Solutions for Precalculus, Enhanced with Graphing Utilities, 9th Edition by Sullivan - 2025 Published (All Chapters included)

Prealgebra & Introductory Algebra,
6th Edition by Elayn Martin-Gay




Complete Chapter Solutions Manual
are included (Ch 1 to 16)




** Immediate Download
** Swift Response
** All Chapters included

,Table of Contents are given below




1. The Whole Numbers

2. Integers and Introduction to Solving Equations

3. Solving Equations and Problem Solving

4. Fractions and Mixed Numbers

5. Decimals

6. Ratio, Proportion, and Percent

7. Graphs, Triangle Applications, and Introduction to Statistics and

Probability

8. Geometry and Measurement

9. Equations, Inequalities, and Problem Solving

10. Exponents and Polynomials

11. Factoring Polynomials

12. Rational Expressions

13. Graphing Equations and Inequalities

14. Systems of Equations

15. Roots and Radicals

16. Quadratic Equations

,Solutions Manual organized in reverse order, with the last chapter displayed first, to ensure that all
chapters are included in this document. (Complete Chapters included Ch16-1)


Chapter 16
Section 16.1 Practice Exercises −1 ± 15
x=
4
1. x 2 − 16 = 0
−1 ± 15
x 2 = 16 The solutions are .
4
x = 16 or x = − 16
x=4 x = −4
7. h = 16t 2
The solutions are 4 and −4.
12,144 = 16t 2
2. 3x 2 = 11 759 = t 2
11 27.5 ≈ t or −27.5 ≈ t
x2 =
3 −27.5 is rejected because time cannot be
negative. The free-falling dive of 12,144 feet
11 11 lasted approximately 27.5 seconds.
x= or x = −
3 3
11 ⋅ 3 11 ⋅ 3 Vocabulary, Readiness & Video Check 16.1
x= x=−
3⋅ 3 3⋅ 3 1. To solve, a becomes the radicand and the square
33 33 root of a negative number is not a real number.
x= x=−
3 3
2. A quadratic equation must be in the form of a
33 33 variable (or polynomial) squared equal to some
The solutions are and − .
3 3 nonnegative number in order for the property to
be used. For Video Notebook 6, we use
3. ( x − 4)2 = 49 h = 16t 2 , so this equation can easily be placed in
x − 4 = 49 or x − 4 = − 49 this form. The negative value is rejected because
x−4 =7 x − 4 = −7 it will not be used in the context of the
application.
x = 11 x = −3
The solutions are 11 and −3. Exercise Set 16.1

4. ( x − 5) 2 = 18 2. k2 − 9 = 0
x − 5 = 18 or x − 5 = − 18 (k − 3)(k + 3) = 0
x −5 = 3 2 x − 5 = −3 2 k − 3 = 0 or k + 3 = 0
x = 5+3 2 x = 5−3 2 k =3 k = −3
The solutions are k = 3 and k = −3.
x = 5±3 2
The solutions are 5 ± 3 2. 4. m 2 + 6m = 7
m 2 + 6m − 7 = 0
5. ( x + 3) 2 = −5 (m + 7)( m − 1) = 0
This equation has no real solution because the m + 7 = 0 or m − 1 = 0
square root of −5 is not a real number. m = −7 m =1
The solutions are m = −7 and m = 1.
6. (4 x + 1)2 = 15
4 x + 1 = 15 or 4 x + 1 = − 15
4 x = −1 + 15 4 x = −1 − 15
−1 + 15 −1 − 15
x= x=
4 4


647

, Chapter 16: Quadratic Equations ISM: Prealgebra and Introductory Algebra


6. 2 x 2 − 98 = 0 20. 5 x 2 = 2
2( x 2 − 49) = 0 2
x2 =
2( x − 7)( x + 7) = 0 5
x − 7 = 0 or x + 7 = 0 2 2
x= or x = −
x=7 x = −7 5 5
The solutions are x = 7 and x = −7. 2 5 5 2
x= ⋅ x=− ⋅
5 5 5 5
8. 7 a 2 − 175 = 0
10 10
7(a 2 − 25) = 0 x= x=−
5 5
7(a − 5)(a + 5) = 0
10
a − 5 = 0 or a + 5 = 0 The solutions are ± .
a=5 a = −5 5
The solutions are a = 5 and a = −5.
22. 2 x 2 = 9
10. x 2 + 10 x = −24 9
x2 =
x 2 + 10 x + 24 = 0 2
( x + 6)( x + 4) = 0 9 9
x= or x = −
x+6 = 0 or x + 4 = 0 2 2
x = −6 x = −4 9 2 92
x= ⋅ x=− ⋅
The solutions are x = −6 and x = −4. 2 2 2 2
3 2 3 2
12. x 2 = 121 x= x=−
2 2
x = 121 or x = − 121 3 2
x = 11 x = −11 The solutions are x = ± .
2
The solutions are x = ±11.
24. 3 x 2 − 45 = 0
14. x 2 = 22
3x 2 = 45
x = 22 or x = − 22
x 2 = 15
The solutions are x = ± 22.
x = 15 or x = − 15
1 The solutions are x = ± 15.
16. x 2 =
16
1 1 26. ( x + 2) 2 = 25
x= or x = −
16 16 x + 2 = 25 or x + 2 = − 25
1 1 x+2=5 x + 2 = −5
x= x=−
4 4 x=3 x = −7
1 The solutions are x = 3 and x = −7.
The solutions are x = ± .
4
28. ( x − 7) 2 = 2
2
18. x = −25 has no real solution because the square x−7 = 2 or x − 7 = − 2
root of −25 is not a real number. x =7+ 2 x =7− 2
The solutions are x = 7 ± 2.




648