9.3E: Exercises
- Page ID
- 30277
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Practice Makes Perfect
Add and Subtract Like Square Roots
In the following exercises, simplify.
\(8\sqrt{2}−5\sqrt{2}\)
- Answer
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\(3\sqrt{2}\)
\(7\sqrt{2}−3\sqrt{2}\)
\(3\sqrt{5}+6\sqrt{5}\)
- Answer
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\(9\sqrt{5}\)
\(4\sqrt{5}+8\sqrt{5}\)
\(9\sqrt{7}−10\sqrt{7}\)
- Answer
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\(−\sqrt{7}\)
\(11\sqrt{7}−12\sqrt{7}\)
\(7\sqrt{y}+2\sqrt{y}\)
- Answer
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\(9\sqrt{y}\)
\(9\sqrt{n}+3\sqrt{n}\)
\(\sqrt{a}−4\sqrt{a}\)
- Answer
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\(−3\sqrt{a}\)
\(\sqrt{b}−6\sqrt{b}\)
\(5\sqrt{c}+2\sqrt{c}\)
- Answer
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\(7\sqrt{c}\)
\(7\sqrt{d}+2\sqrt{d}\)
\(8\sqrt{a}−2\sqrt{b}\)
- Answer
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\(8\sqrt{a}−2\sqrt{b}\)
\(5\sqrt{c}−3\sqrt{d}\)
\(5\sqrt{m}+\sqrt{n}\)
- Answer
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\(5\sqrt{m}+\sqrt{n}\)
\(\sqrt{n}+3\sqrt{p}\)
\(8\sqrt{7}+2\sqrt{7}+3\sqrt{7}\)
- Answer
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\(13\sqrt{7}\)
\(6\sqrt{5}+3\sqrt{5}+\sqrt{5}\)
\(3\sqrt{11}+2\sqrt{11}−8\sqrt{11}\)
- Answer
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\(−3\sqrt{11}\)
\(2\sqrt{15}+5\sqrt{15}−9\sqrt{15}\)
\(3\sqrt{3}−8\sqrt{3}+7\sqrt{5}\)
- Answer
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\(−5\sqrt{3}+7\sqrt{5}\)
\(5\sqrt{7}−8\sqrt{7}+6\sqrt{3}\)
\(6\sqrt{2}+2\sqrt{2}−3\sqrt{5}\)
- Answer
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\(8\sqrt{2}−3\sqrt{5}\)
\(7\sqrt{5}+\sqrt{5}−8\sqrt{10}\)
\(3\sqrt{2a}−4\sqrt{2a}+5\sqrt{2a}\)
- Answer
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\(4\sqrt{2a}\)
\(\sqrt{11b}−5\sqrt{11b}+3\sqrt{11b}\)
\(8\sqrt{3c}+2\sqrt{3c}−9\sqrt{3c}\)
- Answer
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\(\sqrt{3c}\)
\(3\sqrt{5d}+8\sqrt{5d}−11\sqrt{5d}\)
\(5\sqrt{3ab}+\sqrt{3ab}−2\sqrt{3ab}\)
- Answer
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\(4\sqrt{3ab}\
\(8\sqrt{11cd}+5\sqrt{11cd}−9\sqrt{11cd}\)
\(2\sqrt{pq}−5\sqrt{pq}+4\sqrt{pq}\)
- Answer
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\(\sqrt{pq}\)
\(11\sqrt{2rs}−9\sqrt{2rs}+3\sqrt{2rs}\)
In the following exercises, simplify.
\(\sqrt{50}+4\sqrt{2}\)
- Answer
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\(9\sqrt{2}\)
\(\sqrt{48}+2\sqrt{3}\)
\(\sqrt{80}−3\sqrt{5}\)
- Answer
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\(\sqrt{5}\)
\(\sqrt{28}−4\sqrt{7}\)
\(\sqrt{27}−\sqrt{75}\)
- Answer
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\(−2\sqrt{3}\)
\(\sqrt{72}−\sqrt{98}\)
\(\sqrt{48}+\sqrt{27}\)
- Answer
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\(7\sqrt{3}\)
\(\sqrt{45}+\sqrt{80}\)
\(2\sqrt{50}−3\sqrt{72}\)
- Answer
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\(−8\sqrt{2}\)
\(3\sqrt{98}−\sqrt{128}\)
\(2\sqrt{12}+3\sqrt{48}\)
- Answer
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\(16\sqrt{3}\)
\(4\sqrt{75}+2\sqrt{108}\)
\(\frac{2}{3}\sqrt{72}+\frac{1}{5}\sqrt{50}\)
- Answer
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\(5\sqrt{2}\)
\(\frac{2}{5}\sqrt{75}+\frac{3}{4}\sqrt{48}\)
\(\frac{1}{2}\sqrt{20}−\frac{2}{3}\sqrt{45}\)
- Answer
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\(−\sqrt{5}\)
\(\frac{2}{3}\sqrt{54}−\frac{3}{4}\sqrt{96}\)
\(\frac{1}{6}\sqrt{27}−\frac{3}{8}\sqrt{48}\)
- Answer
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\(−\sqrt{3}\)
\(\frac{1}{8}\sqrt{32}−\frac{1}{10}\sqrt{50}\)
\(\frac{1}{4}\sqrt{98}−\frac{1}{3}\sqrt{128}\)
- Answer
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\(−\frac{3}{4}\sqrt{2}\)
\(\frac{1}{3}\sqrt{24}+\frac{1}{4}\sqrt{54}\)
\(\sqrt{72a^5}−\sqrt{50a^5}\)
- Answer
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\(a^2\sqrt{2a}\)
\(\sqrt{48b^5}−\sqrt{75b^5}\)
\(\sqrt{80c^7}−\sqrt{20c^7}\)
- Answer
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\(2c^3\sqrt{5c}\)
\(\sqrt{96d^9}−\sqrt{24d^9}\)
\(9\sqrt{80p^4}−6\sqrt{98p^4}\)
- Answer
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\(36p^2\sqrt{5}−42p^2\sqrt{2}\)
\(8\sqrt{72q^6}−3\sqrt{75q^6}\)
\(2\sqrt{50r^8}+4\sqrt{54r^8}\)
- Answer
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\(10r^4\sqrt{2}+12r^4\sqrt{6}\)
\(5\sqrt{27s^6}+2\sqrt{20s^6}\)
\(3\sqrt{20x^2}−4\sqrt{45x^2}+5x\sqrt{80}\)
- Answer
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\(14x\sqrt{5}\)
\(2\sqrt{28x^2}−6\sqrt{3x^2}+6x\sqrt{7}\)
\(3\sqrt{128y^2}+4y\sqrt{162}−8\sqrt{98y^2}\)
- Answer
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\(−12y\sqrt{2}\)
\(3\sqrt{75y^2}+8y\sqrt{48}−\sqrt{300y^2}\)
Mixed Practice
\(2\sqrt{8}+6\sqrt{8}−5\sqrt{8}\)
- Answer
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\(3\sqrt{8}\)
\(\frac{2}{3}\sqrt{27}+\frac{3}{4}\sqrt{48}\)
\(\sqrt{175k^4}−\sqrt{63k^4}\)
- Answer
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\(2k^2\sqrt{7}\)
\(\frac{5}{6}\sqrt{162}+\frac{3}{16}\sqrt{128}\)
\(2\sqrt{363}−2\sqrt{300}\)
- Answer
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\(2\sqrt{3}\)
\(\sqrt{150}+4\sqrt{6}\)
\(9\sqrt{2}−8\sqrt{2}\)
- Answer
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\(\sqrt{2}\)
\(5\sqrt{x}−8\sqrt{y}\)
\(8\sqrt{13}−4\sqrt{13}−3\sqrt{13}\)
- Answer
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\(\sqrt{13}\)
\(5\sqrt{12c^4}−3\sqrt{27c^6}\)
\(\sqrt{80a^5}−\sqrt{45a^5}\)
- Answer
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\(a^2\sqrt{5a}\)
\(\frac{3}{5}\sqrt{75}−\frac{1}{4}\sqrt{48}\)
\(21\sqrt{19}−2\sqrt{19}\)
- Answer
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\(19\sqrt{19}\)
\(\sqrt{500}+\sqrt{405}\)
\(\frac{5}{6}\sqrt{27}+\frac{5}{8}\sqrt{48}\)
- Answer
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\(5\sqrt{3}\)
\(11\sqrt{11}−10\sqrt{11}\)
\(\sqrt{75}−\sqrt{108}\)
- Answer
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\(−\sqrt{3}\)
\(2\sqrt{98}−4\sqrt{72}\)
\(4\sqrt{24x^2}−\sqrt{54x^2}+3x\sqrt{6}\)
- Answer
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\(8x\sqrt{6}\)
\(8\sqrt{80y^6}−6\sqrt{48y^6}\)
Everyday Math
A decorator decides to use square tiles as an accent strip in the design of a new shower, but she wants to rotate the tiles to look like diamonds. She will use 9 large tiles that measure 8 inches on a side and 8 small tiles that measure 2 inches on a side. Determine the width of the accent strip by simplifying the expression \(9(8\sqrt{2})+8(2\sqrt{2})\). (Round to the nearest tenth of an inch.)
- Answer
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124.5 inches
Suzy wants to use square tiles on the border of a spa she is installing in her backyard. She will use large tiles that have area of 12 square inches, medium tiles that have area of 8 square inches, and small tiles that have area of 4 square inches. Once section of the border will require 4 large tiles, 8 medium tiles, and 10 small tiles to cover the width of the wall. Simplify the expression \(4\sqrt{12}+8\sqrt{8}+10\sqrt{4}\) to determine the width of the wall.
Writing Exercises
Explain the difference between like radicals and unlike radicals. Make sure your answer makes sense for radicals containing both numbers and variables.
- Answer
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Answers will vary.
Explain the process for determining whether two radicals are like or unlike. Make sure your answer makes sense for radicals containing both numbers and variables.
Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?