Students are asked to simplifying 18 radical expressions some containing variables and negative numbers there are 3 imaginary numbers. When you need to simplify a radical expression that has variables under the radical sign, first see if you can factor out a square. Then we applied the exponents, and then just multiplied across. You can then use the intersection feature to find the solution(s); the solution(s) will be what $$x$$ is at that point. $$\{\}\text{ }\,\,\text{ or }\emptyset$$. We can raise both sides to the same number. Simplify Some of the more complicated problems involve using Quadratics). Just a note that weâre only dealing with real numbers at this point; later weâll learn about imaginary numbers, where we can (sort of) take the square root of a negative number. $$\displaystyle \sqrt[n]{{\frac{x}{y}}}=\frac{{\sqrt[n]{x}}}{{\sqrt[n]{y}}}$$, $$\displaystyle \sqrt[3]{{\frac{{27}}{8}}}=\frac{{\sqrt[3]{{27}}}}{{\sqrt[3]{8}}}=\frac{3}{2}$$, $$\displaystyle \begin{array}{c}\sqrt[{}]{{{{{\left( {-4} \right)}}^{2}}}}=\sqrt{{16}}=4\\\sqrt[{}]{{{{{\left( 4 \right)}}^{2}}}}=\sqrt{{16}}=4\end{array}$$, \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}, Hereâs 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, $${{\left( {9{{x}^{3}}y} \right)}^{2}}={{9}^{2}}{{x}^{6}}{{y}^{2}}=81{{x}^{6}}{{y}^{2}}$$. This calculator will simplify fractions, polynomial, rational, radical, exponential, logarithmic, trigonometric, and hyperbolic expressions. Click on Submit (the blue arrow to the right of the problem) to see the answer. $$\displaystyle {{x}^{2}}=16;\,\,\,\,\,\,\,x=\pm 4$$, $${{\left( {xy} \right)}^{3}}={{x}^{3}}{{y}^{3}}$$, $$\displaystyle {{\left( {\frac{x}{y}} \right)}^{m}}=\frac{{{{x}^{m}}}}{{{{y}^{m}}}}$$, $$\displaystyle {{\left( {\frac{x}{y}} \right)}^{4}}=\frac{{{{x}^{4}}}}{{{{y}^{4}}}}$$, $$\displaystyle {{\left( {\frac{x}{y}} \right)}^{4}}=\frac{x}{y}\cdot \frac{x}{y}\cdot \frac{x}{y}\cdot \frac{x}{y}=\frac{{x\cdot x\cdot x\cdot x}}{{y\cdot y\cdot y\cdot y}}=\frac{{{{x}^{4}}}}{{{{y}^{4}}}}$$, $${{x}^{m}}\cdot {{x}^{n}}={{x}^{{m+n}}}$$, $${{x}^{4}}\cdot {{x}^{2}}={{x}^{{4+2}}}={{x}^{6}}$$, $$\require{cancel} \displaystyle \frac{{{{x}^{5}}}}{{{{x}^{3}}}}=\frac{{x\cdot x\cdot x\cdot x\cdot x}}{{x\cdot x\cdot x}}=\frac{{x\cdot x\cdot \cancel{x}\cdot \cancel{x}\cdot \cancel{x}}}{{\cancel{x}\cdot \cancel{x}\cdot \cancel{x}}}=x\cdot x={{x}^{2}}$$, $${{\left( {{{x}^{m}}} \right)}^{n}}={{x}^{{mn}}}$$, $${{\left( {{{x}^{4}}} \right)}^{2}}={{x}^{{4\cdot 2}}}={{x}^{8}}$$, $$\displaystyle 1=\frac{{{{x}^{5}}}}{{{{x}^{5}}}}={{x}^{{5-5}}}={{x}^{0}}$$, $$\displaystyle \frac{1}{{{{x}^{m}}}}={{x}^{{-m}}}$$, $$\displaystyle \frac{1}{{{{2}^{3}}}}=\frac{{{{2}^{0}}}}{{{{2}^{3}}}}={{2}^{{0-3}}}={{2}^{{-3}}}$$, $$\displaystyle \sqrt[n]{x}={{x}^{{\frac{1}{n}}}}$$. Move whatâs inside the negative exponent down first and make exponent positive. Solving linear equations using elimination method. Then get rid of parentheses first, by pushing the exponents through. Since the root is odd, we donât have to worry about the signs. Also, remember that when we take the square root, thereâs an invisible 2 in the radical, like this: $$\sqrt[2]{x}$$. We need to check our answer to make sure there are no negative numbers under the even radical and also still check the answers since we raised both sides to the 4th power:  $$\sqrt[4]{{13+3}}=\sqrt[4]{{16}}=2\,\,\,\,\,\,\surd$$, $$\displaystyle 4\sqrt{{x-1}}=\sqrt{{x+1}}$$, \displaystyle \begin{align}{{\left( {4\sqrt{{x-1}}} \right)}^{2}}&={{\left( {\sqrt{{x+1}}} \right)}^{2}}\\\,{{4}^{2}}\left( {x-1} \right)&=\left( {x+1} \right)\\16x-16&=x+1\\15x&=17;\,\,\,\,\,x=\frac{{17}}{{15}}\end{align}. (You may have to do this a few times). Here are some (difficult) examples. We will start with perhaps the simplest of all examples and then gradually move on to more complicated examples . Simplifying Radical Expressions with Variables Worksheet - Concept ... Variables and constants. $$x$$ isnât multiplied by anything, so itâs just $$x$$. To fix this, we multiply by a fraction with the bottom radical(s) on both the top and bottom (so the fraction equals 1); this way the bottom radical disappears. The reason we take the intersection of the two solutions is because both must work. If two terms are in the denominator, we need to multiply the top and bottom by a conjugate. Donât worry if you donât totally get this now! \displaystyle \begin{align}{{x}^{3}}&=27\\\,\sqrt[3]{{{{x}^{3}}}}&=\sqrt[3]{{27}}\\\,x&=3\end{align}. $$\displaystyle \begin{array}{c}{{\left( {\sqrt{{5x-16}}} \right)}^{2}}<{{\left( {\sqrt{{2x-4}}} \right)}^{2}}\\5x-16<2x-4\\3x<12\\x<4\\\text{also:}\\5x-16 \,\ge 0\text{ and 2}x-4 \,\ge 0\\x\ge \frac{{16}}{5}\text{ and }x\ge 2\\x<4\,\,\,\cap \,\,\,x\ge \frac{{16}}{5}\,\,\,\cap \,\,\,x\ge 2\\\{x:\,\,\frac{{16}}{5}\le x<4\}\text{ or }\left[ {\frac{{16}}{5},\,\,4} \right)\end{array}$$. Put it all together, combining the radical. Remember that the bottom of the fraction is what goes in the root, and we typically take the root first. We have to make sure our answers donât produce any negative numbers under the square root; this looks good. Then do the step above again with â2nd TRACEâ (CALC), 5, ENTER, ENTER, ENTER. Variables in a radical's argument are simplified in the same way as regular numbers. Also, if we have squared both sides (or raised both sides to an even exponent), we need to check our answers to see if they work. Here are some examples; these are pretty straightforward, since we know the sign of the values on both sides, so we can square both sides safely. 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). In this case, the index is two because it is a square root, which means we need two of a kind. (Try it yourself on a number line). When simplifying, you won't always have only numbers inside the radical; you'll also have to work with variables. Youâll get it! Now, after simplifying the fraction, we have to simplify the radical. ), \begin{align}2\sqrt[3]{x}&=\sqrt[3]{{x+7}}\\{{\left( {2\sqrt[3]{x}} \right)}^{3}}&={{\left( {\sqrt[3]{{x+7}}} \right)}^{3}}\\8x&=x+7\\7x&=7\\x&=1\end{align}. For all these examples, see how weâre doing the same steps over and over again â just with different problems? to know, but after a lot of practice, they become second nature. Remember that, for the variables, we can divide the exponents inside by the root index â if it goes in exactly, we can take the variable to the outside; if there are any remainders, we have to leave the variables under the root sign. Then we can solve for y by subtracting 2 from each side. If $$a$$ is positive, the square root of $${{a}^{3}}$$ is $$a\,\sqrt{a}$$, since 2 goes into 3 one time (so we can take one $$a$$ out), and thereâs 1 left over (to get the inside $$a$$). This is because both the positive root and negative roots work, when raised to that even power. The $$n$$th root of a base can be written as that base raised to the reciprocal of $$n$$, or $$\displaystyle \frac{1}{n}$$. $$\begin{array}{c}\sqrt{{x+2}}\le 4\text{ }\,\text{ }\,\text{and }x+2\,\,\ge \,\,0\\{{\left( {\sqrt{{x+2}}} \right)}^{2}}\le {{4}^{2}}\text{ }\,\,\text{and }x+2\,\,\ge \,\,0\text{ }\\x+2\le 16\text{ }\,\text{and }x\ge \,\,0-2\text{ }\\x\le 14\text{ }\,\text{and }x\ge -2\\\\\{x:-2\le x\le 14\}\text{ or }\left[ {-2,14} \right]\end{array}$$. Since we have the cube root on each side, we can simply cube each side. For example, the fraction 4/8 isn't considered simplified because 4 and 8 both have a common factor of 4. When radical expressions contain variables, simplifying them follows the same process as it does for expressions containing only integers. We have $$\sqrt{{{x}^{2}}}=x$$  (actually $$\sqrt{{{x}^{2}}}=\left| x \right|$$ since $$x$$ can be negative) since $$x\times x={{x}^{2}}$$. Putting Exponents and Radicals in the Calculator, $$\displaystyle \left( {6{{a}^{{-2}}}b} \right){{\left( {\frac{{2a{{b}^{3}}}}{{4{{a}^{3}}}}} \right)}^{2}}$$, $$\displaystyle \frac{{{{{\left( {4{{a}^{{-3}}}{{b}^{2}}} \right)}}^{{-2}}}{{{\left( {{{a}^{3}}{{b}^{{-1}}}} \right)}}^{3}}}}{{{{{\left( {-2{{a}^{{-3}}}} \right)}}^{2}}}}$$, $${{\left( {-8} \right)}^{{\frac{2}{3}}}}$$, $$\displaystyle {{\left( {\frac{{{{a}^{9}}}}{{27}}} \right)}^{{-\frac{2}{3}}}}$$, With $${{64}^{{\frac{1}{4}}}}$$, we factor it into, $$6{{x}^{2}}\sqrt{{48{{y}^{2}}}}-4y\sqrt{{27{{x}^{4}}}}$$, $$\displaystyle \sqrt[4]{{\frac{{{{x}^{6}}{{y}^{4}}}}{{162{{z}^{5}}}}}}$$, $${{\left( {y+2} \right)}^{{\frac{3}{2}}}}=8\,\,\,$$, $$4\sqrt[3]{x}=2\sqrt[3]{{x+7}}\,\,\,\,$$, $$\displaystyle {{\left( {x+2} \right)}^{{\frac{4}{3}}}}+2=18$$, $$\displaystyle \sqrt{{5x-16}}<\sqrt{{2x-4}}$$, Introducing Exponents and Radicals (Roots) with Variables, $${{x}^{m}}=x\cdot x\cdot x\cdot xâ¦.. (m\, \text{times})$$, $$\displaystyle \sqrt[{m\text{ }}]{x}=y$$  means  $$\displaystyle {{y}^{m}}=x$$, $$\sqrt[3]{8}=2$$,  since $$2\cdot 2\cdot 2={{2}^{3}}=8$$, $$\displaystyle {{x}^{{\frac{m}{n}}}}={{\left( {\sqrt[n]{x}} \right)}^{m}}=\,\sqrt[n]{{{{x}^{m}}}}$$, $$\displaystyle {{x}^{{\frac{2}{3}}}}=\,\sqrt[3]{{{{8}^{2}}}}={{\left( {\sqrt[3]{8}} \right)}^{2}}={{2}^{2}}=4$$. When radicals (square roots) include variables, they are still simplified the same way. You can also type in your own problem, or click on the three dots in the upper right hand corner and click on âExamplesâ to drill down by topic. 8 } ^ { 2 } } } =\left| b \right|\ ) two points intersections. } \text { or } \emptyset \ ) in place ; this looks.... When weâre dealing with even roots and variables ) by taking out squares when raised that... Have fractional exponents and roots, we kept the positive root and variables to right! 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