grade 11 pre-calculus unit a: quadratic … grade 11 pre-calculus unit a: quadratic equations...
TRANSCRIPT
1
GRADE 11 PRE-CALCULUS
UNIT A: QUADRATIC EQUATIONS (ALGEBRA)
CLASS NOTES
1. A definition of Algebra: A branch of mathematics which describes basic arithmetic
relations using variables.
2. Algebra is just a language. Instead of saying ‘Jason is 6 years older than Karen’; a
mathematician says: J = K + 6. It is a language that is universal and very concise with a select
few but well defined rules. For example we could change the equation to K= J – 6 because of
one of the special rules.
3. To date we have mainly just used Algebra of simple functions like lines. In this unit we
learn to manipulate more complex relationships between variables.
QUADRATIC EQUATION
4. The general form of a quadratic function is of the form y = f(x) = ax2
+ bx + c. When
the function takes on a specific value for y, it becomes an equation. The most common
equation of interest is when the y = 0.
5. The quadratic equation is of the form ax2
+ bx + c = 0
You try to ‘solve’ the equation x2
– 5x – 6 = 0. Put your answer below.
The solution is the ‘root(s)’ of the equation or the x-intercepts of the function graph. We had
avoided finding the roots or x-intercepts in Unit C. Recall also that a quadratic function can
have either 0, 1, or 2 roots or x-intercepts.
SOLVING QUADRATIC EQUATIONS USING GRAPHING TOOLS
6. We already did this in Unit C. Just graph the function, find where it crosses the x-axis.
Using the CALCULATE function of the TI-83 is the easiest method. Notice this often only
gives an approximate but very close answer though! Really close, but not necessarily exact!
gr11precalc_A_QuadraticEqns.doc Revised:20141102
To ‘solve’ means to find a
value of a variable (or
variables), that makes an
equation true.
2
SOLVING QUADRATIC EQUATIONS USING FACTORING
7. Back to Grade 10! Example Solving Quadratic: solve for x given that x2 – 5x – 6 = 0.
Use the zero property that if a*b = 0 then either a, or b, or both must be zero!
Recall how to factor: what two numbers multiply to give –6 but add to give –5.
The result is that x2 – 5x – 6 can be factored as (x – 6)(x + 1). So if
0652 =−− xx is the same as 0)1)(6( =+− xx then it can only be true for x = 6 and x = –1.
8. The factoring method works nicely for some nicely contrived quadratic expressions and
for integer number solutions. You will need to recall your factoring of trinomials and
difference of squares from Grade 9 and 10.
a. Review of Factoring of trinomial expressions: ax2 + bx + c.
Factoring actually seldom works in real life applications unless the question has been nicely
contrived. You will soon learn a way that you never have to factor again!
b. Factoring General Trinomials
ax2 + bx + c, where a = 1
.
Example: x2 + 7x + 6
What two factors multiply to give 6 and add
to give 7. 6 and 1. So
x2 + 7x + 6 is the same as (x + 6)(x + 1)
Has two different roots
c. Factoring Perfect Squares. Special
trinomials where (b/2)2 = c and a =1
btw: ‘Perfect squares’ are numbers like: 1, 4,
9, 16, 25, 36, 49, etc.
Example: x2 + 6x + 9
The first term is a perfect square and so is the
last. Good clue that you may be able to just
write:
(x + 3 )(x + 3 ) or just (x + 3 )2
It has two factors the same, the equation
equated to ‘0’ has only one root: –3.
3
d. Factoring Difference of squares.
They have no b term (it is zero). A typical
form:
ax2 – c
where a is a perfect square and c is a perfect
square.
Example: 4x2 – 81.
Since this is the difference of two perfect
squares it can immediately be written as:
(2x + 9)(2x – 9) (Has two separate factors and therefore two
separate roots of its equation: –4.5 and +4.5)
e. Factoring when a ≠≠≠≠ 1
These are often considerably more difficult
and only work for specially contrived
expressions. Methods:
Try factoring out the ‘a’ first
Try Trial and error The AC Method. See the appendix to
these notes for the AC Method.
Example: 6x2 + x –1
Factoring out the ‘a’.
6x2 + x –1 = 6 (x
2 + 1/6x – 1/6)
What two things multiply to give –1/6 and
add to give +1/6?
Can you see that it is ½ and – 1/3? I don’t
expect that you do! But sometimes factoring
out the ‘a’ does work nicely if it can be done.
Trial and error. You know it should
look something like (mx + p)(nx + q),
so that m*n=6 and p*q = –1 and mq +
np = +1 Notice (2x + 1) (3x – 1) works nicely
9. You try a few; factor each of the following quadratic equations and state the roots
(without using a graph or graphing calculator). Recall that a quadratic has a square on one
unknown. Notice also that we always equate a quadratic equation to zero to solve it!
a. x2 – 5x – 6 = 0
b. x2 – 4x + 4 = 0
c. 4x2 – 4x = 0
d. 4x2 – 16 = 0
e. x2 – 2x = 24
f. 2x2 + 4x – 16 = 0
4
e. x2 – 2x = 24
f. 2x2 + 4x – 16 = 0
10. Now check the solutions (the roots) with a graphing calculator or an on-line grapher to
see that they are correct.
AN APPLICATION
11. An object is shot upwards at a speed of 300 meters per second on earth (almost the speed
of sound). The equation for its height above ground is h(t) = –5t2 + 300t. The h, a function of
t, is height in meters, the t is time in seconds. Find when then the body hits the earth after it is
thrown; ie: when h = 0.
Ans: Solve for 0 = –5t
2 + 300t. So t = 0 or t = 60.
Solving Simple Radical Equations by Unsquaring (‘taking Square Root’)
12. A radical equation involves roots (square roots, cube roots, …) and fractional exponents.
We will only need square roots in Grade 11.
Solve for x: x2 = 4; x = _______; x
2 = 9; x = ________. x
2 = 81; x = _________
13. Notice that we always have ___________ answers when we take a square root (or any
even root for that matter). We use the symbol _________ toshow the answers, it really just
means ________________
Solve Simple Radical Equations by Taking square Root:
Example: (x + 2)2 = 64
Unsquare Both sides: x + 2 = ±±±± 8 Remember there are two answers when unsquare!
Isolate the x: x = 2 ±±±± 8; so x = 10 or x = –6
Solution set: {
–6, 10} (Notice we put the solution set in the ‘{ }’ braces rather than the ‘( )’
parenthesis just so we don’t inadvertently think that it is a point on a coordinate plane)
5
You try; solve for x:
a. (x – 2)2 = 16
Ans: {-2, 6}
b. (x + 3)2 + 2 = 18
Ans: {-7, 1}
c. 702)4(2 2 =−−x
Ans: {-2, 10}
d. 14)1(2 2 =−x
Ans: }71;71{ −−+− or 71±−
SOLVING TRIGONOMETRIC EQUATIONS WITH A QUADRATIC FORM
Example 1: Solve for θ [0,360].
(if you have not done Grade 11 Trig yet omit this)
2cos2(θθθθ) – 1 = 0
Ans: 45°°°°, 135°°°°, 225°°°°, 315°°°°
Example 2: Solve for θθθθ.
9sin2θθθθ = +4
Ans: θθθθ = 41.8°°°°, 138.2°°°°, 221.8°°°°, 318.2°°°°
Note: cos2(θ) means (cos(θ))
2
Notice there are four answers to these
quadratic trigonometric equations!
6
14. If you find trigonometric functions in a quadratic form confusing just make them easier to
understand by substituting something simpler for the trigonometric function.
So 2cos2(θθθθ) – 1 = 0 becomes 2x
2 – 1 = 0. Then don’t forget to turn the x back into cos(θθθθ) at the
end.
IMAGINARY NUMBERS
15. We have been told it is not possible to take the square root of a negative number. That is
not quite true! Several hundred years ago we invented a number system called the ‘imaginary’
numbers.
16. The imaginary number ‘i’ is defined as: i2 = –1 or 1−+=i
It is used to write the square root of any negative number
17. ‘i’ has the property sufficient to solve:
aiaaa =−=−=− *1*1
it also has the property:
( ) aai −=2
Example:
18. Solve x2 = –8.
Solution:
8−±=x
ix 2224 ±=−±=
SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE
19. Not all quadratic equations are easily factored by just looking at them. And not all are in
a simple squared binomial form; (x – h)2 ready to un-square. Often Completing the Square
will help (always in fact!). Recall completing the square from Unit C – Quadratic Functions.
We add and subtract a ‘magic’ number to make a ‘perfect square trinomial’. Completing the
square is the way we can take equations with an x and an x2 in them and just turn them into
something with an just an ‘x + h’ in them that makes them easier to solve.
Don’t panic over the ± . It is just a
mathematicians way of saying two
things at once. It is not really an
actual operation! It means there is a
plus answer and a negative answer
7
EXAMPLE 1: (simple where a in the general form is equal to 1)
20. Solve for x: x2 – 4x + 2 = 0
Solution:
0242 =+− xx
242 −=− xx 22
2
2
42
2
44
−+−=
−+− xx
42442 +−=+− xx
(x - 2)2
= 2
22 ±=−x
22 ±=x
Check the positive answer
(2+√2)2–4(2+√2)+2=
4 + 4√2 + 2 – 8 – 4√2 + 2 = 0 ♫
Check the negative answer:
General form
Isolate the variable terms to one side
Add
2
2
bto both sides. (This is the same a
adding and subtracting it on the same side)
Simplify
Factor and simplify
Check that the answers work by evaluating !
21. Check the answer with a graphing calculator or an on-line graphing tool by graphing now
if you want! But of course, it will only give you a real close answer, not exact like when we
solve it ‘analytically’ using algebra.
22. Check the answer also just by evaluating a very accurate value of your solutions.
Knowing how to use the memory button on your calculator will help to remember those
lengthy approximations of the irrational values.
Remember? The magic number to add is half
of the ‘b’ coefficient of the x, all squared. Or
(b/2)2 if and only if the a is 1.
8
EXAMPLE 2: (where the a in the general form is not equal to 1; a ≠≠≠≠1)
22. Solve: 2x2
– 6x – 8 = 0
Solution:
23. Check the answer with a graphing calculator or other graphing tool.
THE QUADRATIC FORMULA
24. All the previous methods of solving a quadratic are interesting but they have their
limitations. There is a way to solve a quadratic everytime with one formula. It is called the
quadratic formula. It looks like this:
a
acbbx
2
42 −±−=
Memorize it if you plan to be in science or engineering!
It is really just a ‘pre-washed’ version of completing the square!
DERIVATION AND PROOF OF THE QUADRATIC FORMULA
25. The quadratic formula is easily derived from the general form or a trinomial. The
quadratic formula is really just completing the square, with all the algebra already done for you
so all you have to do is ‘plug in’.
(You are not responsible to know the proof, but interesting to see why it is true! It simply falls
out of completing the square!) (See separate notes). If you take Calculus in university you will
do lots of proofs!
The secret to these, where the leading
coefficient on the x2 term is not a 1
(a ≠≠≠≠ 1), is just to divide the entire
equation by that annoying ‘a’
coefficient if you can.
9
EXAMPLES OF SOLVING WITH QUADRATIC FORMULA
26. Solve for x: x2
+ 3x – 9 = 0
27. Solve for x: 2x2
+ 8x +10 = 0
SOLVE EQUATIONS THAT CAN BE RE-WRITTEN AS QUADRATIC EQUATIONS
28. Some equations are not really quadratic, but some can be re-written as such.
Solve: 2x4
– 5x2
– 2 = 0
How many roots does
this have?
How many ‘roots’ does
this have?
This is called a ‘quartic’ equation, its
highest power is x4. A fourth degree
polynomial.To solve, just say ‘something
else’ = x2
10
29. Some trigonometric equations have a quadratic ‘form’. Solve for θθθθ:
2sin2θθθθ + sinθθθθ – 1 = 0 by substituting x = sin(θθθθ) and using the quadratic formula. Keep θθθθ in the
interval domain of [0, 360]
Ans: 30°, 150°, 270° (Omit this idea if you have not yet done Grade 11 Trig. But you will see
it again)
PROPERTIES OF THE QUADRATIC FORMULA
30. Discriminant. The expression acb 42 − under the radical in the Quadratic Formula is
called the discriminant. It separates or ‘discriminates’ how many roots and what type of roots
a quadratic formula has.
If b2 - 4ac is: Then the equation has:
>0 and is a non-perfect square
2 real irrational roots
>0 and is a perfect square
2 real rational roots
=0
1 real number root
<0
0 real roots; the roots are
‘imaginary’
11
31. Graph the following equations, and then check with the ‘discriminant’, (b2– 4ac), to see
how many real number solutions exist. Record the number and type of roots for each eqn:
a) x2
– 2x –12 = 0
Nbr and type of Roots:
b) –x2
+ 2x + 18 = 0
Nbr and type of Roots
c) x2
– 4 = 0
Nbr and type of Roots
d) x2
– x + 2 = 0
Nbr and type of Roots
e) x2
+ 6x +1 = 0
Nbr and type of Roots
f) –x
2 + x - 4=0
Nbr and type of Roots
g) x2
– 6x + 9 = 0
Nbr and type of Roots
h) x2
+ 4x + 4 = 0
Nbr and type of Roots
i) –x
2 + 10x – 25 = 0
Nbr and type of Roots
Often in engineering it is handy to check the discriminant first to make sure there will be a
solution before you start calculating the full solution. You don’t want to build a bridge with
non-real numbers!!!
WRITE THE QUADRATIC EQUATION GIVEN ITS ROOTS
32. If a quadratic equation has two roots: r1 and r2, then the quadratic equation can be written
as the product of two binomials: (x - r1)*(x - r2) = 0
EXAMPLE 1: Say the roots are known to be –3 and +4. Then the quadratic can be found
from the two binomials: (x + 3 )*(x – 4) = 0. or by multiplying:
x2
– x – 12 = 0
12
EXAMPLE 2. The roots (ie: where y = 0) are known to be 3
137 + and
3
137 −
33. The quadratic equation can be re-written as:
03
137*
3
137=
−−
+− xx
03
137*
3
137
3
137
3
1372 =
+
−+
+−
−− xxx by F.O.I.L
09
1349
3
13
3
7
3
13
3
72 =
−+−−+− xxxxx Simplifying
09
36
3
142 =
+− xx Further simplifying
043
142 =+− xx is one quadratic equation with the roots 3
137 + and
3
137 −
34. Notice also though that multiplying by 3; 012143 2 =+− xx is also a quadratic
equation that has those roots. There are an infinite number of equations that have the same
roots. You would need another point (or a vertex) to get a particular equation.
EXAMPLE OF FINDING ROOTS
35. It is known that the equation for the morning rush hour traffic at Higgins and Main can be
closely modeled by:
c = –45t2
+ 750t – 2900 where c is the number of cars in a minute and t is the local time in
hours. The equation is valid only throughout the domain 06:00 to 11:00.
The construction crews don’t like to disrupt traffic when traffic exceeds more than 100 cars in a
minute. Find the approximate times between which the road crews will avoid disrupting
traffic.
Solution:
36. We want to know when 100 = –45t
2 + 750t – 2900. So find the roots of the equation:
0 = –45t2
+ 750t – 3000
Ans: 6.666 and 10 so 06:40am and 10:00am
13
EXAMPLE FROM THE DAVINCI CODE
37. Maybe you watched the movie or read the book the Davinci Code. A key part of that
movie is the ‘Divine Number’ or the ‘Golden Ratio’. The number they give in the movie is
1.618, but they are wrong: it is actually the positive solution to the equation x2 – x – 1 = 0
38. You find the magic ‘Divine Number’ from the movie exactly!!! None of this decimal or
rounding stuff! Only exact numbers! Even sun flowers grow obeying this equation!
A-1
APPENDIX 1
FACTORING USING AC METHOD
1. Factoring trinomial expressions in one unknown is an important skill necessary to
eventually solve quadratic equations. Trinomial expressions are of the form ax2 + bx + c and
they can sometimes be factored into two binomials using ‘real’ numbers. You will learn later
that there are other numbers that are not ‘real’, and that in fact all trinomials can be factored
into two factors. And don’t forget: factoring is just ‘un-multiplying’! In fact, factoring is really
just dividing in other words.
2. You likely have learned several methods already: ‘factoring out common factors’,
‘perfect squares factoring’, ‘difference of squares’ factoring, ‘grouping’, ‘trial and error’, etc.
3. Factoring expressions of the form ax2 + bx + c can often be complicated if the a ≠≠≠≠ 1
(a not equal to one).
4. Example: Try your known methods of factoring the trinomial expression:
4x2 + 5x – 6
a. You cannot do the simple trick of finding two integers that multiply to give –6 and add
to give 5 because the ‘a’ (a = 4) is not equal to one.
b. You might stumble upon the grouping method: and notice that 4x2 + 5x – 6 could be
written as:
4x2 ( – 3x +8x) – 6
so: 4x
2 – 3x + 8x – 6
so: x(4x – 3) + 2(4x – 3) which is x bouquets of (4x – 3) plus 2 bouquets of (4x – 3) so:
(x + 2) bouquets of (4x –3); so (x + 2)(4x – 3).
Notice I like to call bracketed terms ‘bouquets’ because that is what they really are! My wife
loves her bouquets of 4x’s less three. Last week I bought her ‘x’ of them and this week I
bought her 2 of them. So she got (x + 2) bouquets of (4x – 3) or 4x2 + 5x – 6.
A-2
THE AC METHOD OF FACTORING
5. The AC Method will work sometimes for many trinomial polynomial expressions to help
you more easily find the four terms that can be readily factored by grouping.
Let’s try 4x2 + 5x – 6
• Step 1: Multiply the a*c: ac = –24
• Step 2: Find the factors (that multiply together) of –24 that also add to give +5. Notice
that –3 and +8 work.
• Step 3: Rewrite the original expression with the smaller number times x first (without
consideration of its sign; the one with the smaller absolute value as we call it) and the larger
number times x next in place of the original x term;
4x2 – 3x + 8x – 6. Notice that – 3x + 8 x is still 5x.
• Step 4: Factor the first two terms: x(4x – 3) and factor the second two terms 2(4x – 3)
• Step 5: Do the final factoring. x(4x – 3) + 2(4x – 3) both have (4x – 3) in common; so
(4x – 3)(x + 2) is the final answer.
• Step 6: Check your answer by multiplying out again (F.O.I.L)
(4x – 3)(x + 2) = 4x2 + 5x – 6
PRACTICE EXAMPLES:
6. Factor: 6x2 + 17x + 12
• ac = 72 which has factors of 36 and 2 , 9 and 8, etc.
• 9 and 8 add to give 17 though!
• So rewrite as: 6x2 +8x + 9x + 12 (smallest factor: ‘8’ first)
• Factor the first two terms and the last two terms:
2x(3x + 4) + 3(3x + 4)
• Do the final factoring since both parts have a 3x + 4 in common; thus:
(3x + 4)*(2x + 3)
• Check the answer by multiplying!
A-3
7. Example 2: Factor 10x2 –x – 3
• ac = –30 which has factors of 15, –2 and –5, 6 and –6, 5, etc...
• –6 and 5 add to give the ‘b’ of –1 though!
• So rewrite as: 10x2 +5x – 6x –3
• Factor the first two terms and the last two terms:
5x(2x + 1) + -3(2x + 1)
• Do the final factoring since both parts have a 3x + 4 in common; thus:
(5x – 3)*(2x + 1)
• Check the answer by multiplying! It works!
8. One for you to try. Factor: 16x2 + 34x + 4 (hint: 32 and 2 are the ac factors)
The answer is: (8x + 1)(2x + 4) but notice you should have factored out a GCF constant factor
of two first to make the factoring less onerous , so really the fully factored answer is: 2(8x +
1)(x + 2). Multiply it out to check it!
FACTORING IN TWO UNKNOWNS
9. Factoring in two unknowns is also possible using the AC Method with some adaptation.
10. Example. Factor: 2x2 + 5xy + 2y
2. Pretend this is in a form ax
2 + bxy + cy
2
‘ac’ = 4 and factors 4 and 1 add to 5
so: 2x2 + 1xy + 4xy + 2y
2.
So: x(2x + y) + 2y(2x + y)
So: (x + 2y)(2x + y)
A-4
SOME PRACTICE ON YOUR OWN
17. Try these on your own. The answers are given. Factor:
a. 3x2 + 7x + 4
Ans: (3x + 4)(x + 1)
b. 15x2 – 7x – 2
Ans: (3x – 2)(5x + 1)
c. –12x2 + 4x + 21
Ans: (6x + 7)(–2x + 3)
d. 15x2 – 44x – 20
Ans: (5x + 2)(3x –10)
Simplify by factoring the numerator
e. 2
62
−
−+
x
xx
Ans: (x + 3)
f. 3
383 2
−
−−
x
xx
Ans: (3x + 1)
B-1
APPENDIX B
FACTORING USING GEOMETRY (ALGEBRA TILES)
1. Some students like to see a more visual way of factoring. Your nieces and nephews in
Grade 5 and 6 probably learn it this way using Algebra Tiles. In fact this is the original way
that ancient mathematicians factored trinomials. This is just a demonstration of the method,
you can research it further on your own if you want.
2. Factor the trinomial 2x2 + 8x + 6
3. The ancients would draw squares of size x*x (ie: x2) in the dirt, and rectangles of length x
and width one, and tiny squares of length one by one. You have to pretend you do not know
how long a length of x is; it is ‘rubber’.
‘x’ wide
‘x’ lo
ng
‘x’ lo
ng
1
‘x2’ 1
1
1‘x’
‘x’ wide
‘x’ lo
ng
‘x’ lo
ng
1
‘x2’ 1
1
1‘x’
4. So 2x2 + 8x + 6 would be scratched in the dirt as:
x2
1
x
x2 x
x
x
xxxx
11 1 1 1
x2x2
11
xx
x2x2 xx
xx
xx
xxxxxxxx
1111 11 11 11
5. And then they noticed they could group them together like this:
1xx2 x xx
1
1
x2 xxxx
1
1
1
1xx2 x xx
1
1
11xxx2x2 xx xxxx
11
11
x2 xxxx
1
1
1
x2x2 xxxxxxxx
11
11
11
Which is really two groups of x
2 + 4x + 3
B-2
6. But both of these groups could be re-grouped like this:
x2 xxx
x
1 1 1
x + 1 + 1 + 1
x
+ 1
x2 xxx
x
1 1 1
x + 1 + 1 + 1
x
+ 1
x2 xxx
x
1 1 1
x + 1 + 1 + 1
x
+ 1
x2x2 xxxxxx
xx
11 11 11
x + 1 + 1 + 1
x
+ 1
x2 xxx
x
1 1 1
x + 1 + 1 + 1
x
+ 1
x2x2 xxxxxx
xx
11 11 11
x + 1 + 1 + 1
x
+ 1
Which can be expressed mathematically as 2 bunches of (x+3) by (x+1); in other words:
2[(x+3)(x+1)] or 2(x+3)(x+1)
So: 2x
2 + 8x + 6 = 2(x+3)(x+1)
Of course, negative numbers were a bit perplexing for the ancients! So they didn’t bother with
negative numbers. But you may find your nieces and nephews are taught that red tiles are
negatives, and green are positive and that a red cancels a green because they are opposites. It
just complicates things, but it can be done.
C-1
APPENDIX C
PRACTICE PROBLEMS
Exact Solutions only! None of those crazy rounded off decimals earthling things!
Change up the methods if you want!
Solve by Factoring
Don’t forget to equate to zero!
a. 01452 =−+ pp
{-7, 2}
b. 0302 =−− ww
{-5, 6}
c. 60192 −=+ uu
u=-4 or u = -15
d. 22 4151694 ttt −=+−
}
81,1{
e. 0 1 8c 12c2 =++
2
1;
6
1 −=
−= cc
f. 2474 2 =−y
27±=y
Solve by Taking Square Root
a. 91109 2 =+n
}3{±
b. 8543 2 −=− x
22±
C-2
c. 10828 2 −=−t
3±=t
d. 269
102
=+z
{12}
e. 1453 2 =+n
}3{±
f. 12)1( 2 =−z
121±=z or better: 321±=z
Solve by completing the square
Always works!
a. 6582 =− ww
{-13, 5}
b. 153142 =++ hh
717 ±−
c. 3563 2 =++ xx
3
1± or better yet
3
3±
d. 5882 2 −=++ xx
C-3
Solve using the Quadratic Formula
a. 6582 =− ww
b. 0672 2 =++ xx
c. 322 =+− xx
d. 5442 −=++ xx