Appendix A — Answer Key, Chapter 11: Radical Expressions

SOL A.EO.4 (a, b, c, d) · Covers textbook Chapter 11 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 120 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used in every answer below. The principal square root $\sqrt{a}$ is the non-negative number whose square is $a$, so $\sqrt{49} = 7$ and never $\pm 7$. A square root is in simplest form when its radicand has no perfect-square factor other than $1$. The cube root $\sqrt[3]{a}$ of an integer $a$ is the unique real number whose cube is $a$, and it carries the sign of $a$. A cube root is in simplest form when its radicand has no perfect-cube factor other than $\pm 1$. Like radicals share index and radicand, and combine only after simplifying. Roots split products, never sums. Rational exponents in this chapter are only $\tfrac12$ and $\tfrac13$: $a^{1/2} = \sqrt{a}$ for $a \ge 0$, and $a^{1/3} = \sqrt[3]{a}$ for every integer $a$. No answer divides radicals or rationalizes a denominator. No sum, difference, or product uses a variable radicand.

The figures used repeatedly in the chapter:

• Figure 1 is the perfect-square reference table for $n = 1$ through $15$, with four extras ($400$, $625$, $900$, $10\,000$)
• Figure 2 extracts $\sqrt{72} = 6\sqrt{2}$ by marking $36$ as the largest perfect-square factor
• Figure 3 places $\sqrt{2}$, $\sqrt{10}$, $3\sqrt{5}$, and $6\sqrt{2}$ on a number line between consecutive integers
• Figure 4 is the simplest-form checklist on six expressions
• Figure 5 is the perfect-cube reference table in both signs
• Figure 6 extracts $\sqrt[3]{-54} = -3\sqrt[3]{2}$ by marking $27$ as the largest perfect cube
• Figure 7 sorts seven radical terms into three like piles totaling $11\sqrt{2} + 7\sqrt{3} + 6\sqrt[3]{2}$
• Figure 8 kills $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ with $a = 9$, $b = 16$
• Figure 9 forces $a^{1/2} = \sqrt{a}$ and $a^{1/3} = \sqrt[3]{a}$ from the product law
• Figure 10 is the decimal-check table that catches wrong simplifications
• Figure 11 is the blank practice tables for square roots and cube roots

Lesson 11.1 — Square Roots and Simplest Form

Guided practice

1. From the figure: $1$, $4$, $9$, $16$, $25$, $36$, $49$, $64$, $81$, $100$, $121$, $144$, $169$, $196$, $225$. A whole number is a perfect square when it appears in the middle row ($n^2$).
2. $\sqrt{81} = 9$ and $\sqrt{169} = 13$. Neither uses $\pm$ because the radical symbol names the principal (non-negative) square root only; $\pm$ would answer $x^2 = 81$ or $x^2 = 169$, not the radical.
3. Factor pairs: $1 \cdot 72$, $2 \cdot 36$, $3 \cdot 24$, $4 \cdot 18$, $6 \cdot 12$, $8 \cdot 9$. Perfect squares: $1$, $4$, $9$, $36$. Largest: $36$. Simplest form: $6\sqrt{2}$.
4. Pulling $4$ gives $2\sqrt{18}$. The perfect square still hiding is $9$ (since $18 = 9 \cdot 2$).
5. $\sqrt{15}$ passes. $\sqrt{20}$ fails → $2\sqrt{5}$. $2\sqrt{18}$ fails → $6\sqrt{2}$.
6. The test looks only under the bar. In $5\sqrt{7}$ the radicand $7$ is prime, so there is no perfect-square factor left to pull; the coefficient $5$ is irrelevant to the test.

Independent practice

7. a) $6$ b) $10$ c) $12$ d) $15$
8. a) $\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}$ b) $\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}$ c) $\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$ d) $\sqrt{98} = \sqrt{49 \cdot 2} = 7\sqrt{2}$
9. a) $2\sqrt{12} = 2\sqrt{4 \cdot 3} = 2 \cdot 2\sqrt{3} = 4\sqrt{3}$ b) $5\sqrt{8} = 5\sqrt{4 \cdot 2} = 5 \cdot 2\sqrt{2} = 10\sqrt{2}$ c) $3\sqrt{27} = 3\sqrt{9 \cdot 3} = 3 \cdot 3\sqrt{3} = 9\sqrt{3}$ d) $-4\sqrt{50} = -4\sqrt{25 \cdot 2} = -4 \cdot 5\sqrt{2} = -20\sqrt{2}$
10. $\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}$; $\sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2}$
11. Pulling the largest square finishes the job in one step. $\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}$ has a square-free radicand. Pulling only $4$ leaves $2\sqrt{18}$, and $18$ still hides a $9$, so another extraction is required.
12. $\sqrt{128} = \sqrt{64 \cdot 2} = 8\sqrt{2}$ feet.
13. The student treated $50$ as if taking a square root produced the factor $25$ as a coefficient without splitting correctly — equivalently, wrote $25$ where $\sqrt{25} = 5$ belongs. Correct: $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$.
14. The coefficient out front does not clear a perfect square under the bar. Checklist: $18 = 9 \cdot 2$, so it fails. Finished: $2\sqrt{18} = 2 \cdot 3\sqrt{2} = 6\sqrt{2}$.
15. a) Yes — $14 = 2 \cdot 7$ b) No → $3\sqrt{5}$ c) Yes — $3$ is prime d) No → $6\sqrt{2}$
16. a) $16$; $4$ (since $\sqrt{80} = \sqrt{16 \cdot 5} = 4\sqrt{5}$) b) $16$; $4$ (since $\sqrt{112} = \sqrt{16 \cdot 7} = 4\sqrt{7}$)

Exit ticket 11.1

17. $8$; $11$
18. $6\sqrt{2}$; $3\sqrt{20} = 3 \cdot 2\sqrt{5} = 6\sqrt{5}$
19. $\sqrt{15}$ yes; $2\sqrt{18}$ no → $6\sqrt{2}$
20. $\sqrt{49}$ asks for the non-negative number whose square is $49$. That number is $7$. Writing $\pm 7$ would answer the equation $x^2 = 49$, which has two solutions; the radical symbol names only the principal one.

Lesson 11.2 — Estimating Square Roots

Guided practice

21. $\sqrt{2}$: $1 < 2 < 4$, so between $1$ and $2$; $\approx 1.414$. $\sqrt{10}$: $9 < 10 < 16$, so between $3$ and $4$; $\approx 3.162$.
22. $\sqrt{45} = 3\sqrt{5} \approx 6.708$, between $6$ and $7$. $\sqrt{72} = 6\sqrt{2} \approx 8.485$, between $8$ and $9$.
23. Simplifying replaces one name of a number with another name of the same number. $\sqrt{45}$ and $3\sqrt{5}$ are equal, so they occupy the same point.
24. $25 < 30 < 36$, so between $5$ and $6$.
25. $\sqrt{75} = 5\sqrt{3}$. Since $64 < 75 < 81$, it lies between $8$ and $9$.
26. $9$ is larger. Since $80 < 81 = 9^2$, it follows that $\sqrt{80} < 9$. As a check: $\sqrt{80} = 4\sqrt{5} \approx 8.944 < 9$.

Independent practice

27. a) $2$ and $3$ ($4 < 7 < 9$) b) $4$ and $5$ ($16 < 20 < 25$) c) $7$ and $8$ ($49 < 55 < 64$) d) $9$ and $10$ ($81 < 90 < 100$)
28. a) $3\sqrt{2} \approx 4.243$ b) $3\sqrt{5} \approx 6.708$ c) $6\sqrt{2} \approx 8.485$ d) $7\sqrt{2} \approx 9.899$
29. $\sqrt{8} = 2\sqrt{2}$, so those two are equal. Then $3$, then $\sqrt{10}$. Order: $2\sqrt{2} = \sqrt{8} < 3 < \sqrt{10}$.
30. $\sqrt{162} = \sqrt{81 \cdot 2} = 9\sqrt{2}$ feet; between $12$ and $13$ (since $144 < 162 < 169$); $\approx 12.73$.
31. The middle row of the reference table lists the perfect squares. To place $\sqrt{40}$, find consecutive entries sandwiching $40$: $36$ and $49$, whose roots are $6$ and $7$.
32. The student sandwiched $50$ between $25$ and $36$ and read off $5$ and $6$, but $50$ is not between $25$ and $36$. The correct perfect-square sandwich is $49 < 50 < 64$, so $\sqrt{50}$ lies between $7$ and $8$.
33. $5\sqrt{2} > 7$, because $(5\sqrt{2})^2 = 25 \cdot 2 = 50 > 49 = 7^2$. (As a decimal, $5\sqrt{2} \approx 7.071 > 7$.)
34. $\sqrt{128} = 8\sqrt{2}$; $121 < 128 < 144$, so between $11$ and $12$.
35. Side $= \sqrt{52} = 2\sqrt{13} \approx 7.2$ meters.
36. a) Radicand $72$, since $8 < \sqrt{72} < 9$ and $\sqrt{72} = 6\sqrt{2}$ b) $(4\sqrt{3})^2 = 16 \cdot 3 = 48$, and $36 < 48 < 49$, so between $6$ and $7$

Exit ticket 11.2

37. $7$ and $8$ ($49 < 60 < 64$)
38. $3\sqrt{5} \approx 6.708$
39. $\sqrt{20} = 2\sqrt{5}$, so $\sqrt{20}$ and $2\sqrt{5}$ are equal and both less than $4$: $2\sqrt{5} = \sqrt{20} < 4$
40. They are equal numbers — $\sqrt{72} = 6\sqrt{2}$ by the extraction in Figure 2 — so they must mark the same point.

Lesson 11.3 — Cube Roots

Guided practice

41. Positive cubes: $1$, $8$, $27$, $64$, $125$, $216$, $343$, $512$, $729$, $1000$. Negative cubes: $-1$, $-8$, $-27$, $-64$, $-125$, $-216$, $-343$, $-512$, $-729$, $-1000$. Boxed claim: every integer has exactly one real cube root, and it carries the sign of the integer.
42. $4$; $-4$. A cube root can be negative because an odd number of negative factors stays negative, so some real number cubes to a negative; a square root of a negative would need a real number whose square is negative, which is impossible.
43. Pairs: $1 \cdot 54$, $2 \cdot 27$, $3 \cdot 18$, $6 \cdot 9$. Perfect cubes: $1$, $27$. Finished: $-3\sqrt[3]{2}$.
44. $9$ is a perfect square, not a perfect cube, so $\sqrt[3]{9}$ is not an integer and pulling $9$ does not simplify a cube root.
45. $\sqrt[3]{24} = \sqrt[3]{8 \cdot 3} = 2\sqrt[3]{3}$; $\sqrt[3]{-16} = \sqrt[3]{-8 \cdot 2} = -2\sqrt[3]{2}$
46. Cube roots are defined for every integer, positive or negative, so the identity $\sqrt[3]{a}\sqrt[3]{b} = \sqrt[3]{ab}$ has no sign restriction. Square roots of negatives are not real, so both factors in $\sqrt{a}\sqrt{b} = \sqrt{ab}$ must be non-negative.

Independent practice

47. a) $2$ b) $-3$ c) $6$ d) $-10$
48. a) $2\sqrt[3]{3}$ b) $3\sqrt[3]{2}$ c) $\sqrt[3]{128} = \sqrt[3]{64 \cdot 2} = 4\sqrt[3]{2}$ d) $5\sqrt[3]{2}$
49. a) $-2\sqrt[3]{2}$ b) $-3\sqrt[3]{2}$ c) $\sqrt[3]{-128} = \sqrt[3]{-64 \cdot 2} = -4\sqrt[3]{2}$ d) $-3\sqrt[3]{5}$
50. $5\sqrt[3]{16} = 5\sqrt[3]{8 \cdot 2} = 5 \cdot 2\sqrt[3]{2} = 10\sqrt[3]{2}$; $-2\sqrt[3]{54} = -2\sqrt[3]{27 \cdot 2} = -2 \cdot 3\sqrt[3]{2} = -6\sqrt[3]{2}$
51. $\sqrt{-8}$ asks for a real number whose square is $-8$; no such real number exists. $\sqrt[3]{-8}$ asks for a real number whose cube is $-8$, and $-2$ works because $(-2)^3 = -8$.
52. Edge $= \sqrt[3]{250} = 5\sqrt[3]{2}$ feet.
53. The student pulled the factor $27$ but dropped the minus sign, so the answer is positive when $\sqrt[3]{-54}$ must be negative. Correct: $-3\sqrt[3]{2}$, as in the figure.
54. The student pulled a perfect square out of a cube root. Correct: $\sqrt[3]{54} = \sqrt[3]{27 \cdot 2} = 3\sqrt[3]{2}$.
55. a) Yes b) No → $2\sqrt[3]{5}$ c) Yes — $9$ is not a perfect cube d) No → $-4\sqrt[3]{2}$
56. a) $8$; $2$ b) $-27$; $-3$

Exit ticket 11.3

57. $5$; $-4$
58. $2\sqrt[3]{5}$; $-3\sqrt[3]{2}$
59. No; $\sqrt[3]{16} = 2\sqrt[3]{2}$
60. Cubing is a one-to-one function on the real numbers: as $x$ runs through all reals, $x^3$ hits every real exactly once. The unique $x$ with $x^3 = a$ is $\sqrt[3]{a}$, and when $a$ is negative that $x$ is negative.

Lesson 11.4 — Adding, Subtracting, and Multiplying Radicals

Guided practice

61. Simplified terms: $3\sqrt{2}$, $5\sqrt{2}$, $5\sqrt{3}$, $3\sqrt{2}$, $2\sqrt{3}$, $4\sqrt[3]{2}$, $2\sqrt[3]{2}$. Totals: $11\sqrt{2}$, $7\sqrt{3}$, $6\sqrt[3]{2}$. Finished sum: $11\sqrt{2} + 7\sqrt{3} + 6\sqrt[3]{2}$.
62. Like radicals need the same index and the same radicand. $\sqrt{2}$ has index $2$ and $\sqrt[3]{2}$ has index $3$, so they are different numbers and stay separate.
63. $\sqrt{9 + 16} = \sqrt{25} = 5$ and $\sqrt{9} + \sqrt{16} = 7$. Conclusion: $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$.
64. $\sqrt{25 - 9} = 4$ and $\sqrt{25} - \sqrt{9} = 2$. The shaded failing rows are the sum rule and the difference rule.
65. $(5 - 2 + 1)\sqrt{7} = 4\sqrt{7}$
66. $3\sqrt{2} + 5\sqrt{2} - 2\sqrt{2} = 6\sqrt{2}$
67. $2 \cdot 3 \cdot \sqrt{100} = 6 \cdot 10 = 60$
68. $2\sqrt[3]{2} + 5\sqrt[3]{2} = 7\sqrt[3]{2}$

Independent practice

69. a) $11\sqrt{5}$ b) $-3\sqrt{2}$ c) $13\sqrt[3]{3}$ d) $-5\sqrt{11}$
70. a) $2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$ b) $5\sqrt{2} - 2\sqrt{2} = 3\sqrt{2}$ c) $3\sqrt{2} + 2\sqrt{2} + 4\sqrt{2} = 9\sqrt{2}$ d) $4\sqrt{3} - 5\sqrt{3} = -\sqrt{3}$
71. a) $\sqrt{144} = 12$ b) $15\sqrt{16} = 15 \cdot 4 = 60$ c) $8\sqrt{36} = 8 \cdot 6 = 48$ d) $-10\sqrt{36} = -10 \cdot 6 = -60$
72. a) $\sqrt[3]{8} = 2$ b) $8\sqrt[3]{27} = 8 \cdot 3 = 24$ c) $-15\sqrt[3]{8} = -15 \cdot 2 = -30$ d) $-2\sqrt[3]{2} + 3\sqrt[3]{2} = \sqrt[3]{2}$
73. $11\sqrt{2} + 7\sqrt{3} + 6\sqrt[3]{2}$ (matches the figure).
74. $\sqrt{12} = 2\sqrt{3}$, so $\sqrt{12} + \sqrt{3} = 2\sqrt{3} + \sqrt{3} = 3\sqrt{3}$. After simplifying, $\sqrt{12} + \sqrt{2} = 2\sqrt{3} + \sqrt{2}$, and those radicands differ, so nothing combines.
75. $\sqrt{48} = 4\sqrt{3}$ and $\sqrt{75} = 5\sqrt{3}$. Difference: $5\sqrt{3} - 4\sqrt{3} = \sqrt{3}$ feet.
76. The false rule is $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$. Correct value: $\sqrt{25} = 5$. Figure 8 kills the rule.
77. The student added under the radical as if roots split sums. Correct: $\sqrt{18} + \sqrt{8} = 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2}$.
78. $2 \cdot 5\sqrt{2} + 3 \cdot 4\sqrt{2} - 3\sqrt{2} = 10\sqrt{2} + 12\sqrt{2} - 3\sqrt{2} = 19\sqrt{2}$
79. $6\sqrt{3} \cdot \sqrt{12} = 6\sqrt{36} = 6 \cdot 6 = 36$, and $\sqrt{27} = 3\sqrt{3}$, so $36 - 3\sqrt{3}$
80. a) $7$ b) $ab$

Exit ticket 11.4

81. $10\sqrt{5}$
82. $5\sqrt{2} + 4\sqrt{2} = 9\sqrt{2}$
83. $6\sqrt{36} = 6 \cdot 6 = 36$
84. Example: $a = 9$, $b = 16$ gives $5 \neq 7$. Product property that holds: $\sqrt{a}\sqrt{b} = \sqrt{ab}$ for $a \ge 0$, $b \ge 0$.

Lesson 11.5 — Rational Exponents

Guided practice

85. By the product law, $a^{1/2} \cdot a^{1/2} = a^{1/2 + 1/2} = a^1 = a$. So $a^{1/2}$ is a number that squares to $a$. For $a \ge 0$ that number is $\sqrt{a}$. Boxed conclusion: $a^{1/2} = \sqrt{a}$ for $a \ge 0$.
86. $a^{1/3} \cdot a^{1/3} \cdot a^{1/3} = a^{1/3 + 1/3 + 1/3} = a^1 = a$, so $a^{1/3} = \sqrt[3]{a}$. Holds for every integer $a$.
87. $\sqrt{5} = 5^{1/2}$; check $2.23606^2 \approx 5$. $\sqrt[3]{-8} = (-8)^{1/3}$; check $(-2)^3 = -8$.
88. $\sqrt{45} = 3\sqrt{5}$: both sides $\approx 6.70820\ldots$, agrees. $\sqrt{45} = 9\sqrt{5}$: left $\approx 6.708$, right $\approx 20.125$, wrong.
89. If two expressions name the same number, their decimals must match; a mismatch means at least one side is wrong. Matching decimals can still happen by coincidence or rounding, so agreement supports the claim without proving the algebra.
90. Table A. $\sqrt{50}$: $25$; $\sqrt{25 \cdot 2}$; $5\sqrt{2}$. $\sqrt{48}$: $16$; $\sqrt{16 \cdot 3}$; $4\sqrt{3}$. $\sqrt{98}$: $49$; $\sqrt{49 \cdot 2}$; $7\sqrt{2}$. $\sqrt{200}$: $100$; $\sqrt{100 \cdot 2}$; $10\sqrt{2}$. Table B. $\sqrt[3]{24}$: $8$; $\sqrt[3]{8 \cdot 3}$; $2\sqrt[3]{3}$. $\sqrt[3]{-16}$: $-8$; $\sqrt[3]{-8 \cdot 2}$; $-2\sqrt[3]{2}$. $\sqrt[3]{250}$: $125$; $\sqrt[3]{125 \cdot 2}$; $5\sqrt[3]{2}$. $\sqrt[3]{-135}$: $-27$; $\sqrt[3]{-27 \cdot 5}$; $-3\sqrt[3]{5}$.

Independent practice

91. a) $36^{1/2}$ b) $7^{1/2}$ c) $27^{1/3}$ d) $(-64)^{1/3}$
92. a) $\sqrt{81} = 9$ b) $\sqrt[3]{125} = 5$ c) $\sqrt{9} = 3$ d) $\sqrt[3]{-27} = -3$
93. a) $7$ b) $2$ c) $4$ d) $-5$
94. $16^{1/2} \cdot 16^{1/2} = 16^{1} = 16$. Therefore $16^{1/2}$ is the non-negative number whose square is $16$, which is $\sqrt{16} = 4$.
95. $27^{1/3} \cdot 27^{1/3} \cdot 27^{1/3} = 27^{1} = 27$. So $27^{1/3} = \sqrt[3]{27} = 3$.
96. a) Agrees (both $\approx 8.485$) b) Wrong ($\sqrt{72} \approx 8.485$, $4\sqrt{3} \approx 6.928$) c) Agrees (both $\approx -3.780$)
97. The product law was already derived for integer exponents by counting. Extending the notation to $\tfrac12$ without breaking that law requires $a^{1/2} \cdot a^{1/2} = a$. The unique non-negative number with that property is $\sqrt{a}$. So the identification is forced by consistency with Chapter 10, not chosen for convenience.
98. Side $= \sqrt{50} = 50^{1/2} = 5\sqrt{2} \approx 7.07$ meters.
99. The student took the reciprocal instead of the square root. Correct: $9^{1/2} = \sqrt{9} = 3$. Keeping $9^{1/2} \cdot 9^{1/2} = 9$ intact would have ruled out $\tfrac19$, since $\tfrac19 \cdot \tfrac19 \neq 9$.
100. First, $(-8)^{1/2} = \sqrt{-8}$ is not a real number. Second, even if someone meant a cube root, $(-8)^{1/3} = -2$, not $-4$. The exponent $\tfrac12$ is the wrong index for a negative base in the reals.
101. a) $80$; $4$ (since $\sqrt{80} = 80^{1/2} = 4\sqrt{5}$) b) $-3\sqrt[3]{2}$
102. $-27$; $\sqrt[3]{-27 \cdot 5}$; $-3\sqrt[3]{5}$

Exit ticket 11.5

103. $121^{1/2} = 11$; $(-8)^{1/3} = -2$
104. $a^{1/2} \cdot a^{1/2} = a^{1} = a$, so $a^{1/2}$ squares to $a$; for $a \ge 0$ that forces $a^{1/2} = \sqrt{a}$.
105. Left $\approx 6.708$, right $\approx 8.660$, wrong. Correct: $3\sqrt{5}$.
106. $\sqrt[3]{27} = 3$

Chapter 11 Review

107. $9$; $14$. Four larger squares from the figure: $400$, $625$, $900$, $10\,000$.
108. Largest perfect-square factor $36$. Steps: $\sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36} \cdot \sqrt{2} = 6\sqrt{2}$.
109. $2\sqrt{18}$ fails because $18 = 9 \cdot 2$ → $6\sqrt{2}$. $\sqrt{20}$ fails because $20 = 4 \cdot 5$ → $2\sqrt{5}$.
110. a) $4\sqrt{3}$ b) $10\sqrt{2}$ c) $10\sqrt{2}$ d) $-3 \cdot 5\sqrt{3} = -15\sqrt{3}$
111. Between $6$ and $7$; $\sqrt{40} = 2\sqrt{10} \approx 6.325$.
112. $5$; $-10$. Cubing is one-to-one on the reals, so each integer is the cube of exactly one real number, and that number's sign matches the integer's sign.
113. $-3\sqrt[3]{2}$. $9$ is a square, not a cube.
114. a) $2\sqrt[3]{3}$ b) $-2\sqrt[3]{2}$ c) $2 \cdot 4\sqrt[3]{2} = 8\sqrt[3]{2}$ d) $-3\sqrt[3]{5}$
115. $11\sqrt{2} + 7\sqrt{3} + 6\sqrt[3]{2}$
116. a) $5\sqrt{2} + 3\sqrt{2} - 2\sqrt{2} = 6\sqrt{2}$ b) $\sqrt{324} = 18$ c) $6\sqrt{100} = 6 \cdot 10 = 60$ d) $2\sqrt[3]{2} + 4\sqrt[3]{2} = 6\sqrt[3]{2}$
117. $\sqrt{9 + 16} = 5$ and $\sqrt{9} + \sqrt{16} = 7$, so unequal. Product property: $\sqrt{a}\sqrt{b} = \sqrt{ab}$ for $a \ge 0$, $b \ge 0$.
118. $a^{1/2} \cdot a^{1/2} = a$ forces $a^{1/2} = \sqrt{a}$ for $a \ge 0$. $49^{1/2} = \sqrt{49} = 7$; $(-8)^{1/3} = \sqrt[3]{-8} = -2$.
119. Agrees; wrong. Mistakes caught: wrong factor pulled ($5\sqrt{3}$), and square root taken twice ($9\sqrt{5}$).
120. Side $= \sqrt{72} = 6\sqrt{2} = 72^{1/2} \approx 8.49$ meters. Perimeter $= 4 \cdot 6\sqrt{2} = 24\sqrt{2}$ meters.