Section 1.5 Quadratic Equations. Solving Quadratic Equations by Factoring.
SAT Problem of the Day. 5.6 Quadratic Equations and Complex Numbers 5.6 Quadratic Equations and...
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Transcript of SAT Problem of the Day. 5.6 Quadratic Equations and Complex Numbers 5.6 Quadratic Equations and...
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