9.6: Addition and Subtraction of Square Root Expressions
- Page ID
- 49398
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The Logic Behind The Process
Now we will study methods of simplifying radical expressions such as
\(4\sqrt{3} + 8\sqrt{3}\) or \(5\sqrt{2x} - 11\sqrt{2x} + 4(\sqrt{2x} + 1)\)
The procedure for adding and subtracting square root expressions will become apparent if we think back to the procedure we used for simplifying polynomial expressions such as
\(4x + 8x\) or \(5a - 11a + 4(a+1)\)
The variables \(x\) and \(a\) are letters representing some unknown quantities (perhaps \(x\) represents \(\sqrt{3}\) and \(a\) represents \(\sqrt{2x}\)). Combining like terms gives us
\(\begin{array}{flushleft}
4x + 8x = 12x & \text{ or } & 4\sqrt{3} + 8\sqrt{3} = 12\sqrt{3}\\
\text{and}\\
5a - 11a + 4(a + 1) & \text{ or } & 5\sqrt{2x} - 11\sqrt{2x} + 4(\sqrt{2x} + 1)\\
5a - 11a + 4a + 4 && 5\sqrt{2x} - 11\sqrt{2x} + 4\sqrt{2x} + 4\\
-2a && -2\sqrt{2x} + 4
\end{array}\)
The Process
Let's consider the expression \(4\sqrt{3} + 8\sqrt{3}\). There are two ways to look at the simplification process.
We are asking, "How many square roots of \(3\) do we have?"
\(4 \sqrt{3}\) means we have \(4\) "square roots of \(3\)"
Thus, altogether we have \(12\) "square roots of \(3\)."
We can also use the idea of combining like terms. If we recall, the process of combining like terms is based on the distributive property
\(4x + 8x = 12x\) because \(4x + 8x = (4 + 8)x = 12x\)
We could simplify \(4\sqrt{3} + 8\sqrt{3}\) using the distributive property.
4\sqrt{3} + 8\sqrt{3} = (4 + 8)\sqrt{3} = 12\sqrt{3}\)
Both methods will give us the same result. The first method is probably a bit quicker, but keep in mind, however, that the process works because it is based on one of the basic rules of algebra, the distributive property of real numbers.
Sample Set A
Simplify the following radical expressions.
\(-6\sqrt{10} + 11\sqrt{10} = 5\sqrt{10}\)
\(4\sqrt{32} + 5\sqrt{2} \text{ Simplify } \sqrt{32}\)
\(\begin{array}{flushleft}
4\sqrt{16 \cdot 2} + 5\sqrt{2} &= 4\sqrt{16}\sqrt{2} + 5\sqrt{2}\\
&= 4 \cdot 4\sqrt{2} + 5\sqrt{2}\\
&= 16\sqrt{2} + 5\sqrt{2}\\
&=21\sqrt{2}
\end{array}\)
\(-3x\sqrt{75} + 2x\sqrt{48} - x\sqrt{27} \text{ Simplify each of the three radicals}\)
\(\begin{array}{flushleft}
-3x\sqrt{75} + 2x\sqrt{48} - x\sqrt{27} &= -3x\sqrt{25 \cdot 3} + 2x\sqrt{16 \cdot 3} - x\sqrt{9 \cdot 2}\\
&= -15x\sqrt{3} + 8x\sqrt{3} - 3x\sqrt{3}\\
&=(-15x + 8x - 3x)\sqrt{3}\\
&=-10x\sqrt{3}
\end{array}\)
\(5a\sqrt{24a^3} - 7\sqrt{54a^5} + a^2\sqrt{6a} + 6a \text{ Simplify each radical}\)
\(\begin{array}{flushleft}
5a\sqrt{24a^3} - 7\sqrt{54a^5} + a^2\sqrt{6a} + 6a &= 5a\sqrt{4 \cdot 6 \cdot a^2 \cdot a} - 7\sqrt{9 \cdot 6 \cdot a^4 \cdot a} + a^2\sqrt{6a} + 6a\\
&= 10a^2\sqrt{6a} - 21a^2\sqrt{6a} + a^2\sqrt{6a} + 6a\\
&= (10a^2 - 21a^2 + a^2) \sqrt{6a} + 6a\\
&= -10a^2\sqrt{6a} + 6a\\
&= -2a(5a\sqrt{6a} - 3)
\end{array}\)
Practice Set A
Find each sum or difference.
\(4\sqrt{18} - 5\sqrt{8}\)
- Answer
-
\(2\sqrt{2}\)
\(6x\sqrt{48} + 8x\sqrt{75}\)
- Answer
-
\(64x\sqrt{3}\)
\(-7\sqrt{84x} - 12\sqrt{189x} + 2\sqrt{21x}\)
- Answer
-
\(-48\sqrt{21x}\)
\(9\sqrt{6} - 8\sqrt{6} + 3\)
- Answer
-
\(\sqrt{6} + 3\)
\(\sqrt{a^3} + 4a\sqrt{a}\)
- Answer
-
\(5a\sqrt{a}\)
\(4x\sqrt{54x^3} + \sqrt{36x^2} + 3\sqrt{24x^5} - 3x\)
- Answer
-
\(18x^2\sqrt{6x} + 3x\)
Sample Set B
\(\dfrac{3 + \sqrt{8}}{3 - \sqrt{8}}\) We'll rationalize the denominator by multiplying this fraction by \(1\) in the form \(\dfrac{3 + \sqrt{8}}{3 + \sqrt{8}}\).
\(\begin{array}{flushleft}
\dfrac{3+\sqrt{8}}{3-\sqrt{8}} \cdot \frac{3+\sqrt{8}}{3+\sqrt{8}} &=\dfrac{(3+\sqrt{8})(3+\sqrt{8})}{3^{2}-(\sqrt{8})^{2}} \\
&=\dfrac{9+3 \sqrt{8}+3 \sqrt{8}+\sqrt{8} \sqrt{8}}{9-8} \\
&=\dfrac{9+6 \sqrt{8}+8}{1} \\
&=17+6 \sqrt{8} \\
&=17+6 \sqrt{4 \cdot 2} \\
&=17+12 \sqrt{2}
\end{array}\)
\(\dfrac{2 + \sqrt{7}}{4 - \sqrt{3}}\). Rationalize the denominator by multiplying this fraction by \(1\) in the form \(\dfrac{4 + \sqrt{3}}{4 + \sqrt{3}}\).
\(\begin{array}{flushleft}
\dfrac{2+\sqrt{7}}{4-\sqrt{3}} \cdot \dfrac{4+\sqrt{3}}{4+\sqrt{3}} &=\dfrac{(2+\sqrt{7})(4+\sqrt{3})}{4^{2}-(\sqrt{3})^{2}} \\
&=\dfrac{8+2 \sqrt{3}+4 \sqrt{7}+\sqrt{21}}{16-3} \\
&=\dfrac{8+2 \sqrt{3}+4 \sqrt{7}+\sqrt{21}}{13}
\end{array}\)
Practice Set B
Simplify each by performing the indicated operation.
\(\sqrt{5}(\sqrt{6} - 4)\)
- Answer
-
\(\sqrt{30} - 4\sqrt{5}\)
\((\sqrt{5} + \sqrt{7})(\sqrt{2} + \sqrt{8})\)
- Answer
-
\(3\sqrt{10} + 3\sqrt{14}\)
\((3\sqrt{2} - 2\sqrt{3})(4\sqrt{3} + \sqrt{8})\)
- Answer
-
\(8\sqrt{6} - 12\)
\(\dfrac{4 + \sqrt{5}}{3 - \sqrt{8}}\)
- Answer
-
\(12 + 8\sqrt{2} + 3\sqrt{5} + 2\sqrt{10}\)
Exercises
For the following problems, simplify each expression by performing the indicated operation.
\(4\sqrt{5} - 2\sqrt{5}\)
- Answer
-
\(2\sqrt{5}\)
\(10 \sqrt{2} + 8\sqrt{2}\)
\(-3\sqrt{6} - 12\sqrt{6}\)
- Answer
-
\(-15 \sqrt{6}\)
\(-\sqrt{10} - 2\sqrt{10}\)
\(3\sqrt{7x} + 2\sqrt{7x}\)
- Answer
-
\(5\sqrt{7x}\)
\(6\sqrt{3a} + \sqrt{3a}\)
\(2\sqrt{18} + 5\sqrt{32}\)
- Answer
-
\(26\sqrt{2}\)
\(4\sqrt{27} - 3\sqrt{48}\)
\(\sqrt{200} - \sqrt{128}\)
- Answer
-
\(2\sqrt{2}\)
\(4\sqrt{300} + 2\sqrt{500}\)
\(6\sqrt{40} + 8\sqrt{80}\)
- Answer
-
\(12\sqrt{10} + 32\sqrt{5}\)
\(2\sqrt{120} - 5\sqrt{30}\)
\(8\sqrt{60} - 3\sqrt{15}\)
- Answer
-
\(13\sqrt{15}\)
\(\sqrt{a^3} - 3a\sqrt{a}\)
\(\sqrt{4x^3} + x\sqrt{x}\)
- Answer
-
\(3x\sqrt{x}\)
\(2b\sqrt{a^3b^5} + 6a\sqrt{ab^7}\)
\(5xy\sqrt{2xy^3} - 3y^2\sqrt{2x^3y}\)
- Answer
-
\(2xy^2\sqrt{2xy}\)
\(5\sqrt{20} + 3\sqrt{45} - 3\sqrt{40}\)
\(\sqrt{24} - 2\sqrt{54} - 4\sqrt{12}\)
- Answer
-
\(-4\sqrt{6} - 8\sqrt{3}\)
\(6\sqrt{18} + 5\sqrt{32} + 4\sqrt{50}\)
\(-8\sqrt{20} - 9\sqrt{125} + 10\sqrt{180}\)
- Answer
-
\(-\sqrt{5}\)
\(2\sqrt{27} + 4\sqrt{3} - 6\sqrt{12}\)
\(\sqrt{14} + 2\sqrt{56} - 3\sqrt{136}\)
- Answer
-
\(5\sqrt{14} - 6\sqrt{34}\)
\(3\sqrt{2} + 2\sqrt{63} + 5\sqrt{7}\)
\(4ax\sqrt{3x} + 2\sqrt{3a^2x^3} + 7\sqrt{3a^2x^3}\)
- Answer
-
\(13ax\sqrt{3x}\)
\(3by\sqrt{5y} + 4\sqrt{5b^2y^3} - 2\sqrt{5b^2y^3}\)
\(\sqrt{2}(\sqrt{3} + 1)\)
- Answer
-
\(\sqrt{6} + \sqrt{2}\)
\(\sqrt{3}(\sqrt{5} - 3)\)
\(\sqrt{5}(\sqrt{3} - \sqrt{2})\)
- Answer
-
\(\sqrt{15} - \sqrt{10}\)
\(\sqrt{7}(\sqrt{6} - \sqrt{3})\)
\(\sqrt{8}(\sqrt{3} + \sqrt{2})\)
- Answer
-
\(2(\sqrt{6} + 2)\)
\(\sqrt{10}(\sqrt{10} - \sqrt{5})\)
\((1 + \sqrt{3})(2 - \sqrt{3})\)
- Answer
-
\(-1 + \sqrt{3}\)
\((5 + \sqrt{6})(4 - \sqrt{6})\)
\((3 - \sqrt{2})(4 - \sqrt{2})\)
- Answer
-
\(7(2 - \sqrt{2})\)
\((5 + \sqrt{7})(4 - \sqrt{7})\)
\((\sqrt{2} + \sqrt{5})(\sqrt{2} + 3\sqrt{5})\)
- Answer
-
\(17 + 4\sqrt{10}\)
\((2\sqrt{6} - \sqrt{3})(3\sqrt{6} + 2\sqrt{3})\)
\((4\sqrt{5} - 2\sqrt{3})(3\sqrt{5} + \sqrt{3})\)
- Answer
-
\(54 - 2\sqrt{15}\)
\((3\sqrt{8} - 2\sqrt{2})(4\sqrt{2} - 5\sqrt{8})\)
\((\sqrt{12} + 5\sqrt{3})(2\sqrt{3} - 2\sqrt{12})\)
- Answer
-
\(-42\)
\((1 + \sqrt{3})^2\)
\((3 + \sqrt{5})^2\)
- Answer
-
\(14 + 6\sqrt{5}\)
\((2 - \sqrt{6})^2\)
\((2 - \sqrt{7})^2\)
- Answer
-
\(11 - 4\sqrt{7}\)
\((1 + \sqrt{3x})^2\)
\((2 + \sqrt{5x})^2\)
- Answer
-
\(4 + 4\sqrt{5x} + 5x\)
\((3 - \sqrt{3x})^2\)
\((8 - \sqrt{6b})^2\)
- Answer
-
\(64 - 16\sqrt{6b} + 6b\)
\((2a + \sqrt{5a})^2\)
\((3y - \sqrt{7y})^2\)
- Answer
-
\(9y^2 - 6y\sqrt{7y} + 7y\)
\((3 + \sqrt{3})(3 - \sqrt{3})\)
\((2 + \sqrt{5})(2 - \sqrt{5})\)
- Answer
-
\(-1\)
\((8 + \sqrt{10})(8 - \sqrt{10})\)
\((6 + \sqrt{7})(6 - \sqrt{7})\)
- Answer
-
\(29\)
\((\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3}\)
\((\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2})\)
- Answer
-
\(3\)
\((\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})\)
\((\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})\)
- Answer
-
\(x - y\)
\(\dfrac{2}{5 + \sqrt{3}}\)
\(\dfrac{4}{6 + \sqrt{2}}\)
- Answer
-
\(\dfrac{2(6 - \sqrt{2})}{17}\)
\(\dfrac{1}{3 - \sqrt{2}}\)
\(\dfrac{1}{4 - \sqrt{3}}\)
- Answer
-
\(\dfrac{4 + \sqrt{3}}{13}\)
\(\dfrac{8}{2 - \sqrt{6}}\)
\(\dfrac{2}{3 - \sqrt{7}}\)
- Answer
-
\(3 + \sqrt{7}\)
\(\dfrac{\sqrt{5}}{3 + \sqrt{3}}\)
\(\dfrac{\sqrt{3}}{6 + \sqrt{6}}\)
- Answer
-
\(\dfrac{2\sqrt{3} - \sqrt{2}}{10}\)
\(\dfrac{2 - \sqrt{8}}{2 + \sqrt{8}}\)
\(\dfrac{4 + \sqrt{5}}{4 - \sqrt{5}}\)
- Answer
-
\(\dfrac{21 + 8\sqrt{5}}{11}\)
\(\dfrac{1 + \sqrt{6}}{1 - \sqrt{6}}\)
\(\dfrac{8 - \sqrt{3}}{2 + \sqrt{18}}\)
- Answer
-
\(\dfrac{-16 + 2\sqrt{3} + 24\sqrt{2} - 3\sqrt{6}}{14}\)
\(\dfrac{6 - \sqrt{2}}{4 + \sqrt{12}}\)
\(\dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\)
- Answer
-
\(5 - 2\sqrt{6}\)
\(\dfrac{\sqrt{6a} - \sqrt{8a}}{\sqrt{8a} + \sqrt{6a}}\)
\(\dfrac{\sqrt{2b} - \sqrt{3b}}{\sqrt{3b} + \sqrt{2b}}\)
- Answer
-
\(2\sqrt{6} - 5\)
Exercises For Review
Simplify \((\dfrac{x^5y^3}{x^2y})^5\)
Simplify \((8x^3y)^2(x^2y^3)^4\)
- Answer
-
\(64x^{14}y^{14}\)
Write \((x-1)^4(x-1)^{-7}\) so that only positive exponents appear.
Simpify \(\sqrt{27x^5y^{10}z^3}\)
- Answer
-
\(3x^2y^5z\sqrt{3xz}\)
Simplify \(\dfrac{1}{2 + \sqrt{x}}\) by rationalizing the denominator.