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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.
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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