SAT Problem of the Day

5.6 Quadratic Equations and Complex 5.6 Quadratic Equations and Complex NumbersNumbers

Objectives: •Classify and find all roots of a quadratic equation

Solutions of a Quadratic Equation

If b2 – 4ac > 0, then the quadratic equation has 2 distinct real solutions.

Let ax2 + bx + c = 0, where a = 0.

If b2 – 4ac = 0, then the quadratic equation has 1 real solution.If b2 – 4ac < 0, then the quadratic equation has 0 real solutions.

The expression b2 – 4ac is called the discriminant.

Example 1Find the discriminant for each equation. Then determine the number of real solutions.

a) 3x2 – 6x + 4 = 0 b2 – 4ac

= (-6)2 – 4(3)(4) =

36 – 48 =

-12 no real solutions

b) 3x2 – 6x + 3 = 0 b2 – 4ac

= (-6)2 – 4(3)(3) =

36 – 36 =

0 one real solution

c) 3x2 – 6x + 2 = 0 b2 – 4ac

= (-6)2 – 4(3)(2) =

36 – 24 =

12 two real solutions

PracticeIdentify the number of real solutions:

1) -3x2 – 6x + 15 = 0

Imaginary NumbersThe imaginary unit is defined as and i2 = -1.

i 1

If r > 0, then the imaginary number is defined as follows:

r

r 1 r i r

10 1 10 i 10

Example 2Solve 6x2 – 3x + 1 = 0. 2b b 4acx 2a

23 3 4(6)(1)x 2(6)

3 9 24x 12

3 15x 12

3 i 15x 12

PracticeSolve -4x2 + 5x – 3 = 0.

Homework

Lesson 5.6 exercises 19-35 Odd

SAT Problem of the Day

5.6 Quadratic Equations and Complex 5.6 Quadratic Equations and Complex NumbersNumbers

Objectives: •Graph and perform operations on complex numbers

Imaginary NumbersA complex number is any number that can be written as a + bi, where a and b are real numbers and

i 1; a is called the real part and b is called the imaginary part.

3 + 4i

real part

imaginary part

3 4i

Example 1Find x and y such that -3x + 4iy = 21 – 16i.

Real parts Imaginary parts-3x = 21

x = -74y = -16y = -

4x = -7 and y = -4

Example 2Find each sum or difference.

a) (-10 – 6i) + (8 – i) = (-10 +

8) = -2 – 7i

b) (-9 + 2i) – (3 – 4i)= (-9 –

3)= -12 + 6i+ (2i + 4i)

+ (-6i – i)

Example 3Multiply.

(2 – i)(-3 – 4i)= -6- 8i +

3i+ 4i2

= -6- 5i + 4(-1)= -10 – 5i

Conjugate of a Complex Number

The conjugate of a complex number a + bi is a – bi.The conjugate of a + bi is denoted a + bi.

Example 4

multiply by 1, using the conjugate of the denominator

3 2iSimplif y . Write your answer in standard f orm.4 i

3 2i4 i

4 i4 i

= (3 – 2i)(-4 + i)

(-4 – i)(-4 - i)

= -1216

- 3i+ 4i

+ 8i+ 2i2- 4i - i2

= -1216

+ 5i+ 2(-1)- (-1) = -14

17

+ 5i

Practice3 4iSimplif y . Write your answer in standard f orm.2 i

Homework

Lesson 5.6 Exercises 49-57 odd, 65, 67, 71, 75