SAT Problem of the Day

5.6 Quadratic Equations and Complex Numbers 5.6 Quadratic Equations and Complex Numbers Objectives: Classify and find all roots of a quadratic equation

Solutions of a Quadratic Equation If b 2 – 4ac > 0, then the quadratic equation has 2 distinct real solutions. Let ax 2 + bx + c = 0, where a = 0. If b 2 – 4ac = 0, then the quadratic equation has 1 real solution. If b 2 – 4ac < 0, then the quadratic equation has 0 real solutions. The expression b 2 – 4ac is called the discriminant.

Example 1 Find the discriminant for each equation. Then determine the number of real solutions. a) 3x 2 – 6x + 4 = 0 b 2 – 4ac =(-6) 2 – 4(3)(4) =36 – 48 =-12 no real solutions b) 3x 2 – 6x + 3 = 0 b 2 – 4ac =(-6) 2 – 4(3)(3) =36 – 36 =0 one real solution c) 3x 2 – 6x + 2 = 0 b 2 – 4ac =(-6) 2 – 4(3)(2) =36 – 24 =12 two real solutions

Practice Identify the number of real solutions: 1) -3x 2 – 6x + 15 = 0

Imaginary Numbers The imaginary unit is defined as and i 2 = -1. If r > 0, then the imaginary number is defined as follows:

Example 2 Solve 6x 2 – 3x + 1 = 0.

Practice Solve -4x 2 + 5x – 3 = 0.

Homework Lesson 5.6 exercises Odd

SAT Problem of the Day

5.6 Quadratic Equations and Complex Numbers 5.6 Quadratic Equations and Complex Numbers Objectives: Graph and perform operations on complex numbers

Imaginary Numbers A complex number is any number that can be written as a + bi, where a and b are real numbers and a is called the real part and b is called the imaginary part i real part imaginary part 34i

Example 1 Find x and y such that -3x + 4iy = 21 – 16i. Real partsImaginary parts -3x = 21 x = -7 4y = -16 y = -4 x = -7 and y = -4

Practice Find x and y such that 2x + 3iy = i.

Example 2 Find each sum or difference. a) (-10 – 6i) + (8 – i) = ( ) = -2 – 7i b) (-9 + 2i) – (3 – 4i) = (-9 – 3) = i + (2i + 4i) + (-6i – i)

Example 3 Multiply. (2 – i)(-3 – 4i) = -6- 8i+ 3i+ 4i 2 = -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 – 2i) (-4 + i) (-4 – i) (-4 - i) = i + 4i + 8i+ 2i 2 - 4i- i 2 = i+ 2(-1) - (-1) = i

Practice

Homework Lesson 5.6 Exercises odd, 65, 67, 71, 75