There are five main things youâll have to do to simplify exponents and radicals. to know, but after a lot of practice, they become second nature. 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. $$\begin{array}{c}\sqrt[{\text{odd} }]{{{{x}^{{\text{odd}}}}}}=x\\\sqrt[{\text{even} }]{{{{x}^{{\text{even}}}}}}=\left| {\,x\,} \right|\end{array}$$, $$\begin{array}{c}\sqrt{{{{{\left( {-2} \right)}}^{3}}}}=\sqrt{{-8}}=-2\\\sqrt{{{{{\left( {-2} \right)}}^{2}}}}=\sqrt{4}=2\end{array}$$. $${{\left( {-8} \right)}^{{\frac{2}{3}}}}={{\left( {\sqrt{{-8}}} \right)}^{2}}={{\left( {-2} \right)}^{2}}=4$$. We can âundoâ the fourth root by raising both sides to the forth. Note that this works when $$n$$ is even too, if  $$x\ge 0$$. We could have also just put this one in the calculator (using parentheses around the fractional roots). To simplify a numerical fraction, I would cancel off any common numerical factors. Each root had a âperfectâ answer, so we took the roots first. $$\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,=\frac{{5{{{(\sqrt{3})}}^{3}}}}{{2{{{(\sqrt{3})}}^{4}}}}=\frac{{5{{{(\sqrt{3})}}^{3}}}}{{2\cdot 3}}=\frac{{5{{{(\sqrt{3})}}^{3}}}}{6}$$. We can check our answer by trying random numbers in our solution (like $$x=2$$) in the original inequality (which works). Combine like radicals. $$\displaystyle {{\left( {\frac{{{{2}^{{-1}}}+{{2}^{{-2}}}}}{{{{2}^{{-4}}}}}} \right)}^{{-1}}}$$. We have to make sure we square the, We correctly solved the equation but notice that when we plug in. $$\begin{array}{c}\sqrt{{x+2}}\le 4\text{ }\,\text{ }\,\text{and }x+2\,\,\ge \,\,0\\{{\left( {\sqrt{{x+2}}} \right)}^{2}}\le {{4}^{2}}\text{ }\,\,\text{and }x+2\,\,\ge \,\,0\text{ }\\x+2\le 16\text{ }\,\text{and }x\ge \,\,0-2\text{ }\\x\le 14\text{ }\,\text{and }x\ge -2\\\\\{x:-2\le x\le 14\}\text{ or }\left[ {-2,14} \right]\end{array}$$. \begin{align}{{9}^{{x-2}}}\cdot {{3}^{{x-1}}}&={{\left( {{{3}^{2}}} \right)}^{{x-2}}}\cdot {{3}^{{x-1}}}\\&={{3}^{{2(x-2)}}}\cdot {{3}^{{x-1}}}={{3}^{{2x-4}}}\cdot {{3}^{{x-1}}}\\&={{3}^{{2x-4+x-1}}}={{3}^{{3x-5}}}\end{align}, \displaystyle \begin{align}\sqrt[{}]{{45{{a}^{3}}{{b}^{2}}}}&=\left( {\sqrt[{}]{{45}}} \right)\sqrt[{}]{{{{a}^{3}}{{b}^{2}}}}\\&=\left( {\sqrt[{}]{9}} \right)\left( {\sqrt[{}]{5}} \right)\left( {\sqrt[{}]{{{{a}^{3}}}}} \right)\sqrt[{}]{{{{b}^{2}}}}\\&=3\left( {\sqrt[{}]{5}} \right)\left( {\sqrt[{}]{{{{a}^{2}}}}} \right)\left( {\sqrt[{}]{a}} \right)\sqrt[{}]{{{{b}^{2}}}}\\&=3\left( {\sqrt[{}]{5}} \right)\left| a \right|\cdot \sqrt{a}\cdot \left| b \right|\\&=3\left| a \right|\left| b \right|\left( {\sqrt[{}]{{5a}}} \right)\end{align}, Separate the numbers and variables. Letâs check our answer:  $${{\left( {2+2} \right)}^{{\frac{3}{2}}}}={{\left( 4 \right)}^{{\frac{3}{2}}}}={{\left( {\sqrt{4}} \right)}^{3}}={{2}^{3}}=8\,\,\,\,\,\,\surd$$, (Notice in this case, that we have to make sure  is positive since we are taking an even root, but when we work the problem, we can be assured it is, since we are squaring the right-hand side. Here are those instructions again, using an example from above: Push GRAPH. Radicals (which comes from the word ârootâ and means the same thing) means undoing the exponents, or finding out what numbers multiplied by themselves comes up with the number. Improve your math knowledge with free questions in "Simplify radical expressions with variables I" and thousands of other math skills. You can see that we have two points of intersections; therefore, we have two solutions. The $$n$$th root of a base can be written as that base raised to the reciprocal of $$n$$, or $$\displaystyle \frac{1}{n}$$. Factor the number into its prime factors and expand the variable â¦ Here are even more examples. Letâs first try some equations with odd exponents and roots, since these are a little more straightforward. The reason we take the intersection of the two solutions is because both must work. When we solve for variables with even exponents, we most likely will get multiple solutions, since when we square positive or negative numbers, we get positive numbers. Just like we had to solve linear inequalities, we also have to learn how to solve inequalities that involve exponents and radicals (roots). We need to take the intersection (all must work) of the inequalities: $$\displaystyle x<4\text{ and }x\ge \frac{{16}}{5}\text{ and }x\ge 2$$. In this example, we simplify 3â(500x³). As you become more familiar with dividing and simplifying radical expressions, make sure you continue to pay attention to the roots of the radicals that you are dividing. Since we have the cube root on each side, we can simply cube each side. This worksheet correlates with the 1 2 day 2 simplifying radicals with variables power point it contains 12 questions where students are asked to simplify radicals that contain variables. The numerator factors as (2)(x); the denominator factors as (x)(x). Radical Expressions Session 2 . (Note that we could have also raised each side to the $$\displaystyle \frac{1}{3}$$ power.) Decimal representation of rational numbers. Also, all the answers we get may not work, since we canât take the even roots of negative numbers. 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. \displaystyle \begin{align}\sqrt{{\frac{{{{x}^{6}}{{y}^{4}}}}{{162{{z}^{5}}}}}}&=\frac{{\sqrt{{{{x}^{6}}{{y}^{4}}}}}}{{\sqrt{{\left( {81} \right)\left( 2 \right){{z}^{5}}}}}}=\frac{{xy\sqrt{{{{x}^{2}}}}}}{{3z\sqrt{{2z}}}}\\&=\frac{{xy\sqrt{{{{x}^{2}}}}}}{{3z\sqrt{{2z}}}}\cdot \frac{{\sqrt{{{{{\left( {2z} \right)}}^{3}}}}}}{{\sqrt{{{{{\left( {2z} \right)}}^{3}}}}}}\\&=\frac{{xy\sqrt{{{{x}^{2}}}}\sqrt{{8{{z}^{3}}}}}}{{3z\sqrt{{{{{\left( {2z} \right)}}^{4}}}}}}=\frac{{xy\sqrt{{8{{x}^{2}}{{z}^{3}}}}}}{{3z\left( {2z} \right)}}\\&=\frac{{xy\sqrt{{8{{x}^{2}}{{z}^{3}}}}}}{{6{{z}^{2}}}}\end{align}. A worked example of simplifying elaborate expressions that contain radicals with two variables. In math, sometimes we have to worry about âproper grammarâ. We can raise both sides to the same number. Since a negative number times a negative number is always a positive number, you need to remember when taking a square root that the answer will be both a positive and a negative number or â¦ When radical expressions contain variables, simplifying them follows the same process as it does for expressions containing only integers. Using a TI30 XS Multiview Calculator, here are the steps: Notice that when we place a negative on the outside of the 8, it performs the radicals first (cube root of 8, and then squared) and then puts the negative in front of it. Remember that $${{a}^{0}}=1$$. Combination Formula, Combinations without Repetition. Writing and evaluating expressions. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. $${{32}^{{\tfrac{3}{5}}}}\cdot {{81}^{{\tfrac{1}{4}}}}\cdot {{27}^{{-\tfrac{1}{3}}}}$$. Simplifying radicals with variables is a bit different than when the radical terms contain just numbers. Now that we know about exponents and roots with variables, we can solve equations that involve them. This process is called rationalizing the denominator. $$\displaystyle \sqrt[n]{{{{x}^{n}}}}=\,\left| x \right|$$, $$\displaystyle \begin{array}{c}\sqrt{{{{{\left( {\text{neg number }x} \right)}}^{4}}}}=\sqrt{{\text{pos number }{{x}^{4}}}}\\=\text{positive }x=\left| x \right|\end{array}$$, (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!!). Then we can solve for y by subtracting 2 from each side. 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). Here are some (difficult) examples. 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. Then, to rationalize, since we have a 4th root, we can multiply by a radical that has the 3rd root on top and bottom. We canât take the even root of a negative number and get a real number. $$\begin{array}{c}{{x}^{2}}=-4\\\emptyset \text{ or no solution}\end{array}$$, $$\begin{array}{c}{{x}^{2}}=25\\x=\pm 5\end{array}$$, We need to check our answers:    $${{\left( 5 \right)}^{2}}-1=24\,\,\,\,\surd \,\,\,\,\,\,\,\,{{\left( {-5} \right)}^{2}}-1=24\,\,\,\,\surd$$, $$\begin{array}{c}{{\left( {\sqrt{{x+3}}} \right)}^{4}}={{2}^{4}}\\x+3=16\\x=13\end{array}$$. Perfect square roots on both sides, we can put it all together combining. Introduction to multiplying Polynomials â you are ready x x make it positive expression completely ( or find squares. 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