Name ________________________________________ Date __________________ Class__________________
LESSON
2-6
Reteach
The Quadratic Formula
The Quadratic Formula is another way to find the roots of a quadratic
equation or the zeros of a quadratic function.
Find the zeros of f (x) = x2 − 6x − 11.
Step 1
Set f (x) = 0.
Step 2
Write the Quadratic Formula. x =
Step 3
Substitute values for a, b, and c into the Quadratic Formula.
x2 − 6x − 11 = 0
−b ± b 2 − 4ac
2a
a = 1, b = −6, c = −11
−b ± b 2 − 4ac − ( −6 ) ±
x=
=
2a
Step 4
2
Simplify.
x=
Step 5
( −6 ) − 4 (1)( −11)
2 (1)
− ( −6 ) ±
( −6 ) − 4 (1)( −11)
2 (1)
2
=
6 ± 36 + 44 6 ± 80
=
2
2
Write in simplest form.
x=
6 ± 80
80
=3±
=3±
2
2
(16 )( 5 )
2
=3±
4 5
=3±2 5
2
Remember to divide both terms of
the numerator by 2 to simplify.
Find the zeros of each function using the Quadratic Formula.
1. f (x) = x2 + x − 1
2. f (x) = x2 − 6x + 6
x2 + x − 1 = 0
____________________
a = _____, b = _____, c = _____
a = _____, b = _____, c = _____
x=
−b ± b 2 − 4ac
2a
−b ± b 2 − 4ac
2a
( ____ ) − 4 ( ___ )( ____ )
2 ( ___ )
_________________________________
_________________________________________
________________________________________
_________________________________________
________________________________________
x=
− ( ____ ) ±
x=
2
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2-46
Holt McDougal Algebra 2
Name ________________________________________ Date __________________ Class__________________
LESSON
2-6
Reteach
The Quadratic Formula (continued)
The discriminant of ax2 + bx + c = 0 (a ≠ 0) is b2 − 4ac.
Use the discriminant to determine the number of roots of a quadratic equation. A quadratic
equation can have 2 real solutions, 1 real solution, or 2 complex solutions.
Find the type and number of solutions.
x2 + 10x = −25
2x2 − 5x = 3
3x2 − 4x = −2
Write the equation in
standard form:
Write the equation in
standard form:
Write the equation in
standard form:
2x2 − 5x − 3 = 0
x2 + 10x + 25 = 0
3x2 − 4x + 2 = 0
a = 2, b = −5, c = −3
a = 1, b = 10, c = 25
a = 3, b = −4, c = 2
Evaluate the discriminant:
Evaluate the discriminant:
Evaluate the discriminant:
2
2
b − 4ac
b − 4ac
b2 − 4ac
(−5)2 − 4(2)(−3)
(10)2 − 4(1)(25)
(−4)2 − 4(3)(2)
25 + 24
100 − 100
16 − 24
49
0
−8
When b2 − 4ac > 0,
the equation has 2 real
solutions.
When b2 − 4ac = 0, the
equation has 1 real solution.
When b2 − 4ac < 0, the
equation has 2 complex
solutions.
Find the type and number of solutions for each equation.
3. x2 − 12x = −36
4. x2 − 4x = −7
5. x2 − 7x = −3
x2 − 12x + 36 = 0
____________________
____________________
a = ____, b = ____, c = ____
____________________
____________________
b2 − 4ac =
b2 − 4ac =
b2 − 4ac =
________________________
_________________________
Classify solutions:
Classify solutions:
________________________
_________________________
________________________
Classify solutions:
________________________
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2-47
Holt McDougal Algebra 2
Reteach
Challenge
1. a = 1, b = 1, c = −1
1.
Equation Roots Sum of
the
Roots
Product
of the
Roots
x=
−(1) ± (1)2 − 4(1)( −1)
2(1)
x=
−1 ± 1 + 4
2
a.
x2 − 6x +
8=0
4, 2
6
8
x=
−1 ± 5
2
b.
x2 − 7x +
12 = 0
4, 3
7
12
c.
x2 + 2x −
35 = 0
5, −7
2. x2 − 6x + 6 = 0
a = 1, b = −6, c = 6
−( −6) ± ( −6)2 − 4(1)(6)
x=
2(1)
d.
4x2 − 8x +
3=0
1 3
,
2 2
6 ± 36 − 24
x=
2
e.
9x2 + 3x −
2=0
1 2
,−
3 3
x =3± 3
2. a. r1 + r2 = −
3. a = 1, b = −12, c = 36
b. r1r2 =
0
1 real solution
4. x2 − 4x + 7 = 0
−2
−35
3
4
2
−
1
3
−
2
9
b
a
c
a
3. x2 − 4x − 1 = 0
4. k = −3
5. C
6. A
Problem Solving
a = 1, b = −4, c = 7
1. a. t = −0.25, 1.5
−12
b. 30 ft
2 complex solutions
2. a. t = −0.23, 1.61; 35.4 ft
5. x2 − 7x + 3 = 0
b. t = −0.21, 1.77; 44.3 ft
a = 1, b = −7, c = 3
c. t = −0.19, 1.94; 54.3 ft
37
3. C
2 real solutions
4. C
Reading Strategies
1. a = 2, b = −6, c = −9
2. (−6)2 − 4(2)(−9) = 108
3. Since the discriminant is positive, the
equation has two real roots.
4. Yes; since the equation has two real roots,
the related function has two zeros.
5. x =
−( −6) ± 108 3 ± 3 3
=
2(2)
2
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A18
Holt McDougal Algebra 2