# simplifying radical fractions with variables

Be careful to make sure you cube all the numbers (and anything else on that side) too. ], 5 examples of poems in mathematics, free learning maths for year 11, equations to decimal calculators, general aptitude questions ( Example: ) the square root cheaters map, word problems about coin problem with exaples, radical â¦ We can put exponents and radicals in the graphing calculator, using the carrot sign (^) to raise a number to something else, the square root button to take the square root, or the MATH button to get the cube root or $$n$$th root. Keep this in mind: ... followed by multiplying the outer most numbers/variables, ... To simplify this expression, I would start by simplifying the radical on the numerator. Writing and evaluating expressions. $$\displaystyle \frac{{34{{n}^{{2x+y}}}}}{{17{{n}^{{x-y}}}}}$$. Therefore, in this case, $$\sqrt{{{{a}^{3}}}}=\left| a \right|\sqrt{a}$$. 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. 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. Here are those instructions again, using an example from above: Push GRAPH. Writing and evaluating expressions. We can also use the MATH function to take the cube root (4, or scroll down) or nth root (5:). 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!). Now, after simplifying the fraction, we have to simplify the radical. \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}. 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. ], Convert Decimal To Fraction [ Def: A number that names a part of a whole or a part of a group. \displaystyle \begin{align}\left( {6{{a}^{{-2}}}b} \right){{\left( {\frac{{2a{{b}^{3}}}}{{4{{a}^{3}}}}} \right)}^{2}}&=6{{a}^{{-2}}}b\cdot \frac{{4{{a}^{2}}{{b}^{6}}}}{{16{{a}^{6}}}}\\&=\frac{{24{{a}^{0}}{{b}^{7}}}}{{16{{a}^{6}}}}=\frac{{3{{b}^{7}}}}{{2{{a}^{6}}}}\end{align}. 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 weâll see more problems like these, including how to use sign charts with solving radical inequalities here in the Solving Radical Equations and Inequalities section. Here are the rules/properties with explanations and examples. $$\displaystyle \begin{array}{c}{{\left( {\sqrt{{5x-16}}} \right)}^{2}}<{{\left( {\sqrt{{2x-4}}} \right)}^{2}}\\5x-16<2x-4\\3x<12\\x<4\\\text{also:}\\5x-16 \,\ge 0\text{ and 2}x-4 \,\ge 0\\x\ge \frac{{16}}{5}\text{ and }x\ge 2\\x<4\,\,\,\cap \,\,\,x\ge \frac{{16}}{5}\,\,\,\cap \,\,\,x\ge 2\\\{x:\,\,\frac{{16}}{5}\le x<4\}\text{ or }\left[ {\frac{{16}}{5},\,\,4} \right)\end{array}$$. \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. A root âundoesâ raising a number to that exponent. We have $$\sqrt{{{x}^{2}}}=x$$  (actually $$\sqrt{{{x}^{2}}}=\left| x \right|$$ since $$x$$ can be negative) since $$x\times x={{x}^{2}}$$. $$\displaystyle \frac{1}{{{{3}^{2}}}}={{3}^{{-2}}}=\frac{1}{9}$$, $$\displaystyle {{\left( {\frac{2}{3}} \right)}^{{-2}}}={{\left( {\frac{3}{2}} \right)}^{2}}=\frac{9}{4}$$, $$\displaystyle {{\left( {\frac{x}{y}} \right)}^{{-m}}}=\frac{{{{x}^{{-m}}}}}{{{{y}^{{-m}}}}}=\frac{{\frac{1}{{{{x}^{m}}}}}}{{\frac{1}{{{{y}^{m}}}}}}=\frac{1}{{{{x}^{m}}}}\times \frac{{{{y}^{m}}}}{1}=\,{{\left( {\frac{y}{x}} \right)}^{m}}$$, $$\displaystyle \sqrt{8}={{8}^{{\frac{1}{3}}}}=2$$, $$\sqrt[n]{{xy}}=\sqrt[n]{x}\cdot \sqrt[n]{y}$$, $$\displaystyle \begin{array}{l}\sqrt{{72}}=\sqrt{{4\cdot 9\cdot 2}}=\sqrt{4}\cdot \sqrt{9}\cdot \sqrt{2}\\\,\,\,\,\,\,\,\,\,\,\,\,=2\cdot 3\cdot \sqrt{2}=6\sqrt{2}\end{array}$$, ($$\sqrt{{xy}}={{(xy)}^{{\frac{1}{2}}}}={{x}^{{\frac{1}{2}}}}\cdot {{y}^{{\frac{1}{2}}}}=\sqrt{x}\cdot \sqrt{y}$$, (Doesnât work for imaginary numbers under radicals. \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{{27}}} \right)}}^{2}}}}{{{{a}^{{\frac{{18}}{3}}}}}}\\&=\frac{{{{{\left( {\sqrt{{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. By using this website, you agree to our Cookie Policy. With MATH 5 (nth root), select the root first, then MATH 5, then whatâs under the radical. If we donât assume variables under the radicals are non-negative, we have to be careful with the signs and include absolute values for even radicals. Then we can put it all together, combining the radical. Get variable out of exponent, percent equations, how to multiply radical fractions, free worksheets midpoint formula. Since we have to get $${{y}_{2}}$$ by itself, we first have to take the square root of each side (and donât forget to take the plus and the minus). 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. We could also put this in our calculator! We also must make sure our answer takes into account what we call the domain restriction: we must make sure whatâs under an even radical is 0 or positive, so we may have to create another inequality. Here are some (difficult) examples. 1) Factor the radicand (the numbers/variables inside the square root). (Try it yourself on a number line). Watch out for the hard and soft brackets. It gets trickier when we donât know the sign of one of the sides. With $$\sqrt{{64}}$$, we factor 64 into 16 and 4, since $$\displaystyle \sqrt{{16}}=2$$. Since we have the cube root on each side, we can simply cube each side. Some expressions are fractions with and without perfect square roots. I also used âZOOM 3â (Zoom Out) ENTER to see the intersections a little better. $$\displaystyle {{x}^{{-m}}}=\,\frac{1}{{{{x}^{m}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{{{{x}^{{-m}}}}}={{x}^{m}}$$, $$\displaystyle {{\left( {\frac{x}{y}} \right)}^{{-m}}}=\,{{\left( {\frac{y}{x}} \right)}^{m}}$$, $$a\sqrt[{}]{x}\times b\sqrt[{}]{y}=ab\sqrt[{}]{{xy}}$$, (Doesnât work for imaginary numbers under radicals), $$2\sqrt{3}\times \,4\sqrt{5}\,=\,8\sqrt{{15}}$$. $$\sqrt[{\text{even} }]{{\text{negative number}}}\,$$ exists for imaginary numbers, but not for real numbers. 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. With $$\sqrt[{}]{{45}}$$, we factor. If two terms are in the denominator, we need to multiply the top and bottom by a conjugate . 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 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. Then we solve for $${{y}_{2}}$$. This calculator will simplify fractions, polynomial, rational, radical, exponential, logarithmic, trigonometric, and hyperbolic expressions. Assume variables under radicals are non-negative. 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 $$2\cdot 2\cdot 2={{2}^{3}}=8$$, $$\displaystyle {{x}^{{\frac{m}{n}}}}={{\left( {\sqrt[n]{x}} \right)}^{m}}=\,\sqrt[n]{{{{x}^{m}}}}$$, $$\displaystyle {{x}^{{\frac{2}{3}}}}=\,\sqrt{{{{8}^{2}}}}={{\left( {\sqrt{8}} \right)}^{2}}={{2}^{2}}=4$$. We can check our answer by trying random numbers in our solution (like $$x=2$$) in the original inequality (which works). We could have turned the roots into fractional exponents and gotten the same answer â itâs a matter of preference. 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. Simplify the roots (both numbers and variables) by taking out squares. Letâs check our answer:  $${{3}^{3}}-1=27-1=26\,\,\,\,\,\,\surd$$, \displaystyle \begin{align}\sqrt{{x+2}}&=3\\{{\left( {\sqrt{{x+2}}} \right)}^{3}}&={{3}^{3}}\\x+2&=27\\x&=25\end{align}. This shows us that we must plug in our answer when weâre dealing with even roots! In the âproofâ column, youâll notice that weâre using many of the algebraic properties that we learned in the Types of Numbers and Algebraic Properties section, such as the Associate and Commutative properties. Now letâs put it altogether. 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. You should see the second solution at $$x=-10$$. Again, weâll see more of these types of problems in the Solving Radical Equations and Inequalities section. ... Word problems on fractions. You may need to hit âZOOM 6â (ZStandard) and/or âZOOM 0â (ZoomFit) to make sure you see the lines crossing in the graph. We keep moving variables around until we have $${{y}_{2}}$$ on one side. Just like we had to solve linear inequalities, we also have to learn how to solve inequalities that involve exponents and radicals (roots). There are five main things youâll have to do to simplify exponents and radicals. Example 1 Add the fractions: $$\dfrac{2}{x} + \dfrac{3}{5}$$ Solution to Example 1 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). For $$\displaystyle y={{x}^{{\text{even}}}},\,\,\,\,\,\,y=\pm \,\sqrt[{\text{even} }]{x}$$. 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). To do this, weâll set whatâs under the even radical to greater than or equal to 0, solve for $$x$$. Multiply fractions variables calculator, 21.75 decimal to hexadecimal, primary math poems, solving state equation using ode45. Remember that the bottom of the fraction is what goes in the root, and we typically take the root first. \displaystyle \begin{align}{{x}^{3}}&=27\\\,\sqrt{{{{x}^{3}}}}&=\sqrt{{27}}\\\,x&=3\end{align}. Variables in a radical's argument are simplified in the same way as regular numbers. One step equation word problems. $$\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,=\frac{{5{{{(\sqrt{3})}}^{3}}}}{{2{{{(\sqrt{3})}}^{4}}}}=\frac{{5{{{(\sqrt{3})}}^{3}}}}{{2\cdot 3}}=\frac{{5{{{(\sqrt{3})}}^{3}}}}{6}$$. Combine like radicals. By using this website, you agree to our Cookie Policy. Notice that, since we wanted to end up with positive exponents, we kept the positive exponents where they were in the fraction. Remember that exponents, or âraisingâ a number to a power, are just the number of times that the number (called the base) is multiplied by itself. $$\begin{array}{c}{{x}^{2}}=-4\\\emptyset \text{ or no solution}\end{array}$$, $$\begin{array}{c}{{x}^{2}}=25\\x=\pm 5\end{array}$$, We need to check our answers:    $${{\left( 5 \right)}^{2}}-1=24\,\,\,\,\surd \,\,\,\,\,\,\,\,{{\left( {-5} \right)}^{2}}-1=24\,\,\,\,\surd$$, $$\begin{array}{c}{{\left( {\sqrt{{x+3}}} \right)}^{4}}={{2}^{4}}\\x+3=16\\x=13\end{array}$$. 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). Note that we have to remember that when taking the square root (or any even root), we always take the positive value (just memorize this).eval(ez_write_tag([[320,100],'shelovesmath_com-medrectangle-3','ezslot_3',115,'0','0'])); But now that weâve learned some algebra, we can do exponential problems with variables in them! Unless otherwise indicated, assume numbers under radicals with even roots are positive, and numbers in denominators are nonzero. The basic ideas are very similar to simplifying numerical fractions. If you have a base with a negative number thatâs not a fraction, put 1 over it and make the exponent positive. (You can also use the WINDOW button to change the minimum and maximum values of your x and y values.). Find out more here about permutations without repetition. We have to make sure our answers donât produce any negative numbers under the square root; this looks good. Simplifying Radical Expressions with Variables - Concept - Solved Questions. With odd roots, we donât have to worry â we just raise each side that power, and solve! Letâs first try some equations with odd exponents and roots, since these are a little more straightforward. We have a tremendous amount of good reference information on matters ranging from mathematics i to precalculus i $${{\left( {-8} \right)}^{{\frac{2}{3}}}}={{\left( {\sqrt{{-8}}} \right)}^{2}}={{\left( {-2} \right)}^{2}}=4$$. I know this seems like a lot to know, but after a lot of practice, they become second nature. ), $$\displaystyle \sqrt{{\frac{{{{x}^{3}}}}{{{{y}^{3}}}}}}=\sqrt{{\frac{{x\cdot x\cdot x}}{{y\cdot y\cdot y}}}}=\sqrt{{\frac{x}{y}}}\cdot \sqrt{{\frac{x}{y}}}\cdot \sqrt{{\frac{x}{y}}}=\frac{x}{y}=\frac{{\sqrt{{{{x}^{3}}}}}}{{\sqrt{{{{y}^{3}}}}}}$$, $$\displaystyle {{\left( {\sqrt[n]{x}} \right)}^{m}}=\,\sqrt[n]{{{{x}^{m}}}}={{x}^{{\frac{m}{n}}}}$$, $$\displaystyle {{8}^{{\frac{2}{3}}}}=\sqrt{{{{8}^{2}}}}={{\left( {\sqrt{8}} \right)}^{2}}=\,\,{{2}^{2}}\,\,\,=4$$, $$\displaystyle {{\left( {\sqrt[n]{x}} \right)}^{n}}=\sqrt[n]{{{{x}^{n}}}}=\,\,\,x$$, $$\displaystyle \begin{array}{c}{{\left( {\sqrt{{-2}}} \right)}^{3}}=\sqrt{{{{{\left( {-2} \right)}}^{3}}}}\\=\sqrt{{-8}}=-2\end{array}$$, $$\displaystyle {{\left( {\sqrt{x}} \right)}^{5}}=\sqrt{{{{x}^{5}}}}\,\,={{x}^{{\frac{5}{5}}}}={{x}^{1}}=x$$. Note:  You can also check your answers using a graphing calculator by putting in whatâs on the left of the = sign in â$${{Y}_{1}}=$$â and whatâs to the right of the equal sign in â$${{Y}_{2}}=$$â. You can see that we have two points of intersections; therefore, we have two solutions. Finding square root using long division. Rationalizing the denominator An expression with a radical in its denominator should be simplified into one without a radical in its denominator. You have to be a little careful, especially with even exponents and roots (the âevil evensâ), and also when the even exponents are on the top of a fractional exponent (this will become the root part when we solve). Converting repeating decimals in to fractions. In this example, we simplify 3â(500x³). The solutions that donât work when you put them back in the original equation are called extraneous solutions. Solving linear equations using elimination method. 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