Some examples: $$\displaystyle {{x}^{-2}}={{\left( \frac{1}{x} \right)}^{2}}$$  and $$\displaystyle {{\left( \frac{y}{x} \right)}^{-4}}={{\left( \frac{x}{y} \right)}^{4}}$$. Solving linear equations using elimination method. 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. 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. Therefore, in this case, $$\sqrt{{{{a}^{3}}}}=\left| a \right|\sqrt{a}$$. Note also that if the negative were on the outside, like $$-{{8}^{{\frac{2}{3}}}}$$, the answer would be â4. You should see the second solution at $$x=-10$$. Then we can solve for x. Letâs check our answer:  $$2\sqrt{{25+2}}=2(3)=6\,\,\,\,\,\,\surd$$, \begin{align}{{\left( {{{{\left( {y+2} \right)}}^{{\frac{3}{2}}}}} \right)}^{{\frac{2}{3}}}}&={{8}^{{\frac{2}{3}}}}\\{{\left( {y+2} \right)}^{{\frac{3}{2}\times \frac{2}{3}}}}&={{8}^{{\frac{2}{3}}}}\\y+2&={{\left( {\sqrt{8}} \right)}^{2}}={{2}^{2}}\\y+2&=4\\y&=2\end{align}. 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, $$6{{x}^{2}}\sqrt{{48{{y}^{2}}}}-4y\sqrt{{27{{x}^{4}}}}$$, $$\displaystyle \sqrt{{\frac{{{{x}^{6}}{{y}^{4}}}}{{162{{z}^{5}}}}}}$$, $${{\left( {y+2} \right)}^{{\frac{3}{2}}}}=8\,\,\,$$, $$4\sqrt{x}=2\sqrt{{x+7}}\,\,\,\,$$, $$\displaystyle {{\left( {x+2} \right)}^{{\frac{4}{3}}}}+2=18$$, $$\displaystyle \sqrt{{5x-16}}<\sqrt{{2x-4}}$$, 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  $$\displaystyle {{y}^{m}}=x$$, $$\sqrt{8}=2$$,  since $$2\cdot 2\cdot 2={{2}^{3}}=8$$, $$\displaystyle {{x}^{{\frac{m}{n}}}}={{\left( {\sqrt[n]{x}} \right)}^{m}}=\,\sqrt[n]{{{{x}^{m}}}}$$, $$\displaystyle {{x}^{{\frac{2}{3}}}}=\,\sqrt{{{{8}^{2}}}}={{\left( {\sqrt{8}} \right)}^{2}}={{2}^{2}}=4$$. Itâs always easier to simply (for example. The trick is to get rid of the exponents, we need to take radicals of both sides, and to get rid of radicals, we need to raise both sides of the equation to that power. (You can also use the WINDOW button to change the minimum and maximum values of your x and y values.). With MATH 5 (nth root), select the root first, then MATH 5, then whatâs under the radical. Just remember that you have to be really, really careful doing these! Once a To simplify a numerical fraction, I would cancel off any common numerical factors. Eliminate the parentheses with the squared first. To get rid of the radical on the left-hand side, we can cube both sides. See how we could have just divided the exponents inside by the root outside, to end up with the rational (fractional) exponent (sort of like turning improper fractions into mixed fractions in the exponents): $$\sqrt{{{{x}^{5}}{{y}^{{12}}}}}={{x}^{{\frac{5}{3}}}}{{y}^{{\frac{{12}}{3}}}}={{x}^{{\frac{3}{3}}}}{{x}^{{\frac{2}{3}}}}{{y}^{4}}=x\cdot {{x}^{{\frac{2}{3}}}}{{y}^{4}}=x{{y}^{4}}\sqrt{{{{x}^{2}}}}$$? If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!). Since we have the cube root on each side, we can simply cube each side. Remember that when we cube a cube root, we end up with whatâs under the root sign. It gets trickier when we donât know the sign of one of the sides. Simplify This website uses cookies to ensure you get the best experience. Simplify the roots (both numbers and variables) by taking out squares. (You may have to do this a few times). In this example, we simplify â(60x²y)/â(48x). Move all the constants (numbers) to the right. This will give us $$\displaystyle \frac{{16}}{5}\le \,\,x<4$$. To do this, weâll set whatâs under the even radical to greater than or equal to 0, solve for $$x$$. Parentheses are optional around exponents. In algebra, weâll need to know these and many other basic rules on how to handle exponents and roots when we work with them. Unless otherwise indicated, assume numbers under radicals with even roots are positive, and numbers in denominators are nonzero. $$\displaystyle \sqrt[n]{{{{x}^{n}}}}=\,\left| x \right|$$, $$\displaystyle \begin{array}{c}\sqrt{{{{{\left( {\text{neg number }x} \right)}}^{4}}}}=\sqrt{{\text{pos number }{{x}^{4}}}}\\=\text{positive }x=\left| x \right|\end{array}$$, (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!!). Example 1: Add or subtract to simplify radical expression: $2 \sqrt{12} + \sqrt{27}$ Solution: Step 1: Simplify radicals \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}. 1) Factor the radicand (the numbers/variables inside the square root). Then get rid of parentheses first, by pushing the exponents through. Free Radicals Calculator - Simplify radical expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. 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. For this rational expression (this polynomial fraction), I can similarly cancel off any common numerical or variable factors. With odd roots, we donât have to worry about checking underneath the radical sign, since we could have positive or negative numbers as a radicand. We need to take the intersection (all must work) of the inequalities: $$\displaystyle x<4\text{ and }x\ge \frac{{16}}{5}\text{ and }x\ge 2$$. This is accomplished by multiplying the expression by a fraction having the value 1, in an appropriate form. The numerator factors as (2)(x); the denominator factors as (x)(x). $$x$$ isnât multiplied by anything, so itâs just $$x$$. Also, all the answers we get may not work, since we canât take the even roots of negative numbers. There are rules that you need to follow when simplifying â¦ Students simplify radical expressions that include variables with exponents in this activity. The basic ideas are very similar to simplifying numerical fractions. ), \begin{align}2\sqrt{x}&=\sqrt{{x+7}}\\{{\left( {2\sqrt{x}} \right)}^{3}}&={{\left( {\sqrt{{x+7}}} \right)}^{3}}\\8x&=x+7\\7x&=7\\x&=1\end{align}. Here are those instructions again, using an example from above: Push GRAPH. $$\displaystyle \frac{{34{{n}^{{2x+y}}}}}{{17{{n}^{{x-y}}}}}$$. We will start with perhaps the simplest of all examples and then gradually move on to more complicated examples . Youâll see the first point of intersection that it found is where $$x=6$$. When radicals (square roots) include variables, they are still simplified the same way. Donât worry if you donât totally get this now! We canât take the even root of a negative number and get a real number. The âexact valueâ would be the answer with the root sign in it! By using this website, you agree to our Cookie Policy. For example, $$\sqrt{{{{x}^{5}}{{y}^{{12}}}}}={{x}^{1}}{{y}^{4}}\sqrt{{{{x}^{2}}}}=x{{y}^{4}}\sqrt{{{{x}^{2}}}}$$, since 5 divided by 3 is 1, with 2 left over (for the $$x$$), and 12 divided by 3 is 4 (for the $$y$$). 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 just solve for x, just like we would for an equation. Simplifying radicals with variables is a bit different than when the radical terms contain just numbers. When raising a radical to an exponent, the exponent can be on the âinsideâ or âoutsideâ. By using this website, you agree to our Cookie Policy. eval(ez_write_tag([[320,50],'shelovesmath_com-large-mobile-banner-1','ezslot_7',117,'0','0']));eval(ez_write_tag([[320,50],'shelovesmath_com-large-mobile-banner-1','ezslot_8',117,'0','1']));Again, when the original problem contains an even root sign, we need to check our answers to make sure we have end up with no negative numbers under the even root sign (no negative radicands). Also remember that we donât need the parentheses around the exponent in the newer calculator operating systems (but it wonât hurt to have them). $$\begin{array}{c}\sqrt{{x+2}}\le 4\text{ }\,\text{ }\,\text{and }x+2\,\,\ge \,\,0\\{{\left( {\sqrt{{x+2}}} \right)}^{2}}\le {{4}^{2}}\text{ }\,\,\text{and }x+2\,\,\ge \,\,0\text{ }\\x+2\le 16\text{ }\,\text{and }x\ge \,\,0-2\text{ }\\x\le 14\text{ }\,\text{and }x\ge -2\\\\\{x:-2\le x\le 14\}\text{ or }\left[ {-2,14} \right]\end{array}$$. We have to make sure we square the, We correctly solved the equation but notice that when we plug in. $$\begin{array}{c}{{\left( {\sqrt{{x-3}}} \right)}^{3}}>{{4}^{3}}\,\,\,\,\\x-3>64\\x>67\end{array}$$. 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. 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¦ Probably the simplest case is that âx2 x 2 = x x. Add and Subtract Fractions with Variables. Weâll do this pretty much the same way, but again, we need to be careful with multiplying and dividing by anything negative, where we have to change the direction of the inequality sign. Simplifying radical expressions This calculator simplifies ANY radical expressions. 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 $$a$$). Then we can solve for x. Letâs check our answer:   \begin{align}4\sqrt{1}&=2\sqrt{{1+7}}\,\,\,\,\,?\\4\,\,&=\,\,4\,\,\,\,\,\,\surd \end{align}, Now letâs solve equations with even roots. Before we work example, letâs talk about rationalizing radical fractions. To find the other point of intersection, we need to move the cursor closer to that point, so press âTRACEâ and move the cursor closer to the other point of intersection (it should follow along one of the curves). 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. Simplifying Radical Expressions with Variables Worksheet - Concept ... Variables and constants. There are five main things youâll have to do to simplify exponents and radicals. Then we solve for $${{y}_{2}}$$. Then subtract up or down (starting where the exponents are larger) to turn the negative exponents positive. Remember that when we end up with exponential âimproper fractionsâ (numerator > denominator), we can separate the exponents (almost like âmixed fractionsâ) and the move the variables with integer exponents to the outside (see work). ], Convert Decimal To Fraction [ Def: A number that names a part of a whole or a part of a group. Letâs first try some equations with odd exponents and roots, since these are a little more straightforward. You can only do this if the. This oneâs pretty complicated since we have to, $$\begin{array}{l}{{32}^{{\tfrac{3}{5}}}}\cdot {{81}^{{\tfrac{1}{4}}}}\cdot {{27}^{{-\tfrac{1}{3}}}}&={{\left( 2 \right)}^{3}}\cdot {{\left( 3 \right)}^{1}}\cdot {{\left( 3 \right)}^{{-1}}}\\&=8\cdot 3\cdot \tfrac{1}{3}=8\end{array}$$. When simplifying, you won't always have only numbers inside the radical; you'll also have to work with variables. ... Variables and constants. We could have also just put this one in the calculator (using parentheses around the fractional roots). Then do the step above again with â2nd TRACEâ (CALC), 5, ENTER, ENTER, ENTER. We also need to try numbers outside our solution (like $$x=-6$$ and $$x=20$$) and see that they donât work. Example 1 Add the fractions: $$\dfrac{2}{x} + \dfrac{3}{5}$$ Solution to Example 1 Be careful though, because if thereâs not a perfect square root, the calculator will give you a long decimal number thatâs not the âexact valueâ. With $$\sqrt{{64}}$$, we factor 64 into 16 and 4, since $$\displaystyle \sqrt{{16}}=2$$. Notice that when we moved the $$\pm$$ to the other side, itâs still a $$\pm$$. Displaying top 8 worksheets found for - Simplifying Radicals With Fractions. Learn these rules, and practice, practice, practice! We have $$\sqrt{{{x}^{2}}}=x$$  (actually $$\sqrt{{{x}^{2}}}=\left| x \right|$$ since $$x$$ can be negative) since $$x\times x={{x}^{2}}$$. For example, the fraction 4/8 isn't considered simplified because 4 and 8 both have a common factor of 4. Some of the more complicated problems involve using Quadratics). $$\displaystyle {{x}^{2}}=16;\,\,\,\,\,\,\,x=\pm 4$$, $${{\left( {xy} \right)}^{3}}={{x}^{3}}{{y}^{3}}$$, $$\displaystyle {{\left( {\frac{x}{y}} \right)}^{m}}=\frac{{{{x}^{m}}}}{{{{y}^{m}}}}$$, $$\displaystyle {{\left( {\frac{x}{y}} \right)}^{4}}=\frac{{{{x}^{4}}}}{{{{y}^{4}}}}$$, $$\displaystyle {{\left( {\frac{x}{y}} \right)}^{4}}=\frac{x}{y}\cdot \frac{x}{y}\cdot \frac{x}{y}\cdot \frac{x}{y}=\frac{{x\cdot x\cdot x\cdot x}}{{y\cdot y\cdot y\cdot y}}=\frac{{{{x}^{4}}}}{{{{y}^{4}}}}$$, $${{x}^{m}}\cdot {{x}^{n}}={{x}^{{m+n}}}$$, $${{x}^{4}}\cdot {{x}^{2}}={{x}^{{4+2}}}={{x}^{6}}$$, $$\require{cancel} \displaystyle \frac{{{{x}^{5}}}}{{{{x}^{3}}}}=\frac{{x\cdot x\cdot x\cdot x\cdot x}}{{x\cdot x\cdot x}}=\frac{{x\cdot x\cdot \cancel{x}\cdot \cancel{x}\cdot \cancel{x}}}{{\cancel{x}\cdot \cancel{x}\cdot \cancel{x}}}=x\cdot x={{x}^{2}}$$, $${{\left( {{{x}^{m}}} \right)}^{n}}={{x}^{{mn}}}$$, $${{\left( {{{x}^{4}}} \right)}^{2}}={{x}^{{4\cdot 2}}}={{x}^{8}}$$, $$\displaystyle 1=\frac{{{{x}^{5}}}}{{{{x}^{5}}}}={{x}^{{5-5}}}={{x}^{0}}$$, $$\displaystyle \frac{1}{{{{x}^{m}}}}={{x}^{{-m}}}$$, $$\displaystyle \frac{1}{{{{2}^{3}}}}=\frac{{{{2}^{0}}}}{{{{2}^{3}}}}={{2}^{{0-3}}}={{2}^{{-3}}}$$, $$\displaystyle \sqrt[n]{x}={{x}^{{\frac{1}{n}}}}$$. 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. Factor the expression completely (or find perfect squares). When you need to simplify a radical expression that has variables under the radical sign, first see if you can factor out a square. \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}. 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