Download 2.6A The Quadratic Formula

2. 6A The Quadratic Formula
Objectives:
N.CN.7: Solve quadratic equations with real coefficients that have complex solutions.
A.CED.1: Create equations and inequalities in one variable and use them to solve problems.
For the Board: You will be able to solve quadratic equations using the Quadratic Formula.
You will be able to classify roots using the discriminant.
Bell Work 2.6:
Write each function in standard form.
1. f(x) = (x – 4)2 + 3
2. g(x) = 2(x + 6)2 – 11
Evaluate b2 – 4ac for the given values of the variables.
3. a = 2, b = 7, c = 5
4. a = 1, b = 3, c = -3
Anticipatory Set:
By applying completing the square to the generic standard form of a quadratic equation, we can obtain
a formula for solving any quadratic equation using arithmetic.
ax2 + bx + c = 0
ax2 + bx = -c
b
b
-c
-c
x2 + x =
(x2 + x + _____) =
+ _____
a
a
a
a
2
b2
b2
1 b b  b 
b
- c b2
;   = 2
(x2 + x +
)
=
+
 =
2 a 2a  2a  4a
a
a 4a 2
4a 2
- 4ac  b 2
b2
b
- c 4a b 2
- c b2
- 4ac
(x + )2 =
+
=
+
=
+
=

a
a 4a 4a 2
a 4a 2
4a 2
4a 2
4a 2
 4ac  b 2
b
 4ac  b 2  b   4ac  b 2
b

x+ =±
x
=
=
2a
2a
2a
4a 2
a
The Quadratic Formula
If ax2 + bx + c = 0, then the solutions or roots, are x =
 b  b 2  4ac
.
2a
Instruction:
Open the book to page 101 and read example 1.
Example: Find the zeros of each function by using the Quadratic Formula.
1. f(x) = 2x2 – 16x + 27
a = 2, b = -16, c = 27
x=
 (16)  (16) 2  4(2)(27) 16  256  216 16  40 16  2 10
1
=
=
=
= 4
10
2(2)
4
4
4
2
2. f(x) = 2x2 + 3x + 1
a = 2, b = 3, c = 1
 3  3 2  4(2)(1)  3  9  8  3  1
x=
=
=
= (-3 + 1)/4 or (-3 – 1)/4 = -1/2 or -1
2(2)
4
4
White board Activity:
Practice: Find the zeros of each function by using the Quadratic Formula
a. f(x) = x2 + 3x – 7
a = 1, b = 3, c = -7
 3  3 2  4(1)(-7)  3  9  (-28)  3  9  28  3  37
x=
=
=
=
2(1)
2
2
2
2
b. f(x) = x – 8x + 10
a = 1, b = -8, c = 10
x=
 (8)  (8) 2  4(1)(10) 8  64  40 8  24 8  2 6
=
=
=
= 4 6
2(1)
2
2
2
The discriminant of the quadratic equation ax2 + bx + c = 0 (a ≠ 0) is b2 – 4ac.
Value
Graph
Solutions
b2 – 4ac >0
Two distinct real
solutions.
b2 – 4ac = 0
b2 – 4ac < 0
One distinct real solution. No real solution.
Open the book to page 102 and read example 3.
Example: Find the type and number of solutions for each equation.
a. x2 + 36 = 12x
x2 – 12x + 36 = 0
a = 1, b = -12, c = 36
b2 – 4ac = (-12)2 – 4(1)(36) = 144 – 144 = 0
One distinct real solution.
2
b. x + 40 = 12x
x2 – 12x + 40 = 0
a = 1, b = -12, c = 40
2
b – 4ac = (-12)2 – 4(1)(40) = 144 – 160 = -16
No real solution.
c. x2 + 30 = 12x
x2 – 12x + 30 = 0
a = 1, b = -12, c = 30
2
b – 4ac = (-12)2 – 4(1)(30) = 144 – 120 = 24
Two distinct real solutions.
White Board Activity.
Practice: Find the type and number of solutions for each equation.
a. x2 – 4x = -4
x2 – 4x + 4 = 0
a = 1, b = -4, c = 4
b2 – 4ac = (-4)2 – 4(1)(4) = 16 – 16 = 0
One distinct real solution.
b. x2 – 4x = -8
x2 – 4x + 8 = 0
a = 1, b = -4, c = 8
b2 – 4ac = (-4)2 – 4(1)(8) = 16 – 32 = -16
No real solution.
c. x2 – 4x = 2
x2 – 4x - 2 = 0
a = 1, b = -4, c = -2
2
2
b – 4ac = (-4) – 4(1)(-2) = 16 – (-8) = 16 + 8 = 24
Two distinct real solutions.
Assessment:
Question student pairs.
Independent Practice:
Text: pg. 105 – 106 prob. 2 – 7, 14 – 16, 18 – 23, 30 – 35, 38, 39, 41, 42, 45 - 53.
For a Grade:
Text: pg. 105 prob. 2, 18, 38, 50.