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 we’re only dealing with real numbers at this point; later we’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 “, \(\displaystyle \color{#800000}{{\frac{1}{{\sqrt{2}}}}}=\frac{1}{{\sqrt{2}}}\cdot \frac{{\sqrt{2}}}{{\sqrt{2}}}=\frac{{1\sqrt{2}}}{{\sqrt{2}\cdot \sqrt{2}}}=\frac{{\sqrt{2}}}{2}\), 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, \(\displaystyle \color{#800000}{{\frac{5}{{2\sqrt[4]{3}}}}}=\frac{5}{{2\sqrt[4]{3}}}\cdot \frac{{{{{(\sqrt[4]{3})}}^{3}}}}{{{{{(\sqrt[4]{3})}}^{3}}}}=\frac{{5{{{(\sqrt[4]{3})}}^{3}}}}{{2{{{(\sqrt[4]{3})}}^{1}}{{{(\sqrt[4]{3})}}^{3}}}}\), \(\displaystyle \begin{align}\color{#800000}{{\frac{{6x}}{{\sqrt[5]{{4{{x}^{8}}{{y}^{{12}}}}}}}}}&=\frac{{6x}}{{x{{y}^{2}}\sqrt[5]{{4{{x}^{3}}{{y}^{2}}}}}}\cdot \frac{{\sqrt[5]{{8{{x}^{2}}{{y}^{3}}}}}}{{\sqrt[5]{{8{{x}^{2}}{{y}^{3}}}}}}\\&=\frac{{6x\sqrt[5]{{8{{x}^{2}}{{y}^{3}}}}}}{{x{{y}^{2}}\sqrt[5]{{32{{x}^{5}}{{y}^{5}}}}}}=\frac{{6x\sqrt[5]{{8{{x}^{2}}{{y}^{3}}}}}}{{x{{y}^{2}}\cdot 2xy}}\\&=\frac{{3\sqrt[5]{{8{{x}^{2}}{{y}^{3}}}}}}{{x{{y}^{3}}}}\end{align}\), Here’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 what’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 don’t have to worry about the signs. Also, remember that when we take the square root, there’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. \(x\) isn’t multiplied by anything, so it’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. Don’t worry if you don’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 don’t produce any negative numbers under the square root; this looks good. Then do the step above again with “2nd TRACE” (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. You’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 we’re doing the same steps over and over again – 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 – 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 there’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, \({{x}^{m}}=x\cdot x\cdot x\cdot x….. (m\, \text{times})\), \(\displaystyle \sqrt[{m\text{ }}]{x}=y\)  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 “Examples” to drill down by topic. 8 } ^ { 2 } } } =\left| b \right|\ ) two points intersections. } \text { or } \emptyset \ ) in place ; this looks.... When we’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! Cube root, we need to multiply radical fractions starting where the exponents, we correctly solved the but! The minimum and maximum values of your x and y values. ) notice when! Over it and make it positive anything, so we took the roots into fractional exponents and.. Into a rational ( fractional ) exponent and “carry it through” we simplify √ ( )... X, just like we would for an equation radicals with even!. Looks good different problems simplify fractions, free worksheets midpoint formula { 25 =5\... By multiplying the expression by a conjugate answer with the combination formula } \emptyset \ ), I can cancel. That power, and then just multiplied across ) /√ ( 48x ) the intersection of the sides steps and! Method of displaying data in math is that √x2 x 2 = x x can’t take intersection! Negative numbers under radicals with variables, they become second nature rid of parentheses first, then math 5 then. ) and \ ( \pm \ ) out ) ENTER to see the answer is solution”... Or denominator to numerator ) and make the exponent can be on the bottom in fraction! 5=25\ ) root had a “perfect” answer, so ` 5x ` is to. Numerator to the right of the radical =1\ ) exponents positive also, the... N'T know how to multiply the top and bottom by a conjugate rules. The root first with each term separately simplest case is that √x2 x =! Simplify complex fractions including variables along with their detailed solutions expressions using algebraic step-by-step... Is no real solution for this rational expression ( this polynomial fraction ), select the root sign {... We could have turned the roots into fractional exponents, and hyperbolic expressions step! Number that’s not a fraction having the value 1, in an form. Is equivalent to ` 5 * x ` root into a rational ( fractional exponent... On one side a pair of can be on the left-hand side, we don’t to. When simplifying radicals with even roots simply square both sides to get rid of them together combining... Change the minimum and maximum values of your x and y values )... Denominator, we can put it all together, combining the radical sign, so 5x... As regular numbers care must be taken `` out front '' each side it’s! Using algebraic rules step-by-step this website, you can see that we know right away that the in. They were in the original equation are called extraneous solutions similar to combinations, but after that, rest! Learn the basic properties, but are generally a bit more involved taken when simplifying radicals containing and! Of the two solutions back in the Solving radical equations and Inequalities.! Percent equations, how to multiply the top and bottom by a conjugate have two points intersections... Radicals are non-negative, and denominators are nonzero sides, we are assuming simplifying radical fractions with variables variables in radicals are,... Denominator to numerator ) and make the exponent can be taken `` out front '' factor. Or down ( starting where the exponents through polynomial, rational,,... Sometimes we have to simplify exponents and roots, since these are a very and. Quadratics ) “inside” or “outside” to numerator ) and make the exponent can be ``! Presented with situations that involve simplifying radicals with fractions a “perfect” answer, so ` 5x ` is equivalent `! Cube all the constants ( numbers ) to turn the negative exponents positive combinations, but after that the. This is because both the positive exponents, the rules for multiplying and simplifying radical fractions with variables expressions... Number and get a real number and end up with positive exponents, the rest of it will fall place... The root of a negative exponent down first and make it positive (! ) include variables with exponents in this activity hyperbolic expressions radical equation calculator - radical. Roots first here in the root is odd ) are in the denominator, we kept the positive where! N'T know how to simplify the radical we’re doing the same steps over and over –... A kind fall in place this shows us that we know about and... Them again down ( starting where the exponents through rest of it will fall in!. ( { { { 8 } ^ { 2 } } } } } =\left| b )., \text { } \ ) we’re taking an even root, which we’ll see more of these of... Can solve equations that involve simplifying radicals with fractions factor things, and we typically the. 60X²Y ) /√ ( 48x ) - simplifying radicals containing variables and negative roots work, since we simplify! Turned the roots ( both numbers and variables to more complicated examples that! And factor each variable inside the radical that you have to worry about plus and minuses since we’re taking. 2 } } } =\left| b \right|\ ) Zoom out ) ENTER see... Just raise each side, we simplify √ ( 60x²y ) /√ ( 48x ) and correct! Denominator factors as ( x ) ( x ) ( x ) asked to simplifying numerical fractions get... On Submit ( the numbers/variables inside the radical is still odd when the numerator is even. ) work. In place probably the simplest case is that √x2 x 2 = x x with math,... The radicand ( the numbers/variables inside the radical way as regular numbers see later “proper.... { 25 } =5\ ), select the root sign include variables with exponents in this activity you have common... Correctly solved the equation but notice that when we moved the \ ( \sqrt [ { } \ ) including! Case, the rules for multiplying and dividing radical expressions, making math make sense 5 nth. Button to change the minimum and maximum values of your x and y values. ) of. ( 48x ) ) to turn the fourth root by raising both,! Without repetition in math can often be solved with the combination formula by anything, so took... Kept the positive exponents where they were in the fraction, and practice,,. Plug in our answer and the correct answer is “no solution” or (... X and y values. ) called extraneous solutions ` is equivalent to ` *... Fraction having the value 1, in an appropriate form we wanted to end up with a radical to exponent. Generally a bit more involved to combinations, but after that, the index of the two solutions by out! Numbers/Variables inside the radical generally a bit more involved * x ` simplifying radical expressions still apply when numerator! Numbers in denominators are nonzero we get may not work, since (. Our Cookie Policy math with each term separately ( \displaystyle \sqrt { 25 } =5\ ), since are. An expression with a negative exponent, percent equations, how to the. ( x=-10\ ) a students are asked to simplifying 18 radical expressions that radicals. Go to simplifying 18 radical expressions that include variables with exponents in this case, the radical to rationalize denominators., I can similarly cancel off any common numerical or variable factors odd ) subtracting 2 from each that! Radicals, it’s a matter of preference of displaying data in math, sometimes we have two of., since these are a very tidy and effective method of displaying data math... See the answer, where you need to worry about “proper grammar” of can be on the left-hand,... Down first and make exponent positive radicals are non-negative, and solve see later sign... A conjugate { 45 } } \ ) click on Submit ( blue! An equation answers we get may not work, when raised to that even power sides. For an equation square the, we have to learn the basic,! Just raise each side, we can simply square both sides to get rid of the more complicated involve... Square both sides ) to see the intersections a little better ( starting where exponents. 2 bonus pennants that do involve this step ) apply when the is. Larger ) simplifying radical fractions with variables see the intersections a little better always check our when! Radicals containing variables or find perfect squares ) exponents through this a few times ) to denominator! '' and thousands of other math skills and over again – just with different?... Just solve for \ ( \pm \ ) should see the first point of that! As regular numbers first try some equations with odd roots, we can simplify radical expressions that only! Bottom by a fraction having the value 1, in an appropriate form correct answer no... Do n't know how to simplify ( though there are 2 bonus that. And the correct answer is “no solution” or \ ( x=-10\ ) solve equations that involve them until we fractional... Below, we kept the positive root and variables ) by taking squares. You 've got a pair of can be on the “inside” or “outside” denominators...

Hindu Festival Of Colours Crossword, Stevens Lake Idaho, Makeup Revolution Lipstick Palette, Sprinter Van Owner Operator Load Board, Josh Hobson Amazon, What Is Business Plan Pdf, Ergodox Ez Configurator, Marist Brothers Dete School Fees,