Factor the expression completely (or find perfect squares). The same general rules and approach still applies, such as looking to factor where possible, but a bit more attention often needs to be paid. 1) Factor the radicand (the numbers/variables inside the square root). \(\displaystyle \sqrt[n]{{\frac{x}{y}}}=\frac{{\sqrt[n]{x}}}{{\sqrt[n]{y}}}\), \(\displaystyle \sqrt[3]{{\frac{{27}}{8}}}=\frac{{\sqrt[3]{{27}}}}{{\sqrt[3]{8}}}=\frac{3}{2}\), \(\displaystyle \begin{array}{c}\sqrt[{}]{{{{{\left( {-4} \right)}}^{2}}}}=\sqrt{{16}}=4\\\sqrt[{}]{{{{{\left( 4 \right)}}^{2}}}}=\sqrt{{16}}=4\end{array}\), \(\displaystyle \begin{align}\frac{x}{{\sqrt{y}}}&=\frac{x}{{\sqrt{y}}}\cdot \frac{{\sqrt{y}}}{{\sqrt{y}}}\\&=\frac{{x\sqrt{y}}}{y}\end{align}\), \(\displaystyle \begin{align}\frac{4}{{\sqrt{2}}}&=\frac{4}{{\sqrt{2}}}\cdot \frac{{\sqrt{2}}}{{\sqrt{2}}}\\&=\frac{{{}^{2}\cancel{4}\sqrt{2}}}{{{}^{1}\cancel{2}}}=2\sqrt{2}\end{align}\), \(\displaystyle \begin{align}\frac{x}{{x+\sqrt{y}}}&=\frac{x}{{x+\sqrt{y}}}\cdot \frac{{x-\sqrt{y}}}{{x-\sqrt{y}}}\\&=\frac{{x\left( {x-\sqrt{y}} \right)}}{{{{x}^{2}}-y}}\\\frac{x}{{x-\sqrt{y}}}&=\frac{x}{{x-\sqrt{y}}}\cdot \frac{{x+\sqrt{y}}}{{x+\sqrt{y}}}\\&=\frac{{x\left( {x+\sqrt{y}} \right)}}{{{{x}^{2}}-y}}\end{align}\), \(\displaystyle \begin{align}\frac{{\sqrt{3}}}{{1-\sqrt{3}}}&=\frac{{\sqrt{3}}}{{1-\sqrt{3}}}\cdot \frac{{1+\sqrt{3}}}{{1+\sqrt{3}}}\\&=\frac{{\sqrt{3}\left( {1+\sqrt{3}} \right)}}{{\left( {1-\sqrt{3}} \right)\left( {1+\sqrt{3}} \right)}}\\&=\frac{{\sqrt{3}+\sqrt{3}\cdot \sqrt{3}}}{{{{1}^{2}}-{{{\left( {\sqrt{3}} \right)}}^{2}}}}=\frac{{\sqrt{3}+3}}{{-2}}\end{align}\), More rationalizing: when there are two terms in the denominator, we need to multiply both the numerator and denominator by the, To put a radical in the calculator, we can type â, \(\displaystyle \color{#800000}{{\frac{1}{{\sqrt{2}}}}}=\frac{1}{{\sqrt{2}}}\cdot \frac{{\sqrt{2}}}{{\sqrt{2}}}=\frac{{1\sqrt{2}}}{{\sqrt{2}\cdot \sqrt{2}}}=\frac{{\sqrt{2}}}{2}\), Since the \(\sqrt{2}\) is on the bottom, we need to get rid of it by multiplying by, \(\require{cancel} \displaystyle \color{#800000}{{\frac{4}{{2\sqrt{3}}}}}=\frac{4}{{2\sqrt{3}}}\cdot \frac{{\sqrt{3}}}{{\sqrt{3}}}=\frac{{4\sqrt{3}}}{{2\sqrt{3}\cdot \sqrt{3}}}=\frac{{{}^{2}\cancel{4}\sqrt{3}}}{{{}^{1}\cancel{2}\cdot 3}}=\frac{{2\sqrt{3}}}{3}\), Since the \(\sqrt{3}\) is on the bottom, we need to multiply by, \(\displaystyle \color{#800000}{{\frac{5}{{2\sqrt[4]{3}}}}}=\frac{5}{{2\sqrt[4]{3}}}\cdot \frac{{{{{(\sqrt[4]{3})}}^{3}}}}{{{{{(\sqrt[4]{3})}}^{3}}}}=\frac{{5{{{(\sqrt[4]{3})}}^{3}}}}{{2{{{(\sqrt[4]{3})}}^{1}}{{{(\sqrt[4]{3})}}^{3}}}}\), \(\displaystyle \begin{align}\color{#800000}{{\frac{{6x}}{{\sqrt[5]{{4{{x}^{8}}{{y}^{{12}}}}}}}}}&=\frac{{6x}}{{x{{y}^{2}}\sqrt[5]{{4{{x}^{3}}{{y}^{2}}}}}}\cdot \frac{{\sqrt[5]{{8{{x}^{2}}{{y}^{3}}}}}}{{\sqrt[5]{{8{{x}^{2}}{{y}^{3}}}}}}\\&=\frac{{6x\sqrt[5]{{8{{x}^{2}}{{y}^{3}}}}}}{{x{{y}^{2}}\sqrt[5]{{32{{x}^{5}}{{y}^{5}}}}}}=\frac{{6x\sqrt[5]{{8{{x}^{2}}{{y}^{3}}}}}}{{x{{y}^{2}}\cdot 2xy}}\\&=\frac{{3\sqrt[5]{{8{{x}^{2}}{{y}^{3}}}}}}{{x{{y}^{3}}}}\end{align}\), Hereâs another way to rationalize complicated radicals: simplify first, and then multiply by, \(\displaystyle \begin{align}\color{#800000}{{\frac{{3\sqrt{3}}}{{2-2\sqrt{3}}}}}&=\frac{{3\sqrt{3}}}{{2-2\sqrt{3}}}\cdot \frac{{2+2\sqrt{3}}}{{2+2\sqrt{3}}}\\&=\frac{{3\sqrt{3}\left( {2+2\sqrt{3}} \right)}}{{{{2}^{2}}-{{{\left( {2\sqrt{3}} \right)}}^{2}}}}=\frac{{6\sqrt{3}+18}}{{4-12}}\\&=\frac{{6\sqrt{3}+18}}{{-8}}=-\frac{{3\sqrt{3}+9}}{4}\end{align}\), When there are two terms in the denominator (one a radical), multiply both the numerator and denominator by the, \({{\left( {9{{x}^{3}}y} \right)}^{2}}={{9}^{2}}{{x}^{6}}{{y}^{2}}=81{{x}^{6}}{{y}^{2}}\). ], Convert Decimal To Fraction [ Def: A number that names a part of a whole or a part of a group. With \({{64}^{{\frac{1}{4}}}}\), we factor it into 16 and 4, since \({{16}^{{\frac{1}{4}}}}\) is 2. We have to âthrow awayâ our answer and the correct answer is âno solutionâ or \(\emptyset \). Then we solve for \({{y}_{2}}\). We want to raise both sides to the. With odd roots, we donât have to worry about checking underneath the radical sign, since we could have positive or negative numbers as a radicand. âCarry throughâ the exponent to both the top and bottom of the fraction and remember that the cube root of, \(\require{cancel} \displaystyle \begin{align}{{\left( {\frac{{{{2}^{{-1}}}+{{2}^{{-2}}}}}{{{{2}^{{-4}}}}}} \right)}^{{-1}}}&=\frac{{{{2}^{{-4}}}}}{{{{2}^{{-1}}}+{{2}^{{-2}}}}}=\frac{{\frac{1}{{{{2}^{4}}}}}}{{\frac{1}{2}+\frac{1}{4}}}\\&=\frac{{\frac{1}{{{{2}^{4}}}}}}{{\frac{2}{4}+\frac{1}{4}}}=\frac{{\frac{1}{{{{2}^{4}}}}}}{{\frac{3}{4}}}\\&=\frac{1}{{{}_{4}\cancel{{16}}}}\cdot \frac{{{{{\cancel{4}}}^{1}}}}{3}=\frac{1}{{12}}\end{align}\), \(\displaystyle \begin{align}&{{\left( {\frac{{{{2}^{{-1}}}+{{2}^{{-2}}}}}{{{{2}^{{-4}}}}}} \right)}^{{-1}}}\\&=\frac{{{{2}^{{-4}}}}}{{{{2}^{{-1}}}+{{2}^{{-2}}}}}\,\,\times \,\,\frac{{{{2}^{4}}}}{{{{2}^{4}}}}\\&=\frac{{\left( {{{2}^{{-4}}}} \right)\left( {{{2}^{4}}} \right)}}{{{{2}^{{-1}}}\left( {{{2}^{4}}} \right)+{{2}^{{-2}}}\left( {{{2}^{4}}} \right)}}=\frac{1}{{{{2}^{3}}+{{2}^{2}}}}=\frac{1}{{12}}\end{align}\), \(\displaystyle \begin{align}{l}\sqrt[4]{{64{{a}^{7}}{{b}^{8}}}}&=\left( {\sqrt[4]{{64}}} \right)\sqrt[4]{{{{a}^{7}}{{b}^{8}}}}\\&=\left( {\sqrt[4]{{16}}} \right)\left( {\sqrt[4]{4}} \right)\left( {\sqrt[4]{{{{a}^{7}}}}} \right)\sqrt[4]{{{{b}^{8}}}}\\&=2\left( {\sqrt[4]{4}} \right){{a}^{1}}\sqrt[4]{{{{a}^{3}}}}{{b}^{2}}\\&=2a{{b}^{2}}\sqrt[4]{{4{{a}^{3}}}}\end{align}\), \(\displaystyle \begin{align}\sqrt[4]{{64{{a}^{7}}{{b}^{8}}}}&={{\left( {64{{a}^{7}}{{b}^{8}}} \right)}^{{\frac{1}{4}}}}\\&={{\left( {64} \right)}^{{\frac{1}{4}}}}{{\left( {{{a}^{7}}} \right)}^{{\frac{1}{4}}}}{{\left( {{{b}^{8}}} \right)}^{{\frac{1}{4}}}}\\&={{\left( {16} \right)}^{{\frac{1}{4}}}}{{\left( 4 \right)}^{{\frac{1}{4}}}}{{a}^{{\frac{7}{4}}}}{{b}^{{\frac{8}{4}}}}\\&=2{{\left( 4 \right)}^{{\frac{1}{4}}}}{{a}^{{\frac{4}{4}}}}{{a}^{{\frac{3}{4}}}}{{b}^{2}}\\&=2a{{b}^{2}}\sqrt[4]{{4{{a}^{3}}}}\end{align}\), \(\begin{align}6{{x}^{2}}\sqrt{{48{{y}^{2}}}}-4y\sqrt{{27{{x}^{4}}}}\\=6{{x}^{2}}y\sqrt{{16\cdot 3}}-4{{x}^{2}}y\sqrt{{9\cdot 3}}\\=6\cdot 4\cdot {{x}^{2}}y\sqrt{3}-3\cdot 4{{x}^{2}}y\sqrt{3}\\=24\sqrt{3}{{x}^{2}}y-12\sqrt{3}{{x}^{2}}y\\=12\sqrt{3}{{x}^{2}}y\end{align}\). Students simplify radical expressions that include variables with exponents in this activity. \(\begin{array}{c}{{\left( {\sqrt[3]{{x-3}}} \right)}^{3}}>{{4}^{3}}\,\,\,\,\\x-3>64\\x>67\end{array}\). Note also that if the negative were on the outside, like \(-{{8}^{{\frac{2}{3}}}}\), the answer would be â4. But, if we can have a negative \(a\), when we square it and then take the square root, it turns into positive again (since, by definition, taking the square root yields a positive). Free Radicals Calculator - Simplify radical expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. If we donât assume variables under the radicals are non-negative, we have to be careful with the signs and include absolute values for even radicals. In this example, we simplify â(60x²y)/â(48x). Radicals (which comes from the word ârootâ and means the same thing) means undoing the exponents, or finding out what numbers multiplied by themselves comes up with the number. To raise 8 to the \(\displaystyle \frac{2}{3}\), we can either do this in a calculator, or take the cube root of 8 and square it. For all these examples, see how weâre doing the same steps over and over again â just with different problems? Note that weâll see more radicals in the Solving Radical Equations and Inequalities section, and weâll talk about Factoring with Exponents, and Exponential Functions in the Exponential Functions section. From counting through calculus, making math make sense! Now, after simplifying the fraction, we have to simplify the radical. Converting repeating decimals in to fractions. And donât forget that there are many ways to arrive at the same answers! \(\displaystyle \sqrt[n]{{{{x}^{n}}}}=\,\left| x \right|\), \(\displaystyle \begin{array}{c}\sqrt[4]{{{{{\left( {\text{neg number }x} \right)}}^{4}}}}=\sqrt[4]{{\text{pos number }{{x}^{4}}}}\\=\text{positive }x=\left| x \right|\end{array}\), (If negative values are allowed under the radical sign, when we take an even root of a number raised to an even power, and the result is raised to an odd power (like 1), we have to use absolute value!!). Note that when we take the even root (like the square root) of both sides, we have to include the positive and the negative solutions of the roots. If you donât get them at first, donât worry; just try to go over them again. Free radical equation calculator - solve radical equations step-by-step. Therefore, in this case, \(\sqrt{{{{a}^{3}}}}=\left| a \right|\sqrt{a}\). Some examples: \(\displaystyle {{x}^{-2}}={{\left( \frac{1}{x} \right)}^{2}}\) and \(\displaystyle {{\left( \frac{y}{x} \right)}^{-4}}={{\left( \frac{x}{y} \right)}^{4}}\). Similarly, \(\displaystyle \sqrt{{{{b}^{2}}}}=\left| b \right|\). \(\displaystyle \begin{align}\frac{{34{{n}^{{2x+y}}}}}{{17{{n}^{{x-y}}}}}&=2{{n}^{{\left( {2x+y} \right)\,-\,\left( {x-y} \right)}}}\\&=2{{n}^{{2x-x+y-\left( {-y} \right)}}}=2{{n}^{{x+2y}}}\end{align}\), \(\displaystyle \begin{align}&\frac{{{{{\left( {4{{a}^{{-3}}}{{b}^{2}}} \right)}}^{{-2}}}{{{\left( {{{a}^{3}}{{b}^{{-1}}}} \right)}}^{3}}}}{{{{{\left( {-2{{a}^{{-3}}}} \right)}}^{2}}}}\\&=\frac{{{{{\left( {{{a}^{3}}{{b}^{{-1}}}} \right)}}^{3}}}}{{{{{\left( {4{{a}^{{-3}}}{{b}^{2}}} \right)}}^{2}}{{{\left( {-2{{a}^{{-3}}}} \right)}}^{2}}}}\\&=\frac{{{{a}^{9}}{{b}^{{-3}}}}}{{\left( {16{{a}^{{-6}}}{{b}^{4}}} \right)\left( {4{{a}^{{-6}}}} \right)}}=\frac{{{{a}^{9}}{{b}^{{-3}}}}}{{64{{a}^{{-12}}}{{b}^{4}}}}\\&=\frac{{{{a}^{{9-\left( {-12} \right)}}}}}{{64{{b}^{{4-\left( {-3} \right)}}}}}=\frac{{{{a}^{{21}}}}}{{64{{b}^{7}}}}\end{align}\). You can only do this if the. \(\displaystyle {{\left( {\frac{{{{2}^{{-1}}}+{{2}^{{-2}}}}}{{{{2}^{{-4}}}}}} \right)}^{{-1}}}\). You can see that we have two points of intersections; therefore, we have two solutions. By using this website, you agree to our Cookie Policy. Radical Expressions Session 2 . eval(ez_write_tag([[320,50],'shelovesmath_com-large-mobile-banner-2','ezslot_9',139,'0','0']));eval(ez_write_tag([[320,50],'shelovesmath_com-large-mobile-banner-2','ezslot_10',139,'0','1']));Note again that weâll see more problems like these, including how to use sign charts with solving radical inequalities here in the Solving Radical Equations and Inequalities section. With \(\sqrt[{}]{{45}}\), we factor. \(\displaystyle \begin{align}{{\left( {\frac{{{{a}^{9}}}}{{27}}} \right)}^{{-\frac{2}{3}}}}&=\,\,\,{{\left( {\frac{{27}}{{{{a}^{9}}}}} \right)}^{{\frac{2}{3}}}}=\frac{{{{{27}}^{{\frac{2}{3}}}}}}{{{{{\left( {{{a}^{9}}} \right)}}^{{\frac{2}{3}}}}}}=\frac{{{{{\left( {\sqrt[3]{{27}}} \right)}}^{2}}}}{{{{a}^{{\frac{{18}}{3}}}}}}\\&=\frac{{{{{\left( {\sqrt[3]{{27}}} \right)}}^{2}}}}{{{{a}^{6}}}}=\frac{{{{3}^{2}}}}{{{{a}^{6}}}}=\frac{9}{{{{a}^{6}}}}\end{align}\), Flip fraction first to get rid of negative exponent. Prime factorization of the radical on the âinsideâ or âoutsideâ 48x ) a. That âx2 x 2 = x x problem ) to the same way both numbers and variables, practice practice. Simplifies any radical expressions on that side ) too and roots with variables - Concept... variables constants. On how to approach drawing Pie Charts, and practice, practice, practice its denominator should simplified. 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