Appendix A — Answer Key, Chapter 15: Solving Quadratic Equations

SOL A.EI.3 (a, b, c) · Covers textbook Chapter 15 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 132 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used in every answer below: solutions are over the real numbers only; a negative discriminant means no real solutions (never a complex pair). Irrational answers are left in simplest radical form unless a decimal is requested. In context, every algebraically valid root is tested for reasonableness, and a physically meaningless root is rejected with a stated reason. Verifying means substituting into the original equation; graphical and technology checks confirm the same zeros.

The figures used repeatedly in the chapter, for reference:

• Figure 1 is $y = x^2 - x - 6$ meeting the $x$-axis at $(-2, 0)$ and $(3, 0)$
• Figure 2 is the anatomy of $ax^2 + bx + c = 0$, with the example $3x^2 - 5x + 2 = 0$ giving $a = 3$, $b = -5$, $c = 2$
• Figure 3 pairs $x^2 = 9$ (two real solutions) with $x^2 = -9$ (no real solutions)
• Figure 4 factors $x^2 + 2x - 15 = 0$ as $(x + 5)(x - 3) = 0$, with solutions $x = -5$ and $x = 3$
• Figure 5 completes the square for $x^2 + 6x - 7 = 0$, solutions $x = 1$ and $x = -7$
• Figure 6 names the parts of the quadratic formula
• Figure 7 shows $D = 16$, $D = 0$, and $D = -8$ with two, one, and no $x$-intercepts
• Figure 8 is $x^2 - 2x - 4 = 0$ with exact roots $1 \pm \sqrt{5}$ and decimals about $-1.24$ and $3.24$
• Figure 9 is $h = -16t^2 + 32t + 48$, keeping $t = 3$ and rejecting $t = -1$
• Figure 10 is the method-selection flow
• Figure 11 is four blank verification grids

Lesson 15.1 — Standard Form and What a Solution Means

Guided practice

1. $a = 3$, $b = -5$, $c = 2$. The minus sign in front of $5x$ is part of the linear coefficient, so $b$ is $-5$, not $5$.
2. a) $x^2 - 6x + 8 = 0$; $a = 1$, $b = -6$, $c = 8$. b) $2x^2 - 3x + 5 = 0$; $a = 2$, $b = -3$, $c = 5$. c) $0 = x^2 - 4x + 3$, or $x^2 - 4x + 3 = 0$; $a = 1$, $b = -4$, $c = 3$. (From $4x = x^2 + 3$, subtract $4x$: $0 = x^2 - 4x + 3$.)
3. The solutions are $x = -2$ and $x = 3$. The graph names them as the $x$-intercepts $(-2, 0)$ and $(3, 0)$ — where the curve meets the $x$-axis.
4. $(-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0$ ✓. $3^2 - 3 - 6 = 9 - 3 - 6 = 0$ ✓.
5. $4^2 - 5(4) + 4 = 16 - 20 + 4 = 0$ ✓. Yes, $x = 4$ is a solution.
6. A solution $r$ makes $ar^2 + br + c = 0$, so the point $(r, 0)$ lies on the graph of $y = ax^2 + bx + c$. That point is an $x$-intercept, and $r$ is a zero of the function — three names for the same number.

Independent practice

7. a) $a = 1$, $b = 9$, $c = -2$ b) $a = 5$, $b = 0$, $c = -3$ c) $a = -2$, $b = 4$, $c = 1$ d) $x^2 + 0x + 0 = 0$; $a = 1$, $b = 0$, $c = 0$
8. a) $x^2 + 3x - 10 = 0$; $a = 1$, $b = 3$, $c = -10$ b) $2x^2 - x - 7 = 0$; $a = 2$, $b = -1$, $c = -7$ c) $x^2 + 3x - 4 = 0$; $a = 1$, $b = 3$, $c = -4$
9. a) $4 - 10 + 6 = 0$ ✓ — yes b) $9 - 15 + 6 = 0$ ✓ — yes c) $1 + 5 + 6 = 12 \neq 0$ ✗ — no d) $36 - 30 + 6 = 12 \neq 0$ ✗ — no
10. Solutions: $x = -5$ and $x = 3$. Check $x = -5$: $25 - 10 - 15 = 0$ ✓. (Check $x = 3$: $9 + 6 - 15 = 0$ ✓.)
11. Standard form requires every term on one side and $0$ on the other. Correct form: $x^2 - 9 = 0$, so $a = 1$, $b = 0$, $c = -9$.
12. The student dropped the minus sign on the middle term. Correct: $a = 3$, $b = -5$, $c = 2$.
13. $-16t^2 + 32t + 48 = 0$; $a = -16$, $b = 32$, $c = 48$. (Equivalently, divide by $-16$: $t^2 - 2t - 3 = 0$ with $a = 1$, $b = -2$, $c = -3$.)
14. If $a = 0$, the $x^2$ term vanishes and the equation is linear (degree at most $1$), not quadratic.
15. Intercepts at $(-2, 0)$ and $(2, 0)$; solutions $x = -2$ and $x = 2$.
16. Window containing both zeros works (e.g. $-10 \le x \le 10$, $-10 \le y \le 10$). The zero/root feature should report $x = -2$ and $x = 3$.

Exit ticket 15.1

17. $0 = 2x^2 - 5x - 3$, or $2x^2 - 5x - 3 = 0$; $a = 2$, $b = -5$, $c = -3$.
18. $(-2)^2 + 3(-2) + 2 = 4 - 6 + 2 = 0$ ✓. Yes.
19. $x = -3$ and $x = 3$.
20. Algebraically: substitute into the original equation and get $0$. Graphically: confirm the related curve meets the $x$-axis at that input.

Lesson 15.2 — Solving by Square Roots

Guided practice

21. $x = \pm 3$; the left graph crosses the $x$-axis at $(-3, 0)$ and $(3, 0)$. For $x^2 = -9$: no real solutions; the right graph never meets the $x$-axis (lowest point at $(0, 9)$).
22. $x = -7$ or $x = 7$.
23. $x^2 = 25$; $x = -5$ or $x = 5$.
24. $x^2 = 16$; $x = -4$ or $x = 4$.
25. $x - 4 = \pm 3$; $x = 7$ or $x = 1$.
26. $x^2 = -7$. No real solutions, because no real number squares to a negative.

Independent practice

27. a) $x = -9$ or $x = 9$ b) $x = -8$ or $x = 8$ c) $x^2 = 4$; $x = -2$ or $x = 2$ d) $x^2 = 25$; $x = -5$ or $x = 5$
28. a) $x + 5 = \pm 4$; $x = -1$ or $x = -9$ b) $x - 3 = \pm 6$; $x = 9$ or $x = -3$ c) $2x = \pm 10$; $x = 5$ or $x = -5$ d) $x + 1 = 0$; $x = -1$ (one real solution)
29. a) No real solutions b) No real solutions c) No real solutions d) $4x^2 = -12$; no real solutions
30. $(x - 1)^2 = 4$; $x - 1 = \pm 2$; $x = 3$ or $x = -1$.
31. The student kept only the principal square root. Complete solution: $x = -4$ or $x = 4$.
32. $\sqrt{x^2} = |x|$, not $x$. So $x^2 = 9$ means $|x| = 3$, hence $x = \pm 3$. Writing only the positive root drops half of the solution set.
33. $s^2 = 196$; $s = \pm 14$. Keep $s = 14$ feet. Reject $s = -14$ because a side length cannot be negative.
34. $(-7 + 2)^2 = (-5)^2 = 25$ ✓; $(3 + 2)^2 = 5^2 = 25$ ✓.
35. $y = x^2 - 36$ meets the axis at $\pm 6$, matching two real solutions of $x^2 = 36$. $y = x^2 + 16$ never meets the axis, matching no real solutions for $x^2 = -16$.
36. $x + 3 = \pm \sqrt{12} = \pm 2\sqrt{3}$; $x = -3 + 2\sqrt{3}$ or $x = -3 - 2\sqrt{3}$.

Exit ticket 15.2

37. $x^2 = 16$; $x = -4$ or $x = 4$.
38. $x - 5 = \pm 7$; $x = 12$ or $x = -2$.
39. No real solutions ($x^2 = -25$).
40. No real number has a negative square, so $x^2 = k$ cannot hold for any real $x$ when $k < 0$.

Lesson 15.3 — Solving by Factoring

Guided practice

41. $(x + 5)(x - 3) = 0$; $x + 5 = 0$ or $x - 3 = 0$; $x = -5$ or $x = 3$.
42. $(-5)^2 + 2(-5) - 15 = 25 - 10 - 15 = 0$ ✓; $9 + 6 - 15 = 0$ ✓.
43. The zero product property applies only when a product equals $0$. The number $7$ has many factor pairs, so setting one factor equal to $7$ does not force the product to be $7$ in a unique way and does not solve the equation.
44. $(x - 2)(x - 3) = 0$; $x = 2$ or $x = 3$.
45. $(x + 3)(x + 4) = 0$; $x = -3$ or $x = -4$.
46. $(2x + 1)(x - 3) = 0$; $x = -\tfrac{1}{2}$ or $x = 3$.

Independent practice

47. a) $(x - 4)(x - 5) = 0$; $x = 4$ or $x = 5$ b) $(x + 2)(x + 4) = 0$; $x = -2$ or $x = -4$ c) $(x - 7)(x + 4) = 0$; $x = 7$ or $x = -4$ d) $(x - 7)(x + 7) = 0$; $x = 7$ or $x = -7$
48. a) $(2x - 1)(x + 4) = 0$; $x = \tfrac{1}{2}$ or $x = -4$ b) $(3x + 2)(x - 4) = 0$; $x = -\tfrac{2}{3}$ or $x = 4$ c) $5x(x - 1) = 0$; $x = 0$ or $x = 1$ d) $(2x - 3)(2x + 3) = 0$; $x = \tfrac{3}{2}$ or $x = -\tfrac{3}{2}$
49. a) $x^2 - 5x = 0$; $x(x - 5) = 0$; $x = 0$ or $x = 5$ b) $x^2 + 4x - 21 = 0$; $(x + 7)(x - 3) = 0$; $x = -7$ or $x = 3$ c) $2x^2 + 3x - 2 = 0$; $(2x - 1)(x + 2) = 0$; $x = \tfrac{1}{2}$ or $x = -2$
50. Error: used zero product with right side $6$. Correct: $(x - 3)(x + 2) - 6 = 0$ → $x^2 - x - 6 - 6 = 0$ → $x^2 - x - 12 = 0$ → $(x - 4)(x + 3) = 0$; $x = 4$ or $x = -3$.
51. The property says a product is $0$ only when a factor is $0$; for a nonzero right side, many factor pairs work. Counterexample: $(x)(x) = 4$ does not imply $x = 4$.
52. $n(n + 1) = 72$ → $n^2 + n - 72 = 0$ → $(n + 9)(n - 8) = 0$; $n = -9$ or $n = 8$. Pairs: $-9$ and $-8$; $8$ and $9$.
53. $(x - 4)^2 = 0$; $x = 4$ only — one distinct real solution (repeated root).
54. For $x = -1$: $3(1) - 1 - 2 = 0$ ✓. For $x = \tfrac{2}{3}$: $3(\tfrac{4}{9}) + \tfrac{2}{3} - 2 = \tfrac{4}{3} + \tfrac{2}{3} - 2 = 2 - 2 = 0$ ✓.
55. Zeros at $x = -5$ and $x = 3$, matching the factored solutions.
56. $(3x + 1)(2x - 3) = 0$; $x = -\tfrac{1}{3}$ or $x = \tfrac{3}{2}$.

Exit ticket 15.3

57. $(x - 4)(x + 3) = 0$; $x = 4$ or $x = -3$.
58. $(2x + 1)(x - 4) = 0$; $x = -\tfrac{1}{2}$ or $x = 4$.
59. $x^2 + 5x - 24 = 0$; $(x + 8)(x - 3) = 0$; $x = -8$ or $x = 3$.
60. If a product of factors equals zero, then at least one of the factors equals zero.

Lesson 15.4 — Completing the Square

Guided practice

61. The dashed corner is the $3 \times 3$ square of area $9 = \left(\tfrac{6}{2}\right)^2$ needed to finish $(x + 3)^2$. It is added to both sides to keep the equation balanced (equality preserved).
62. Steps as in the figure; checks: $1 + 6 - 7 = 0$ ✓; $49 - 42 - 7 = 0$ ✓.
63. $x^2 + 4x = 12$; $+4$ → $(x + 2)^2 = 16$; $x + 2 = \pm 4$; $x = 2$ or $x = -6$.
64. $x^2 - 8x = -7$; $+16$ → $(x - 4)^2 = 9$; $x - 4 = \pm 3$; $x = 7$ or $x = 1$.
65. $x^2 + 10x = -21$; $+25$ → $(x + 5)^2 = 4$; $x + 5 = \pm 2$; $x = -3$ or $x = -7$.
66. $\left(\tfrac{6}{2}\right)^2 = 9$; perfect-square trinomial $x^2 + 6x + 9 = (x + 3)^2$.

Independent practice

67. a) $(x + 1)^2 = 16$; $x = 3$ or $x = -5$ b) $(x - 3)^2 = 16$; $x = 7$ or $x = -1$ c) $(x + 6)^2 = 4$; $x = -4$ or $x = -8$ d) $(x - 5)^2 = 1$; $x = 6$ or $x = 4$
68. a) $(x + 2)^2 = 5$; $x = -2 \pm \sqrt{5}$ b) $(x - 1)^2 = 5$; $x = 1 \pm \sqrt{5}$ c) $(x + 3)^2 = 7$; $x = -3 \pm \sqrt{7}$
69. $(x - 2)^2 = 3$; $x = 2 \pm \sqrt{3}$.
70. When $b$ is even, $\tfrac{b}{2}$ is an integer, so the added square is an integer and the binomial has integer coefficients. When $b$ is odd, $\tfrac{b}{2}$ is a half-integer and fractions appear; the formula is often cleaner then.
71. Error: added $9$ to one side only. Correct: $(x + 3)^2 = 16$; $x + 3 = \pm 4$; $x = 1$ or $x = -7$.
72. $w(w + 6) = 16$ → $w^2 + 6w - 16 = 0$. Complete: $(w + 3)^2 = 25$; $w = 2$ or $w = -8$. Keep $w = 2$ m. Reject $w = -8$ because a width cannot be negative.
73. $(-9)^2 + 8(-9) - 9 = 81 - 72 - 9 = 0$ ✓; $1 + 8 - 9 = 0$ ✓.
74. $x^2 + 6x - 7 = (x + 3)^2 - 16$; set equal to $0$: $(x + 3)^2 = 16$; $x = 1$ or $x = -7$.
75. Zeros at $x = -7$ and $x = 1$.
76. $(x + 7)^2 = 9$; $x + 7 = \pm 3$; $x = -4$ or $x = -10$.

Exit ticket 15.4

77. $(x + 4)^2 = 36$; $x + 4 = \pm 6$; $x = 2$ or $x = -10$.
78. $(x - 1)^2 = 6$; $x = 1 \pm \sqrt{6}$.
79. $x^2 - 12x + 36 = (x - 6)^2$.
80. The missing corner's area is $\left(\dfrac{b}{2}\right)^2$ — the amount that turns $x^2 + bx$ into a perfect square.

Lesson 15.5 — The Quadratic Formula and the Discriminant

Guided practice

81. Acceptable: $\pm$ produces two candidates; $b^2 - 4ac$ is the discriminant whose sign decides the count; $2a$ divides the entire numerator.
82. $D = 4 + 12 = 16$; $x = \dfrac{2 \pm 4}{2}$; $x = 3$ or $x = -1$. Matches the left panel (crosses twice).
83. $D = 0$; $x = 3$. Matches the middle panel (touches once).
84. $D = 4 - 12 = -8 < 0$; no real solutions. Matches the right panel (misses).
85. $D = 16 + 4 = 20$; $x = \dfrac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$.
86. Exact: $x = 1 - \sqrt{5}$ or $x = 1 + \sqrt{5}$. Decimals: about $-1.24$ and $3.24$.

Independent practice

87. a) $D = 1$; $x = 2$ or $x = 3$ b) $D = 25$; $x = \dfrac{-3 \pm 5}{4}$; $x = \tfrac{1}{2}$ or $x = -2$ c) $D = 16$; $x = \dfrac{-6 \pm 4}{2}$; $x = -1$ or $x = -5$ d) $D = 16$; $x = \dfrac{2 \pm 4}{6}$; $x = 1$ or $x = -\tfrac{1}{3}$
88. a) $D = 0$ — one real solution b) $D = 4 - 20 = -16$ — no real solutions c) $D = 9 + 40 = 49$ — two real solutions ($x = 5$, $x = -2$) d) $D = 144 - 144 = 0$ — one real solution ($x = \tfrac{3}{2}$)
89. a) $x = 1 \pm \sqrt{5}$ b) $x = -2 \pm \sqrt{3}$ c) $x = \dfrac{-1 \pm \sqrt{6}}{5}$ d) $x = 1 \pm \dfrac{\sqrt{6}}{2}$ (equivalently $\dfrac{2 \pm \sqrt{6}}{2}$)
90. The left panel has $D > 0$, so two crossings; the middle has $D = 0$, so one touch; the right has $D < 0$, so the curve misses the axis — three discriminant signs, three pictures.
91. The student read $b$ as $5$ instead of $-5$, so $-b$ became $-5$ instead of $5$. Correct: $a = 1$, $b = -5$, $c = 2$; $D = 25 - 8 = 17$; $x = \dfrac{5 \pm \sqrt{17}}{2}$.
92. $w(w + 2) = 48$ → $w^2 + 2w - 48 = 0$; $(w + 8)(w - 6) = 0$; $w = 6$ or $w = -8$. Keep $w = 6$ m. Reject $w = -8$ (width cannot be negative).
93. $D = 36 - 48 = -12 < 0$; no real solutions, because the discriminant is negative (graph misses the $x$-axis).
94. $D = 0$; $x = \tfrac{3}{2}$. One distinct real solution because the discriminant is zero — the graph touches the axis once.
95. Approximate zeros near $-1.24$ and $3.24$, matching $1 \pm \sqrt{5}$.
96. $(1 + \sqrt{5})^2 - 2(1 + \sqrt{5}) - 4 = 1 + 2\sqrt{5} + 5 - 2 - 2\sqrt{5} - 4 = 0$ ✓.

Exit ticket 15.5

97. $D = 9 + 16 = 25$; $x = \dfrac{3 \pm 5}{4}$; $x = 2$ or $x = -\tfrac{1}{2}$.
98. $D = 4 - 32 = -28 < 0$; no real solutions (discriminant negative).
99. $D = 36 - 16 = 20$; $x = \dfrac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5}$.
100. $D > 0$: crosses twice; $D = 0$: touches once; $D < 0$: misses entirely.

Lesson 15.6 — Choosing a Method, Contexts, and Three-Way Verification

Guided practice

101. a) Square roots ($b = 0$) b) Factoring (factors over the integers) c) Completing the square ($a = 1$, $b$ even) — or factoring d) Quadratic formula (or factoring if noticed)
102. Roots $t = -1$ and $t = 3$. Keep $t = 3$ (lands after $3$ s). Reject $t = -1$ because it is before the throw / time after the throw cannot be negative.
103. $-16(9) + 32(3) + 48 = -144 + 96 + 48 = 0$ ✓.
104. Sketch should show intercepts near $(-2, 0)$ and $(3, 0)$ for $y = x^2 - x - 6$.
105. $(x + 5)(x - 3) = 0$; $x = -5$ or $x = 3$. Technology: graph $y = x^2 + 2x - 15$ and use the zero feature; it should report the same two values.
106. The formula always works, but square roots and factoring (when available) are faster and less error-prone; completing the square is clean when $a = 1$ and $b$ is even. The formula is the reliable last resort, not the default first move.

Independent practice

107. a) Square roots (or factoring): $x = \pm 9$ b) Factoring: $(x - 2)(x - 7) = 0$; $x = 2$ or $x = 7$ c) Completing the square or factoring: $(x + 8)(x - 2) = 0$; $x = -8$ or $x = 2$ d) Factoring or formula: $(2x - 1)(x + 3) = 0$; $x = \tfrac{1}{2}$ or $x = -3$
108. $t = -1$ or $t = 3$. The ball hits the ground at $t = 3$ seconds. Reject $t = -1$ (before the throw). Check: $-16(9) + 32(3) + 48 = 0$ ✓.
109. $x^2 + 4x - 45 = 0$; $(x + 9)(x - 5) = 0$; $x = 5$ or $x = -9$. Keep width $5$ m (length $9$ m). Reject $x = -9$ (width cannot be negative).
110. $D = 4 - 20 = -16 < 0$; no real solutions. Graph of $y = x^2 - 2x + 5$ stays above the $x$-axis (vertex at $(1, 4)$).
111. $x = 1 \pm \sqrt{5}$ ≈ $-1.24$ and $3.24$; marks on a grid should sit near those intercepts as in figure 8.
112. Algebra: $(-5)^2 + 2(-5) - 15 = 0$ ✓; $9 + 6 - 15 = 0$ ✓. Graph: intercepts at $(-5, 0)$ and $(3, 0)$. Technology: zero feature reports $x = -5$ and $x = 3$.
113. The student kept a negative time. Reject $t = -2$ as before the motion starts; report $t = 5$ only (with units).
114. Factoring is usually faster when it works, avoids formula arithmetic errors, and makes the zero-product logic visible.
115. $-x^2 + 40x = 300$ → $x^2 - 40x + 300 = 0$ → $(x - 10)(x - 30) = 0$; $x = 10$ or $x = 30$. Both are possible counts of items in this model (revenue $\$300$ at either quantity). Neither is rejected on sign grounds; interpret: revenue is $\$300$ when $10$ or $30$ items are sold.
116. Square roots: $x - 3 = \pm \sqrt{7}$; $x = 3 \pm \sqrt{7}$.

Exit ticket 15.6

117. Factoring (or completing the square): $(x + 3)(x + 7) = 0$; $x = -3$ or $x = -7$.
118. Keep $w = 3$. Reject $w = -8$ because a garden width cannot be negative.
119. Algebraically, graphically, and with technology.
120. No real solutions, because no real number squares to $-9$ (equivalently $D < 0$ for $x^2 + 9 = 0$).

Chapter 15 Review

121. $x = -6$ or $x = 6$.
122. $(x - 2)(x - 6) = 0$; $x = 2$ or $x = 6$.
123. $(x + 3)^2 = 16$; $x = 1$ or $x = -7$.
124. $D = 25 + 24 = 49$; $x = \dfrac{5 \pm 7}{4}$; $x = 3$ or $x = -\tfrac{1}{2}$.
125. $D = 16 + 8 = 24$; $x = \dfrac{4 \pm 2\sqrt{6}}{2} = 2 \pm \sqrt{6}$.
126. Completing the square or formula: $x = -2 \pm \sqrt{3}$. Method named accordingly.
127. $D = 0$; one real solution ($x = 3$). Discriminant zero means the graph touches the $x$-axis once.
128. $D = 4 - 20 = -16 < 0$; no real solutions. The graph misses the $x$-axis.
129. $D = 9 + 40 = 49 > 0$; two real solutions: $x = 5$ and $x = -2$.
130. Algebra: $(-2)^2 - (-2) - 6 = 0$ ✓; $9 - 3 - 6 = 0$ ✓. Graph: intercepts at $(-2, 0)$ and $(3, 0)$ as in figure 1. Technology: zero feature on $y = x^2 - x - 6$ reports the same two inputs.
131. Roots $t = -1$, $t = 3$. Lands at $t = 3$ s. Reject $t = -1$ (before the throw). Method: factor after dividing by $-16$, or use the formula / zero product on $t^2 - 2t - 3 = 0$.
132. $w(w + 4) = 96$ → $w^2 + 4w - 96 = 0$; $(w + 12)(w - 8) = 0$; $w = 8$ or $w = -12$. Keep width $8$ ft (length $12$ ft). Reject $w = -12$ (width cannot be negative). Technology: graph $y = w^2 + 4w - 96$ and confirm a zero at $w = 8$.