Math 3 Honors Module 3

Honors Topics

You learned how to factor the difference of two perfect squares:

Example:

But what if the quadratic is ? You learned that it was not factorable. It is factorable, but with complex factors.

How does it work? Let’s solve an equation. If you solve the equation , you get solutions x =1,-1. If you solve the equation , you get the solutions x = i ,-i.

Since a quadratic factors into (x – zero)(x-zero):

Complex Factors of Quadratics

2 2 ( )( )x y x y x y 2 1 ( 1)( 1)x x x

2 1x

2 1 0x 2 1 0x

2 1 ( )( )x x i x i

Factor each quadratic expression.

Examples of Complex Factors

2

3 3

9

x x

x

i i

2

( 8

64

)( 8 )x i

x

x i

2

( 10 )( 10

1

)

00

x i

x

x i 2

2 2

7

4

49

7x i

x

x i

You must always simplify powers of the imaginary unit, i. You learned the properties:

i² = -1 and

But how would you simplify ? You need to find out how many pairs/sets of imaginary units, i, you have and simplify them.

Since every i² = -1,

Simplify Powers of i

1 i

8i

48 2 4( 1) 1i i i i i i i i i i

Simplify each imaginary number.

Simplifying Powers of i- Examples

1428 2 14( 1) 1i i

2244 2 22( 1) 1i i

3978 2 39( 1) 1i i

714 2 7( 1) 1i i

817 2 8( 1) 1i i i i i i

2449 2 24( 1) 1i i i i i i

1735 2 17( 1) 1i i i i i i

4183 2 41( 1) 1i i i i i i

even 1 or 1i odd or i i i

Recall that you must rationalize the denominator of a fraction. You cannot leave a radical on the bottom of a fraction. You also cannot leave a complex number on the bottom of a fraction.

Conjugates:4-6i, 4+6i ½ + i, ½ - I 7i, -7i

Division of Complex Numbers

Divide by a Complex Number a + bi Multiply numerator and denominator by

the conjugate of the bottom, a – bi

Divide.

Complex Number Division Examples

2

1 (2 )

2 (2 )

2 2

4 4 ( 1)

2 2 1 or

5 5

1

5

2i

i i

i i

i

i

i

i

2

2

3 (1 5 )

1 5 (1 5 )

3 15 3 15( 1)

1 25 1 25( 1)

15 3 15 3 or

26 26 2

3

5

6

1i i

i

i i

i i i

i

i

ii

2

2

(2 3 ) (4 2 )

(4 2 ) (4 2 )

8 4 12 6

16 48 16 6( 1)

16 4( 1)

8 16 6 2 16

16 4 202(1 8 ) 1 8 1 4

20 10 10 5

2 3

4 2i i

i i

i i i

i

i

i i

i

i

i

i

i

Solve a quadratic equation in standard form with leading coefficient of 1

Sum and Product of Roots

2 0x bx c 2 3 2 0

( 2)( 1) 0

2 0 or 1 0

2 or 1

x x

x x

x x

x x

Factored form is (x-m)(x-n) = 0Therefore x = m, x = n are the roots.

m + n = -2 + -1 = -3, which is -bm · n = (-2)(-1) = 2, which is c

If a = 1 and , then:Sum of the roots: m + n = -bProduct of the roots: m · n = c

or Note: if a≠1, sum = -b/a and product = c/a

2 ( ) ( ) 0x m n x m n

2 0x bx c

Given the zeros of the quadratic , write an equation in standard form.

x = 5, -4 {-6, 0} x = 5±3i 5+-4 = 1 -6+0 = -6 (5+3i)+(5-3i) = 10 So -b = 1 So –b = -6 So –b = 10 b = -1 b = 6 b = -10 5(-4) = -20 (-6)(0) = 0 (5+3i)(5-3i)= 34 So c = -20 So c = 0 So c = 34

Sum and Product Examples2 0x bx c

2 20 0x x 2 6 0x x 2 10 34 0x x

Recall the standard form of quadratic equation: .

The discriminant, , of the quadratic indicates the nature of the roots.

2 real roots 1 real root 2 complex roots

Q: How would you find the value of a, b, or c in a quadratic equation given that there are 2 real roots? 1 real root? 2 complex roots?

A: Set up an equation or inequality and solve!

Solve Quadratic Functions Given Nature of the Roots

2 0ax bx c 2 4b ac

2 4 0b ac 2 4 0b ac 2 4 0b ac

Given , find the value(s) of c if there is/are:

2 real roots 1 real root 2 complex roots

*C is less than 1 *C is equal to1 *C is greater than 1

Nature of Roots - Examples2 2 0x x c

2

2

4 0

2 4(1) 0

4 4 0

4 4

1

b ac

c

c

c

c

2

2

4 0

2 4(1) 0

4 4 0

4 4

1

b ac

c

c

c

c

2

2

4 0

2 4(1) 0

4 4 0

4 4

1

b ac

c

c

c

c

You learned how to solve quadratic equations using different methods and you learned how to solve quadratic inequalities by graphing. Now let’s look at solving inequalities algebraically.

Solve Quadratic Inequalities Algebraically

Solving Quadratic Inequalities Using Algebra1. Write the inequality as an equation .2. Solve the quadratic equation by factoring.3. Graph each solution on the same number line.4. Test a number in all three sections of the number line by substituting it into the original inequality. If the inequality is True, each number in that section is a solution.5. Write your solutions as an inequality. If m and n are the zeros for the quadratic equation, the solutions will be: solution 1: m < x < n OR solution 2: x< m or x > n

2 0ax bx c

Examples of Solving Quadratic Inequalities Algebraically

Solve the inequality Write it as an equation and find the roots Graph the roots -2, 2 on the number line Use a closed circle since it is ≤ Check values in each section x = -2, 0, and 3

Solution: -1 ≤ x ≤ 2

Note: if the inequality is , then the False sections would now be true. The solution would be x < -1 or x > 2.

2 2 0x x 2 2 0

( 2)( 1) 0

2 0 or 1 0

2 or 1

x x

x x

x x

x

2( 2) ( 2) 2 0

4 2 2 0

False, -2 is not shaded

2(0) (0) 2 0

0 0 2 0

True, 0 is shaded

2(3) (3) 2 0

9 3 2 0

False, 3 is not shaded

2 2 0x x