ≤ 2π, find the values of x for which .

Solution = 0

sin 2x =

If 0 ≤ x ≤ 2π, then 0 ≤ 2x ≤ 4π.

So we need to solve for 2x, with 0 ≤ 2x ≤ 4π.

cos x 1

2--------– 0 x 4π≤ ≤,=

S

T

A

Ccos

π4--- 1

2--------=

x π π4---–

3π4

------= =

x π π4---+

5π4

------= =

3π4

------ 2π+5π4

------ 2π+

11π4

--------- 13π4

---------

3π4

------ 5π4

------ 11π4

--------- 13π4

---------, , ,

2 sin 2x 3+ 0=

2 sin 2x 3+32

--------–

sin 2x 32

--------–=



43

exercise 1.48 Solve each of the following equations on the interval specified. Give exact answers where

possible. Otherwise, give answers correct to 3 decimal places.

a b

c d

e f

9 Solve for x, giving exact values.

a b

c d

e f

10 Solve for x: for 0 ≤ x ≤ 360. Give exact values for x.

11 Solve for x, 0 ≤ x ≤ 2π. Give exact values for x.

a b c

12 Solve for θ : .

Find the solutions for 2x between 0 and 2π first.


The value of sin 2x is negative so there is a solution in the third quadrant and a solution in the fourth quadrant.


By symmetry, the solution in the third quadrant is

.

Similarly, the solution in the fourth quadrant is .

Additional solutions can be found by adding 2π to those already found.

2x = 2x =

= =

The solutions for 2x are:

2x =

The solutions for x are found by dividing by 2:

x =

S

T

A

Csinπ3--- 3

2--------=

2x π π3---+

4π3

------= =

2x 2π π3---–

5π3

------= =

Warning Order!The addition of 2π must be carried out before the division by 2.

2π 4π3

------+ 2π 5π3

------+

10π3

--------- 11π3

---------

4π3

------ 5π3

------ 10π3

--------- 11π3

---------, , ,

2π3

------ 5π6

------ 5π3

------ 11π6

---------, , ,

continued

cos θ 12--- 0 θ 4π≤ ≤,= sin x 2

2--------– 0 x 4π≤ ≤,=

tan θ 3– 0 θ 4π≤ ≤,= sin x 32

--------– π x π≤ ≤–,=

sin θ 0.3 0 θ 3π≤ ≤,= tanx12--- 2π x 2π≤ ≤–,=

tan 2x1

3-------- 0 x 2π≤ ≤,= 2 sin 2x 3 0 x 2π≤ ≤,=

cos 3x 1

2--------– 0 x 2π≤ ≤,= sin 3x 1 π x π≤ ≤–,–=

2 cosπx2

------ 1 2 x 2≤ ≤–,= tan 3x° 0 180 x 180≤ ≤–,=

cos x 50–( )° 32

--------=

cos 2x π4---–⎝ ⎠

⎛ ⎞ 12---= sin 2x π

4---+⎝ ⎠

⎛ ⎞ 12---= tan 3x π

3---–⎝ ⎠

⎛ ⎞ 1–=

sin 2θ π6---–⎝ ⎠

⎛ ⎞ cos 2θ π6---–⎝ ⎠

⎛ ⎞– 0 0 θ 2π≤ ≤,=

1.4



44

General solutionsDue to the periodic nature of circular functions, a trigonometric equation on the open interval R will have an infinite number of solutions. In this case, solving the equation gives a general solution.

Example 9Find the general solution to each of the following equations.

a b c

Solutiona First solve for x on [0, 2π].

Identify the quadrants in which the solutions lie.The value of sin x is positive so there is a solution in the first quadrant and a solution in the second quadrant.

Find the reference angle in the first quadrant:

So the solution in the first quadrant is .

By symmetry the solution in the second quadrant is .

Additional solutions can be obtained by adding or subtracting integer multiples of 2π.

or

So the general solution is or , where n is an integer, i.e. .

b First solve for 3x on [0, 2π].


The value of cos 3x is positive so there is a solution in the first quadrant and a solution in the fourth quadrant.


So the solution in the first quadrant is .

By symmetry, the solution in the fourth quadrant is .

Additional solutions can be obtained by adding or subtracting integer multiples of 2π.

So the general solution is or , where n ∈ Z.

Dividing by 3 gives or .

sin x1

2--------= cos 3x

32

--------= tan x 1

3--------–=

S

T

A

C

sinx1

2-------=

sinπ4--- 1

2-------=

xπ4---=

x π π4--- 3π

4------=–=

xπ4--- π

4--- 2π π

4--- 4π π

4--- 6π …,±,±,±,=

x3π4

------ 3π4

------ 2π 3π4

------ 4π 3π4

------ 6π …,±,±,±,=

x π4--- 2nπ+= x 3π

4------ 2nπ+= n Z∈

S

T

A

C

cos 3x32

--------=

cosπ6--- 3

2--------=

3xπ6---=

3x 2π π6---–

11π6

---------= =

3xπ6--- 2nπ+= 3x

11π6

--------- 2nπ+=

xπ18------ 2nπ

3----------+= x

11π18

--------- 2nπ3

---------- n Z∈,+=

continued �



45

Finding general solutions with a CASConsider the equation in example 9 part a, , where .

Using the Solve command of a CAS gives the general solution.

The TI-Nspire (see screenshot) expresses the solution as

or . The symbol n1 is used to

denote a general integer.

However, the general solution can be written in a simpler, equivalent way in this case.

Rather than giving the solution in the fourth quadrant as , we can just

as easily give the solution as , since the negative angle is also in the fourth quadrant.

The general solution then becomes or , which can be more

simply written as .

Dividing by 3 then gives , n ∈ Z, and this is equivalent to the solution above.

c First solve for x on [0, 2π].


The value of tan x is negative, so there is a solution in the second quadrant and a solution in the fourth quadrant.

Note that these solutions are exactly π apart, so we can use a slightly different method of solution from that in parts a and b.


By symmetry, the solution in the second quadrant is .

Additional solutions can be obtained by adding or subtracting integer multiples of π. This will automatically give the solution in the fourth quadrant and all other solutions to the equation over R.

So the general solution is .

3x 2π π6--- 11π

6---------=–=

3x π6---–=

3xπ6--- 2nπ+= 3x π

6---– 2nπ+=

3xπ6---± 2nπ+=

xπ18------± 2nπ

3----------+=

tan x 1

3--------–=

S

T

A

Ctan

π6--- 1

3--------=

x π– π6---–

5π6

------=

x5π6

------ nπ n Z∈,+=

TipWhen solving equations involving tan x, the solutions are always π apart. Recall that the graph of y = tan x has period π. A general solution can be obtained by initially finding one solution to the equation and then addinginteger multiples of π to this. For equations involvingtan ax, add integer multiples of .

πa---

TI-Nspire 2.1

ClassPad 2.1

sinx1

2-------=

x R∈

x 2nlπ 3π4

------+= x 2nlπ π4---+=

1.4



46

The ClassPad (see screenshot) expresses the solution as

. The symbols

constn(1) and constn(2) are used to denote general integers.

Although the symbols and style are different, the solution in either case is equivalent to the one found by hand in example 9 part a, where we used the symbol n to denote a general integer.

Example 10Find the first 6 positive solutions to the equation .

SolutionThe general solution is or , , found in example 8 part a and

confirmed above by the CAS solution. The first 6 positive solutions can easily be obtained without technology by successively substituting the values 0, 1 and 2 for n and simplifying.

Alternatively, if CAS has been used to find the general solution, the substitution can be done directly or by copying, pasting and editing as shown in the screenshots.

Hence the first 6 positive solutions are , .

Example 11Find the first 4 positive solutions to the equation cos x = −0.3.

SolutionUsing the Solve command of a CAS gives the general solution

, .In this case, substituting 0 and 1 for n gives one negative and three positive solutions. To get the first 4 positive solutions, substitute 1 and 2 for n in the second case (refer to the screenshot opposite).

Hence the first 4 positive solutions are 1.8755, 4.4077, 8.1589, 10.6909, correct to 4 decimal places.

x 2πconstn 1( ) π4--- x 2πconstn 2( ) 3π

4------+=,+=

⎩ ⎭⎨ ⎬⎧ ⎫

sin x1

2-------=

xπ4--- 2nπ+= x

3π4

------ 2nπ+= n Z∈

π4--- 3π

4------ 9π

4------ 11π

4---------, , , 17π

4--------- 19π

4---------,

x 1.87549 2nπ+±= n Z∈



47

exercise 1.413 Find the general solutions to the following equations. Give exact answers.

a 2 sin x = 1 b c cos 2x = 1

d e 2 cos 3x + 1 = 0 f tan πx = −1

14 Find the first 6 positive solutions for each of the following equations. Give exact answers where possible, otherwise give answers correct to 4 decimal places.

a b c

d sin x = 0.4 e cos x = 0.55 f tan x = 2

15 Find the general solution to the equation sin = and hence find all solutions

on the interval −4π ≤ x ≤ 4π.

continued

sin x 3 cos x– 0=

sin 2x22

--------=

tan x1

3--------= sin x 3

2--------–= cos x 2

2--------–=

x π+3

------------⎝ ⎠⎛ ⎞ 3

2-------–

A calculator uses series to compute the values of sin, cos and tan.

, where x is measured in radians, and the nth term is

(−1)n − 1 .

a Calculate using a calculator, and write down the answer correct to 6 decimal places.

b Use to find an approximate value of , giving the answer to

6 decimal places. How does this compare with the answer obtained in question a?

c Use to find an approximate value of , giving the answer to

6 decimal places. How does this compare with the answer obtained in question a?

d i Write down the 4th term in the series for sin x.

ii Use the first four terms of the series to approximate the value of , giving the answer to 6 decimal places.

iii How does this compare with the answer obtained in question a?

e Determine the number of terms required in the series to calculate the value of , correct to 6 decimal place accuracy. (Compare your value with that in question a.)

f Cos x is given by the series , where x is measured in radians.

Find the number of terms required to approximate the value of cos(2), correct to 6 decimal places.

g Tan x can be approximated by . Using the three terms shown,

to what decimal place accuracy does this approximate the value of ?

sin x xx3

3!----- x5

5!----- …–+–=

12n 1–( )!

-----------------------⎝ ⎠⎛ ⎞ x2n 1–

sin π4---

sin x x x3

3!-----–≈ sin

π4---

sin x x x3

3!-----–

x5

5!-----+≈ sin

π4---

sinπ4---

sinπ4---

cos x 1x2

2!----- x4

4!----- …–+–=

tan x x2x3

3!-------- 16x5

5!-----------+ +≈

tanπ6---

Analysis task 3—approximating values of sin, cos and tan

SAC

Approxim

ating values of sin, cos and tanSAC analysis task

1.3

1.4



48

Further solving techniques

Literal equationsA literal equation is one in which the solutions are given in terms of pronumerals rather than as numerical values. We use the techniques that have already been covered in this chapter to solve literal equations.

exercise 1.51 Solve for x in each of the following.

a 2x − 3y = 5 b a = c + cxt c c(2x + b) = x − 3b

d e f

g 2ekx = 4 h e2kx − 6ekx + 8 = 0 i logx a = 3

j x2 − 4ax − 5a2 = 0 k ax3 − 2bx = 0 l x2 − 4nx + n2 = 0

2 Solve each of the following pairs of simultaneous equations for x and y.

a x − y = 1 b mx + ny = mmx + y = m nx − my = n

Example 1Solve each of the following for x.

a ax − b = 5 b c Aekx = b d x3 − cx = 0

Solutiona ax − b = 5

ax = b + 5

x =

b =

= a

= a

c − 2 = 2ax

x =

c2x------ 1

x--- a+=

WarningIt is essential to type a multiplication sign between a and x in order to solve this equation correctly using the Solve command of a CAS. If the multiplication sign is omitted, ax is assumed to be one variable and so no solution can be found for x.

b 5+a

------------

c Aekx = b

ekx =

kx =

x =

bA----

logebA----

1k---loge

bA----

d x3 − cx = 0

x(x2 − c) = 0

x(x − )(x + ) = 0

x = 0, , −c c

c c

c2x------ 1

x--- a+

c2x------ 1

x---–

c 2–2x

-----------

c 2–2a

-----------

xp--- x

q---+ 2= px r+

rx p–--------------- r

p---= 1

x--- 1

a--- 1

b---+=

1.5



49

Non-linear simultaneous equationsIn section 1.3 we reviewed the techniques for solving linear simultaneous equations. In this section we will consider simultaneous equations where one or more of the equations is non-linear.

Example 2Solve for x and y.

a x + 2y = 4 b y = x + 3

x2 + y2 = 16 xy = 10

Solutiona x + 2y = 4 (1)

x2 + y2 = 16 (2)

Make x the subject in equation (1).

x = 4 − 2ySubstitute for x in equation (2) and solve for y.

(4 − 2y)2 + y2 = 1616 − 16y + 4y2 + y2 = 16

5y2 − 16y = 0y(5y − 16) = 0

y =

Substitute into the linear equation to find x:

x = 4 − 2yWhen y = 0, x = 4.

When .

The solutions are x = 4, y = 0 or

.

b y = x + 3 (1)xy = 10 (2)

Substitute x + 3 for y in equation (2)and solve for x.

x(x + 3) = 10x2 + 3x − 10 = 0

(x + 5)(x − 2) = 0x = −5, 2

Substitute into equation (1):

y = x + 3

When x = −5, y = −2.

When x = 2, y = 5.

The solutions are x = −5, y = −2 or x = 2, y = 5.

0165

------,

y165

------ x, 4 325

------– 125

------–= = =

x125

------ y,–165

------= =

WarningOnce answers have been found for one variable, substitute into the linear equation to find the corresponding values of the second variable. Not only is it easier, but if we substitute in the non-linear equation, we may get extra, invalid answers.

For example, in example 2 part a, one of the answers for y was y = 0. If we substitute into the non-linear equation, we get x2 = 16, giving x = ±4. But the value −4 is invalid (check in the first equation).

1.5



50

Example 3If y = Aekt, y = 10 when t = 1, and y = 25 when t = 3, find the values of A and k, giving exact answers. Also give the answers correct to 2 decimal places.

Solutiony = Aekt

y = 10 when t = 1: 10 = Aek (1)y = 25 when t = 3: 25 = Ae3k (2)

Equation (2) divided by equation (1) gives:

=

= e2k

ln = 2k (writing in logarithm form)

k =

= 0.46 (correct to 2 decimal places)

Substitute into equation (1):

10 = Aek

=

=

= (using elnx = x)

A =

=

= = 6.32 (correct to 2 decimal places)

TI-Nspire 10.2

ClassPad 10.2

TipThe Solve command of a CAS can be used to solve non-linear simultaneous equations. The screenshot opposite shows the solution in example 2 part a.

2510------ Ae3k

Aek-----------

52---

52---

12---ln

52---

k12---ln

52---=

Ae12--- ln5

2---

Aeln 5

2----

Warning Beware roundingWhen substituting the value of k into (1) to find the value of A, use the exact value or, if necessary, a more accurate numerical approximation for k. If 0.46 is used, the value of A comes out to be 6.31, which is not correct to 2 decimal places.

52----A

10 2

5---------------

10 105

------------------

2 10



51

exercise 1.53 Solve for x and y.

a y = x + 1 b x + y = 7 c 3x − 2y = 4

y = x2 − 4x + 7 x2 + y2 = 65 xy = 2

d 3x + 2y = 2 e x + 2y = 3 f x + 2y = 1

4x + xy + y = 2 y2 − x2 = 3 y2 = x + 2

4 If y = Aekt, y = 16 when t = 1, and y = 32 when t = 3, find the exact values of A and k.

5 A = A0e−kt, where A0 and k are real value constants. Find the values of A0 and k, correct to 2 decimal places, if A = 30 when t = 3 and A = 60 when t = −2.

continued

A group of polynomials known as Chebyshev polynomials were first investigated by Russian mathematician Pafnuty Chebyshev (1821–1894). They have many properties that make them very useful in higher level mathematical analysis. The first five Chebyshev polynomials are listed here.

T0(x) = 1

T1(x) = x

T2(x) = 2x2 − 1

T3(x) = 4x3 − 3x

T4(x) = 8x4 − 8x2 + 1

More Chebyshev polynomials can be found by using the rule Tn + 1(x) = 2xTn(x) − Tn − 1(x). For example

T4(x) = 2xT3(x) − T2(x)

= 2x(4x3 − 3x) − (2x2 − 1)

= 8x4 − 6x2 − 2x2 + 1

= 8x4 − 8x2 + 1

a Find T5(x), T6(x) and T7(x).

b Show that .

c Express x3 as a sum or difference of Chebyshev polynomials.

d Any cubic polynomial can be expressed in terms of Chebyshev polynomials. Express ax3 + bx2 + cx + d in terms of Chebyshev polynomials.

e Use question d to express each of the following in terms of Chebyshev polynomials.

i x3 + 2x2 − 5x + 4 ii (x − 1)3 iii 2x3 + x2 − 5

f Show that any polynomial of degree 4 can be written in terms of Chebyshev polynomials.

x2 12---T2 x( ) 1

2---T0 x( )+=

Analysis task 4—Chebyshev polynomials

SAC

Chebyshev polynomials

SAC analysis task

1.4

1.5


Chapter review

52


SummaryAlgebra of polynomials. Polynomials are algebraic expressions of the form P(x) = anxn + an − 1xn − 1 + … + a1x + a0,

where n is a positive integer or zero and the coefficients an, an − 1, …, a1 and a0 are real numbers. The polynomial with leading term anxn is said to have degree n.

. The remainder theorem states that if a polynomial P(x) is divided by (x − a) then the remainder is given by P(a).

. The factor theorem states that if P(a) = 0 then (x − a) is a factor of P(x). When dividing

by (bx − c) the remainder is given by .

. Two polynomials P1(x) and P2(x)

P1(x) = anxn + an − 1xn −1 + an − 2xn − 2 + … + a1x + a0

P2(x) = bnxn + bn − 1xn −1 + bn − 2xn − 2 + … + b1x + b0

are equal for all values of x if and only if an = bn, an − 1 = bn − 1, …, a1 = b1, a0 = b0.

Solution of polynomial equations. Methods for solving quadratic equations include

a factorisation

a completing the square

a using the quadratic formula

a numerical or graphical approximation.

. For cubic and higher order polynomial equations it is necessary to use either factorisation or technology to find solutions.

Pascal’s triangle and binomial expansions. To generate a new row in Pascal’s triangle assign the number 1 to either end of the row

and then all numbers in between are found by adding the two elements diagonally above.

. When expanding (x + a)n, the coefficients are given by the nth row of Pascal’s triangle. Alternatively,

(a + b)n = an + nC1an − 1b + … + nCran − rbr + … + nCn –1abn − 1 + bn

where n ∈ N.

The (r + 1)th term, or general term, is given by nCr an − rbr, r = 0, 1, 2, …, n.

Systems of simultaneous linear equations. Simultaneous equations can be solved algebraically by substitution or elimination. The rref

or Solve commands of a CAS can be used.

. A system of linear equations can have infinitely many solutions. The solutions can be expressed in terms of a parameter.

. An inconsistent system has no solutions. In the case of two equations with two unknowns, this occurs when the two equations represent parallel lines.

Pcb---⎝ ⎠

⎛ ⎞



53

Algebra of exponential and logarithmic functions. Definition of a logarithm: If a > 0 and ax = y then loga y = x.

. Laws of logarithms

a loga xy = loga x + loga y

a

a loga xn = n loga x

a

a loga 1 = 0

. Change of base rule: .

. loga x is negative if 0 < x < 1.

. loge x is referred to as the natural logarithm and is also denoted by ln x.

Algebra of circular functions. Symmetry properties of the trigonometric functions.

. Negative anglessin (−θ) = −sin θcos (−θ) = cos θtan (−θ) = −tan θ

. The signs of the circular functions in each quadrant can be remembered by the CAST rule diagram:

. Special angles

Radians Degrees

Quadrant 2 sin (π − θ ) = sin θcos (π − θ ) = −cos θtan (π − θ ) = −tan θ

sin (180 − θ )° = sin θ °cos (180 − θ )° = −cos θ °tan (180 − θ )° = −tan θ °

Quadrant 3 sin (π + θ ) = −sin θcos (π + θ ) = −cos θtan (π + θ ) = tan θ

sin (180 + θ )° = −sin θ °cos (180 + θ )° = −cos θ °tan (180 + θ )° = tan θ °

Quadrant 4 sin (2π − θ ) = −sin θcos (2π − θ ) = cos θtan (2π − θ ) = −tan θ

sin (360 − θ )° = −sin θ °cos (360 − θ )° = cos θ °tan (360 − θ )° = −tan θ °

θ 0 π 2π

sin θ 0 1 0 −1 0

cos θ 1 0 −1 0 1

tan θ 0 1 undefined 0 undefined 0

log axy--- log a x log a y–=

log a1x--- log a x–=

log a xlog b xlog b a---------------=

x

y

S

T

A

C

π6--- π

4--- π

3--- π

2--- 3π

2------

12---

1

2------- 3

2-------

32

-------1

2------- 1

2---

1

3------- 3


54


. Steps for solving a trigonometric equation:

a Use algebra to get the equation in the form f(bx) = c, where f is one of sine, cosine or tangent.

a Identify the quadrants in which the solutions lie.

a Find the reference angle in the first quadrant.

a Find other solutions using symmetry properties.

. General solutions to equations involving sin x or cos x are found by adding integer multiples of 2π to the solutions on the interval [0, 2π].

. General solutions to equations involving tan x are found by adding integer multiples of π to the solution on the interval [0, π].

Miscellaneous equations. Literal equations are those for which the solutions are given in terms of pronumerals

rather than numerical values.

. Simultaneous equations where one or more of the equations is non-linear are usually solved by substitution. Dividing equations is a useful technique when solving simultaneous exponential equations.

Revision questionsShort answer technology-free questions1 a Find the remainder when the polynomial P(x) = kx3 − 3x2 + 5x + k is divided by (2x + 1).

b For what value(s) of k is (2x + 1) a factor of P(x)?

2 Show that (x + 2a) is a factor of P(x) = x3 + (2a − 3)x2 − 2(3a − 1)x + 4a and hence solve the equation P(x) = 0.

3 Expand:

a (3x − 2)4 b (1 − 2x)5

4 Solve for x in each of the following:

a x4 − 3x2 + 2 = 0 b x3 − 7x − 6 = 0 c x3 − 2x2 − 9x + 18 = 0

5 Solve each of the following systems of linear equations.

a 3x + 2y − 4z = 9 b x + y − z = 4

x + y + z = −1 2x + y − 3z = 7

2x − y + z = −6 4x + 3y − 5z = 15

6 a Find the equation of the quadratic polynomial that passes through the points (1, 2), (2, 1) and (3, −4).

b Find the equation of the cubic polynomial that passes through the points (−1, 14), (0, 5), (1, 2) and (2, −7).

7 Solve for x, giving exact answers.

a 2ex = 24 b 3x + 2 = 92x − 1

c log3 x + log3 (x + 2) = log3 (x + 6) d ln(3x) = ln(x2) + ln(x − 2)



55

8 Find exact values for x that satisfy the following equations.

a b e2x − 8ex + 15 = 0

9 Solve for x.

a log2 (x + 5) = 3 + log2 (x − 2) b loge (x + 1) + loge (x − 1) = loge 8

10 Find the exact values of:

a sin 120° b c tan 300°

d e f

11 Find exact solutions for x if:

a b

c d

12 Solve for x.

a b

13 Find general solutions to each of the following equations.

a b cos 2x = −1 c

14 Solve for x and y.

a 3x − 7y + 6 = 0 b y = 1 + 2e−x c x2 + y2 = 4

x2 − y2 = 16 y = ex y = 2x2 − 2

15 If , express y in terms of x.

Extended response1 The height of a moving body above the ground is given by , where

h metres is the height of the body measured at time t seconds, u metres per second is the initial velocity of the body, a is a constant and h0 metres is the initial height of the body.

a Find the values of a, u and h0 if h = 19 at 1 second, h = 24 at 2 seconds and h = 4 at 4 seconds.

b After how many seconds does the object hit the ground?

c Find the exact times that the body is 10 metres above the ground.

2x 1– 18---=

cos π4---–⎝ ⎠

⎛ ⎞

cos5π6

------ tan5π4

------ sin3π2

------

sin x°12--- 0 x 360≤ ≤,–= cos x

1

2-------- 0 x 2π≤ ≤,=

sin x32

-------- 0 x 4π≤ ≤,–= tan x 1 2π x 2π≤ ≤–,–=

2sinπx6

------ 1 0 x 24≤ ≤,= cos 3x12--- 0 x π≤ ≤,–=

sin x 1

2--------–= tan x

1

3--------=

xlog e alog e y---------------=

h12---at2 ut h0 t 0≥,+ +=


56


2 The depth of water at the entrance to a harbour in a 24 hour period after midnight on a

particular day can be modelled by the equation , where

d metres is the depth of water at time t hours after midnight.

a Find the depth of water at:

i midnight ii 4 am iii 6 pm

b Low tide occurs at 6 am.

i What is the depth of water at low tide?

ii When is the next low tide?

c At what times is the depth of water at the harbour entrance 6 metres?

3 a If P(x) = ax3 + bx2 + cx + 3, write down an expression for P(1) in terms of a, b and c.

b i If (x − 1) is a factor of P(x) and P(2) = −7 and P(−1) = 14, write down a system of three simultaneous equations that could be used to find the values of a, b and c.

ii Hence find the values of a, b and c.

c If (x − 1) is a factor of P(x) = ax3 + bx2 + cx + 3 and P(2) = −7 (but P(−1) is not known),

i express a and b in terms of c.

ii write P(x) with coefficients in terms of c.

d Show that .

e For what value(s) of c does the equation P(x) = 0 have:

i one real solution?

ii two real solutions?

iii three real solutions?

4 Joshua fell over and has a large wound on his shin. As his wound heals, the surface area of the scab decreases. The surface area of the scab, A square centimetres, t days after the wound was received, can be modelled by the equation A = 10 × 2−0.1t.

a What was the original surface area of Joshua’s wound?

b What is the surface area of the wound after 3 days?

c How many days after the wound was received has its surface area reduced by

i 50%? ii 95%?

d The equation for A can be re-expressed in the form A = b × e−kt, where b and k are constants and e is the natural base.

i Write down the value of b.

ii Find the value of k, correct to 4 decimal places.

d 2 cosπt6----- 7 0 t 24≤ ≤,+=

P x( ) 12--- x 1–( ) c 1+( )x2 2 c 3+( )x– 6–( )=

Practice quizStudent CD


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