10.1: Simplify Radicals
- Page ID
- 45134
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Not all radicands are perfect squares, where when we take the square root, we obtain a positive integer. For example, if we input \(\sqrt{8}\) in a calculator, the calculator would display \(2.828427124746190097603377448419\cdots\) and even this number is a rounded approximation of the square root. To be as accurate as possible, we will leave all answers in exact form, i.e., answers contain integers and radicals- no decimals.
When we say to simplify an expression with radicals, the simplified expression should have
- a radical, unless the radical reduces to an integer
- a radicand with no factors containing perfect squares
- no decimals
Following these guidelines ensures the expression is in its simplest form.
Simplify Radicals
If \(a\), \(b\) are any two positive real numbers, then \[\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\nonumber\] In general, if \(a\), \(b\) are any two positive real numbers, then \[\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b},\nonumber\] where \(n\) is a positive integer and \(n\geq 2\).
Simplify: \(\sqrt{75}\)
Solution
We can apply the product rule for radicals to simplify this number. We need to find the largest factor of \(75\) that is a perfect square (since we have a square root) and rewrite the radicand as a product of this perfect square and its other factor. The largest factor of radicand \(75\) that is a perfect square is \(25\).
\[\begin{array}{rl}\sqrt{75}&\text{Rewrite radicand as a product of }25\text{ and }3 \\ \sqrt{25\cdot 3}&\text{Apply product rule for radicals} \\ \sqrt{25}\cdot\sqrt{3}&\text{Simplify each square root} \\ 5\cdot\sqrt{3}&\text{Rewrite} \\ 5\sqrt{3}&\text{Simplified expression}\end{array}\nonumber\]
If the radicand is not a perfect square, we leave as is; hence, we left \(\sqrt{3}\) as is.
Simplify: \(\sqrt{72}\)
Solution
We can apply the product rule for radicals to simplify this number. We need to find the largest factor of \(72\) that is a perfect square (since we have a square root) and rewrite the radicand as a product of this perfect square and its other factor. The largest factor of radicand \(72\) that is a perfect square is \(36\).
\[\begin{array}{rl}\sqrt{72}&\text{Rewrite radicand as a product of }36\text{ and }2 \\ \sqrt{36\cdot 2}&\text{Apply product rule for radicals} \\ \sqrt{36}\cdot\sqrt{2}&\text{Simplify each square root} \\ 6\cdot\sqrt{2}&\text{Rewrite} \\ 6\sqrt{2}&\text{Simplified expression}\end{array}\nonumber\]
If the radicand is not a perfect square, we leave as is; hence, we left \(\sqrt{2}\) as is.
Simplify Radicals with Coefficients
Simplify: \(5\sqrt{63}\)
Solution
We can apply the product rule for radicals to simplify this number and multiply coefficients in the last steps. We need to find the largest factor of \(63\) that is a perfect square (since we have a square root) and rewrite the radicand as a product of this perfect square and its other factor. The largest factor of radicand \(63\) that is a perfect square is \(9\).
\[\begin{array}{rl}5\sqrt{63}&\text{Rewrite radicand as a product of }9\text{ and }7 \\ 5\sqrt{9\cdot 7}&\text{Apply product rule for radicals} \\ 5\cdot\sqrt{9}\cdot\sqrt{7}&\text{Simplify each square root} \\ 5\cdot 3\cdot\sqrt{7}&\text{Rewrite and simplify coefficients} \\ 15\sqrt{7}&\text{Simplified expression}\end{array}\nonumber\]
If the radicand is not a perfect square, we leave as is; hence, we left \(\sqrt{7}\) as is.
Rational Exponents
When we simplify radicals, we extract roots of factors with exponents in which are multiples of the root (index). For example, \(\sqrt{x^4}=\sqrt[2]{x^4}=x^2\), but notice we just divided the power on \(x\) by the root. Let’s look at the example again, but now as division of exponents:
\[\sqrt{x^4}=\color{black}{\sqrt[\color{blue}{3}]{x\color{red}{^4}}=}x^{\dfrac{\color{red}{4}}{\color{blue}{2}}}=x^2\nonumber\]
Division with exponents, or fraction exponents, are called rational exponents.
Let \(a\) be the base, and \(m\) and \(n\) be real real numbers. Then
\[a^{\dfrac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m\nonumber\]
The denominator of a rational exponent is the root on the radical and vice versa
Rewrite each radical with its corresponding rational exponent.
- \((\sqrt[5]{x})^3\)
- \((\sqrt[6]{3x})^5\)
- \(\dfrac{1}{\left(\sqrt[7]{a}\right)^3}\)
- \(\dfrac{1}{\left(\sqrt[3]{xy}\right)^2}\)
Solution
- For the expression \((\sqrt[5]{x})^3\), we see the root is \(5\). This means that the denominator of the rational exponent is \(5\). Hence, the numerator is the exponent \(3\): \((\sqrt[5]{x})^3=x^{\dfrac{3}{5}}\).
- For the expression \((\sqrt[6]{3x})^5\), we see the root is \(6\). This means that the denominator of the rational exponent is \(6\). Hence, the numerator is the exponent \(5\): \((\sqrt[6]{3x})^5=(3x)^{\dfrac{5}{6}}\).
- For the expression \(\dfrac{1}{(\sqrt[7]{a})^3}\), we see the root is \(7\). This means that the denominator of the rational exponent is \(7\). Hence, the numerator is the exponent \(3\). Furthermore, since the expression with the radical is in the denominator, we can rewrite the expression using a negative exponent: \(\dfrac{1}{(\sqrt[7]{a})^3}=(a)^{-\dfrac{3}{7}}\).
- For the expression \(\dfrac{1}{(\sqrt[3]{xy})^2}\), we see the root is \(3\). This means that the denominator of the rational exponent is \(3\). Hence, the numerator is the exponent \(2\). Furthermore, since the expression with the radical is in the denominator, we can rewrite the expression using a negative exponent: \(\dfrac{1}{(\sqrt[3]{xy})^2}=(xy)^{-\dfrac{2}{3}}\).
Rewrite each expression in its equivalent radical form.
- \(a^{\dfrac{5}{3}}\)
- \((2mn)^{\dfrac{2}{7}}\)
- \(x^{-\dfrac{4}{5}}\)
- \((xy)^{-\dfrac{2}{9}}\)
Solution
- From the definition, we know that the denominator of the rational exponent is the root making the numerator the power: \(a^{\dfrac{5}{3}}=\sqrt[3]{a^5}\) or \((\sqrt[3]{a})^5\).
- From the definition, we know that the denominator of the rational exponent is the root making the numerator the power: \((2mn)^{\dfrac{2}{7}}=\sqrt[7]{(2mn)^2}\) or \((\sqrt[7]{2mn})^2\).
- From the definition, we know that the denominator of the rational exponent is the root making the numerator the power: \(x^{−\dfrac{4}{5}} = (\sqrt[5]{x})^{-4}\). Notice that the expression still contains a negative exponent. Hence, we need to reciprocate the radical to rewrite the expression with only positive exponents: \[x^{-\dfrac{4}{5}}=\dfrac{1}{(\sqrt[5]{x})^4}\nonumber\]
- From the definition, we know that the denominator of the rational exponent is the root making the numerator the power: \((xy)^{−\dfrac{2}{9}} = (\sqrt[9]{x})^{−2}\). Notice that the expression still contains a negative exponent. Hence, we need to reciprocate the radical to rewrite the expression with only positive exponents: \[(xy)^{-\dfrac{2}{9}}=\dfrac{1}{(\sqrt[9]{xy})^2}\nonumber\]
Nicole Oresme, a Mathematician born in Normandy was the first to use rational exponents. He used the notation \(\dfrac{1}{3} • 9^{p}\) to represent \(9^{\dfrac{1}{3}}\). However, his notation went largely unnoticed
The ability to change between rational exponential expressions and radical expressions allows us to evaluate expressions.
Evaluate \(27^{-\dfrac{4}{3}}\).
Solution
We first rewrite the expression with only positive exponents, then evaluate the exponen
\[\begin{array}{rl}27^{-\dfrac{4}{3}}&\text{Rewrite the expression with positive exponents} \\ \dfrac{1}{27^{\dfrac{4}{3}}}&\text{Rewrite in radical form} \\ \dfrac{1}{(\sqrt[3]{27})^4}&\text{Evaluate radical }\sqrt[3]{27}=3 \\ \dfrac{1}{(3)^4}&\text{Evaluate exponent }3^4=81 \\ \dfrac{1}{81}&\text{Result} \end{array}\nonumber\]
Thus, \(27^{−\dfrac{4}{3}} = \dfrac{1}{81}\). This result should emphasize the fact that negative exponents means reciprocals, and not negative numbers.
Simplify Radicals with Variables
Commonly, radicands can contain variables. When taking the square roots of variables, we know the root is \(2\); we do not always write it, but we know it’s there. Hence, we apply the product rule of radicals by rewriting the variable’s exponent and rewrite the exponents so that one of the exponents is the largest even number.
Simplify: \(\sqrt{x^6 y^5}\)
Solution
We can apply the product rule for radicals to simplify by rewriting the variable’s exponent and rewrite the exponents so that one of the exponents is the largest even number.
\[\begin{array}{rl}\sqrt{x^6y^5}&\text{Rewrite radicand} \\ \sqrt{x^6\cdot y^4\cdot y^1}&\text{Apply product rule for radicals} \\ \sqrt{x^6}\cdot\sqrt{y^4}\cdot\sqrt{y}&\text{Simplify each square root} \\ x^3\cdot y^2\cdot\sqrt{y}&\text{Rewrite and simplify coefficients} \\ x^3y^2\sqrt{y}&\text{Simplified expression}\end{array}\nonumber\]
Notice that \((x^3)^2\) and \((y^2)^2=y^4\); hence, we extract the perfect squares of the variables and leave the \(\sqrt{y}\) as is.
Recall, when taking a square root of a number, the radicand must be greater than or equal to zero. So, when we are applying the square root to variables, the variables must also be greater than or equal to zero.
Notice, we are essentially dividing the exponents on the variables by two and the factor that remains in the radicand has exponent \(1\).
Simplify: \(-5\sqrt{18x^4y^6z^{10}}\). Assume all variables are positive.
Solution
We can apply the product rule for radicals to simplify by rewriting the variable’s exponent and rewrite the exponents so that one of the exponents is the largest even number.
\[\begin{array}{rl}-5\sqrt{18x^4y^6z^{10}}&\text{Rewrite radicand} \\ -5\cdot\sqrt{9\cdot 2\cdot x^4\cdot y^6\cdot x^{10}}&\text{Apply product rule for radicals} \\ -5\cdot\sqrt{9}\cdot\sqrt{2}\cdot\sqrt{x^4}\cdot\sqrt{y^6}\cdot\sqrt{z^{10}}&\text{Simplify each square root} \\ -5\cdot 3\cdot\sqrt{2}\cdot x^2\cdot y^3\cdot z^5 &\text{Rewrite and simplify coefficients} \\ -15x^2y^3z^5\sqrt{2}&\text{Simplified expression}\end{array}\nonumber\]
Simplify: \(\sqrt{20x^5y^9z^6}\). Assume all variables are positive.
Solution
We can apply the product rule for radicals to simplify by rewriting the variable’s exponent and rewrite the exponents so that one of the exponents is the largest even number.
\[\begin{array}{rl}\sqrt{20x^5y^9z^6}&\text{Rewrite radicand} \\ \sqrt{4\cdot 5\cdot x^4\cdot x\cdot y^8\cdot y\cdot z^6}&\text{Apply product rule for radicals} \\ \sqrt{4}\cdot\sqrt{5}\cdot\sqrt{x^4}\cdot\sqrt{x}\cdot\sqrt{y^8}\cdot\sqrt{y}\cdot\sqrt{z^6}&\text{Simplify each square root} \\ 2\cdot\sqrt{5}\cdot x^2\cdot\sqrt{x}\cdot y^4\cdot\sqrt{y}\cdot z^3&\text{Rewrite and simplify coefficients} \\ 2x^2y^4z^3\sqrt{5xy}&\text{Simplified expression}\end{array}\nonumber\]
Simplify Radicals Homework
Simplify. Assume all variables are positive.
\(\sqrt{245}\)
\(\sqrt{36}\)
\(\sqrt{12}\)
\(3\sqrt{12}\)
\(6\sqrt{128}\)
\(-8\sqrt{392}\)
\(\sqrt{192n}\)
\(\sqrt{196v^2}\)
\(\sqrt{252x^2}\)
\(-\sqrt{100k^4}\)
\(-7\sqrt{64x^4}\)
\(-5\sqrt{36m}\)
\(\sqrt{45x^2y^2}\)
\(\sqrt{16x^3y^3}\)
\(\sqrt{320x^4y^4}\)
\(6\sqrt{80xy^2}\)
\(5\sqrt{245x^2y^3}\)
\(-2\sqrt{180u^3v}\)
\(-8\sqrt{180x^4y^2z^4}\)
\(2\sqrt{80hj^4k}\)
\(-4\sqrt{54mnp^2}\)
\(\sqrt{125}\)
\(\sqrt{196}\)
\(\sqrt{338}\)
\(5\sqrt{32}\)
\(7\sqrt{128}\)
\(-7\sqrt{63}\)
\(\sqrt{343b}\)
\(\sqrt{100n^3}\)
\(\sqrt{200a^3}\)
\(-4\sqrt{175p^4}\)
\(-2\sqrt{128n}\)
\(8\sqrt{112p^2}\)
\(\sqrt{72a^3b^4}\)
\(\sqrt{512a^4b^2}\)
\(\sqrt{512m^4n^3}\)
\(8\sqrt{98mn}\)
\(2\sqrt{72x^2y^2}\)
\(-5\sqrt{72x^3y^4}\)
\(6\sqrt{50a^4bc^2}\)
\(-\sqrt{32xy^2z^3}\)
\(-8\sqrt{32m^2p^4q}\)
Write each expression in radical form with only positive exponents.
\(m^{\dfrac{3}{5}}\)
\((7x)^{\dfrac{3}{2}}\)
\((10r)^{-\dfrac{3}{4}}\)
\((6b)^{-\dfrac{4}{3}}\)
Write each expression in exponential form.
\(\dfrac{1}{(\sqrt{6x})^3}\)
\(\dfrac{1}{(\sqrt[4]{n})^7}\)
\(\sqrt{v}\)
\(\sqrt{5a}\)
Evaluate without using a calculator.
\(8^{\dfrac{2}{3}}\)
\(4^{\dfrac{3}{2}}\)
\(16^{\dfrac{1}{4}}\)
\(100^{-\dfrac{3}{2}}\)