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. Writing and evaluating expressions. This will give us \(\displaystyle \frac{{16}}{5}\le \,\,x<4\). Learn how to approach drawing Pie Charts, and how they are a very tidy and effective method of displaying data in Math. Remember that exponents, or “raising” a number to a power, are just the number of times that the number (called the base) is multiplied by itself. Then we can solve for y by subtracting 2 from each side. To find the other point of intersection, we need to move the cursor closer to that point, so press “TRACE” and move the cursor closer to the other point of intersection (it should follow along one of the curves). For the purpose of the examples below, we are assuming that variables in radicals are non-negative, and denominators are nonzero. Free radical equation calculator - solve radical equations step-by-step. You will have to learn the basic properties, but after that, the rest of it will fall in place! We remember that \(\sqrt{25}=5\), since \(5\times 5=25\). Note:  You can also check your answers using a graphing calculator by putting in what’s on the left of the = sign in “\({{Y}_{1}}=\)” and what’s to the right of the equal sign in “\({{Y}_{2}}=\)”. The numerator factors as (2)(x); the denominator factors as (x)(x). You can then use the intersection feature to find the solution(s); the solution(s) will be what \(x\) is at that point. Remember that when we cube a cube root, we end up with what’s under the root sign. \(x\) isn’t multiplied by anything, so it’s just \(x\). \(\{\}\text{ }\,\,\text{ or }\emptyset \). Then subtract up or down (starting where the exponents are larger) to turn the negative exponents positive. We can check our answer by trying \(x=3.5\) in the original inequality (which works) and \(x=3\) or \(x=5\) (which don’t work). Also note that what’s under the radical sign is called the radicand (\(x\) in the previous example), and for the \(n\)th root, the index is \(n\) (2, in the previous example, since it’s a square root). 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. This one’s pretty complicated since we have to, \(\begin{array}{l}{{32}^{{\tfrac{3}{5}}}}\cdot {{81}^{{\tfrac{1}{4}}}}\cdot {{27}^{{-\tfrac{1}{3}}}}&={{\left( 2 \right)}^{3}}\cdot {{\left( 3 \right)}^{1}}\cdot {{\left( 3 \right)}^{{-1}}}\\&=8\cdot 3\cdot \tfrac{1}{3}=8\end{array}\). We need to take the intersection (all must work) of the inequalities: \(\displaystyle x<4\text{ and }x\ge \frac{{16}}{5}\text{ and }x\ge 2\). Just like we had to solve linear inequalities, we also have to learn how to solve inequalities that involve exponents and radicals (roots). ... Word problems on fractions. The steps in adding and subtracting Radical are: Step 1. 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. By using this website, you agree to our Cookie Policy. 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. Solving linear equations using elimination method. Some of the worksheets for this concept are Grade 9 simplifying radical expressions, Grade 5 fractions work, Radical workshop index or root radicand, Dividing radical, Radical expressions radical notation for the n, Simplifying radical expressions date period, Reducing fractions work 2, Simplifying … We’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. \(\displaystyle \begin{align}{{x}^{3}}&=27\\\,\sqrt[3]{{{{x}^{3}}}}&=\sqrt[3]{{27}}\\\,x&=3\end{align}\). Each root had a “perfect” answer, so we took the roots first. You should see the second solution at \(x=-10\). Don’t worry if you don’t totally get this now! \(\displaystyle \frac{{34{{n}^{{2x+y}}}}}{{17{{n}^{{x-y}}}}}\). When raising a radical to an exponent, the exponent can be on the “inside” or “outside”. Add and Subtract Fractions with Variables. Learn these rules, and practice, practice, practice! 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. Let’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. 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 we’ve learned some algebra, we can do exponential problems with variables in them! Then combine variables and add or subtract exponents. 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. Decimal representation of rational numbers. Move all the constants (numbers) to the right. \(\sqrt[{\text{even} }]{{\text{negative number}}}\,\) exists for imaginary numbers, but not for real numbers. ), \(\begin{align}2\sqrt[3]{x}&=\sqrt[3]{{x+7}}\\{{\left( {2\sqrt[3]{x}} \right)}^{3}}&={{\left( {\sqrt[3]{{x+7}}} \right)}^{3}}\\8x&=x+7\\7x&=7\\x&=1\end{align}\). Just remember that you have to be really, really careful doing these! Then we just solve for x, just like we would for an equation. There are five main things you’ll have to do to simplify exponents and radicals. Then we applied the exponents, and then just multiplied across. With odd roots, we don’t have to worry – we just raise each side that power, and solve! You will have to learn the basic properties, but after that, the rest of it will fall in place! When radicals (square roots) include variables, they are still simplified the same way. Then we solve for \({{y}_{2}}\). Simplifying radicals with variables is a bit different than when the radical terms contain just numbers. Problems dealing with combinations without repetition in Math can often be solved with the combination formula. The 4th root of \({{a}^{7}}\) is  \(a\,\sqrt[4]{{{{a}^{3}}}}\), since 4 goes into 7 one time (so we can take one \(a\) out), and there’s 3 left over (to get the \({{a}^{3}}\)). In the “proof” column, you’ll notice that we’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. We could have turned the roots into fractional exponents and gotten the same answer – it’s a matter of preference. 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. Some expressions are fractions with and without perfect square roots. \(\displaystyle \begin{align}\left( {6{{a}^{{-2}}}b} \right){{\left( {\frac{{2a{{b}^{3}}}}{{4{{a}^{3}}}}} \right)}^{2}}&=6{{a}^{{-2}}}b\cdot \frac{{4{{a}^{2}}{{b}^{6}}}}{{16{{a}^{6}}}}\\&=\frac{{24{{a}^{0}}{{b}^{7}}}}{{16{{a}^{6}}}}=\frac{{3{{b}^{7}}}}{{2{{a}^{6}}}}\end{align}\). Click on Submit (the blue arrow to the right of the problem) to see the answer. You have to be a little careful, especially with even exponents and roots (the “evil evens”), and also when the even exponents are on the top of a fractional exponent (this will become the root part when we solve). We can check our answer by trying random numbers in our solution (like \(x=2\)) in the original inequality (which works). \(\displaystyle \begin{align}\frac{{34{{n}^{{2x+y}}}}}{{17{{n}^{{x-y}}}}}&=2{{n}^{{\left( {2x+y} \right)\,-\,\left( {x-y} \right)}}}\\&=2{{n}^{{2x-x+y-\left( {-y} \right)}}}=2{{n}^{{x+2y}}}\end{align}\), \(\displaystyle \begin{align}&\frac{{{{{\left( {4{{a}^{{-3}}}{{b}^{2}}} \right)}}^{{-2}}}{{{\left( {{{a}^{3}}{{b}^{{-1}}}} \right)}}^{3}}}}{{{{{\left( {-2{{a}^{{-3}}}} \right)}}^{2}}}}\\&=\frac{{{{{\left( {{{a}^{3}}{{b}^{{-1}}}} \right)}}^{3}}}}{{{{{\left( {4{{a}^{{-3}}}{{b}^{2}}} \right)}}^{2}}{{{\left( {-2{{a}^{{-3}}}} \right)}}^{2}}}}\\&=\frac{{{{a}^{9}}{{b}^{{-3}}}}}{{\left( {16{{a}^{{-6}}}{{b}^{4}}} \right)\left( {4{{a}^{{-6}}}} \right)}}=\frac{{{{a}^{9}}{{b}^{{-3}}}}}{{64{{a}^{{-12}}}{{b}^{4}}}}\\&=\frac{{{{a}^{{9-\left( {-12} \right)}}}}}{{64{{b}^{{4-\left( {-3} \right)}}}}}=\frac{{{{a}^{{21}}}}}{{64{{b}^{7}}}}\end{align}\). Turn the fourth root into a rational (fractional) exponent and “carry it through”. This website uses cookies to ensure you get the best experience. Again, we’ll see more of these types of problems in the Solving Radical Equations and Inequalities section. You’ll get it! Learn more ... Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent … (Notice when we have fractional exponents, the radical is still odd when the numerator is odd). The reason we take the intersection of the two solutions is because both must work. Then, to rationalize, since we have a 4th root, we can multiply by a radical that has the 3rd root on top and bottom. Then we can solve for x. Let’s check our answer:   \(\begin{align}4\sqrt[3]{1}&=2\sqrt[3]{{1+7}}\,\,\,\,\,?\\4\,\,&=\,\,4\,\,\,\,\,\,\surd \end{align}\), Now let’s solve equations with even roots. Step 2. If the negative exponent is on the outside parentheses of a fraction, take the reciprocal of the fraction (base) and make the exponent positive. Writing and evaluating expressions. With \({{64}^{{\frac{1}{4}}}}\), we factor it into 16 and 4, since \({{16}^{{\frac{1}{4}}}}\) is 2. Also remember that we don’t need the parentheses around the exponent in the newer calculator operating systems (but it won’t hurt to have them). The basic ideas are very similar to simplifying numerical fractions. Example 1: to simplify $(\sqrt{2}-1)(\sqrt{2}+1)$ type (r2 - 1)(r2 + 1) . \(\displaystyle \sqrt[n]{{{{x}^{n}}}}=\,\left| x \right|\), \(\displaystyle \begin{array}{c}\sqrt[4]{{{{{\left( {\text{neg number }x} \right)}}^{4}}}}=\sqrt[4]{{\text{pos number }{{x}^{4}}}}\\=\text{positive }x=\left| x \right|\end{array}\), (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!!). The solutions that don’t work when you put them back in the original equation are called extraneous solutions. Students will not need to rationalize the denominators to simplify (though there are 2 bonus pennants that do involve this step). “Push through” the exponent when eliminating the parentheses. Similarly, the rules for multiplying and dividing radical expressions still apply when the expressions contain variables. \(\begin{array}{c}{{x}^{2}}=-4\\\emptyset \text{ or no solution}\end{array}\), \(\begin{array}{c}{{x}^{2}}=25\\x=\pm 5\end{array}\), We need to check our answers:    \({{\left( 5 \right)}^{2}}-1=24\,\,\,\,\surd \,\,\,\,\,\,\,\,{{\left( {-5} \right)}^{2}}-1=24\,\,\,\,\surd \), \(\begin{array}{c}{{\left( {\sqrt[4]{{x+3}}} \right)}^{4}}={{2}^{4}}\\x+3=16\\x=13\end{array}\). If two terms are in the denominator, we need to multiply the top and bottom by a conjugate . ], 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 … 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. Simplifying Radical Expressions with Variables. For example, while you can think of as equivalent to since both the numerator and the denominator are square roots, notice that you cannot express as . \(\displaystyle \frac{1}{{{{3}^{2}}}}={{3}^{{-2}}}=\frac{1}{9}\), \(\displaystyle {{\left( {\frac{2}{3}} \right)}^{{-2}}}={{\left( {\frac{3}{2}} \right)}^{2}}=\frac{9}{4}\), \(\displaystyle {{\left( {\frac{x}{y}} \right)}^{{-m}}}=\frac{{{{x}^{{-m}}}}}{{{{y}^{{-m}}}}}=\frac{{\frac{1}{{{{x}^{m}}}}}}{{\frac{1}{{{{y}^{m}}}}}}=\frac{1}{{{{x}^{m}}}}\times \frac{{{{y}^{m}}}}{1}=\,{{\left( {\frac{y}{x}} \right)}^{m}}\), \(\displaystyle \sqrt[3]{8}={{8}^{{\frac{1}{3}}}}=2\), \(\sqrt[n]{{xy}}=\sqrt[n]{x}\cdot \sqrt[n]{y}\), \(\displaystyle \begin{array}{l}\sqrt{{72}}=\sqrt{{4\cdot 9\cdot 2}}=\sqrt{4}\cdot \sqrt{9}\cdot \sqrt{2}\\\,\,\,\,\,\,\,\,\,\,\,\,=2\cdot 3\cdot \sqrt{2}=6\sqrt{2}\end{array}\), (\(\sqrt{{xy}}={{(xy)}^{{\frac{1}{2}}}}={{x}^{{\frac{1}{2}}}}\cdot {{y}^{{\frac{1}{2}}}}=\sqrt{x}\cdot \sqrt{y}\), (Doesn’t work for imaginary numbers under radicals. If you have a base with a negative number that’s not a fraction, put 1 over it and make the exponent positive. It gets trickier when we don’t know the sign of one of the sides. Simplifying radicals containing variables. Flip the fraction, and then do the math with each term separately. Example 1 Add the fractions: \( \dfrac{2}{x} + \dfrac{3}{5} \) Solution to Example 1 Then we take the intersection of both solutions. Let’s first try some equations with odd exponents and roots, since these are a little more straightforward. We keep moving variables around until we have \({{y}_{2}}\) on one side. With a negative exponent, there’s nothing to do with negative 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. \(\begin{array}{c}\sqrt{{x+2}}\le 4\text{ }\,\text{ }\,\text{and }x+2\,\,\ge \,\,0\\{{\left( {\sqrt{{x+2}}} \right)}^{2}}\le {{4}^{2}}\text{ }\,\,\text{and }x+2\,\,\ge \,\,0\text{ }\\x+2\le 16\text{ }\,\text{and }x\ge \,\,0-2\text{ }\\x\le 14\text{ }\,\text{and }x\ge -2\\\\\{x:-2\le x\le 14\}\text{ or }\left[ {-2,14} \right]\end{array}\). If a root is raised to a fraction (rational), the numerator of the exponent is the power and the denominator is the root. You may need to hit “ZOOM 6” (ZStandard) and/or “ZOOM 0” (ZoomFit) to make sure you see the lines crossing in the graph. To do this, we’ll set what’s under the even radical to greater than or equal to 0, solve for \(x\). (You can also use the WINDOW button to change the minimum and maximum values of your x and y values.). Assume variables under radicals are non-negative. Eliminate the parentheses with the squared first. Remember that \({{a}^{0}}=1\). Remember that the bottom of the fraction is what goes in the root, and we typically take the root first. With \(\sqrt[4]{{64}}\), we factor 64 into 16 and 4, since \(\displaystyle \sqrt[4]{{16}}=2\). (Notice when we have fractional exponents, the radical is still even when the numerator is even.). We can raise both sides to the same number. For all these examples, see how we’re doing the same steps over and over again – just with different problems? Simplifying radicals with variables is a bit different than when the radical terms contain just numbers. 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). For this rational expression (this polynomial fraction), I can similarly cancel off any common numerical or variable factors. In algebra, we’ll need to know these and many other basic rules on how to handle exponents and roots when we work with them. Here are those instructions again, using an example from above: Push GRAPH. Radical Expressions Session 2 . Move what’s inside the negative exponent down first and make exponent positive. To simplify a numerical fraction, I would cancel off any common numerical factors. You’ll see the first point of intersection that it found is where \(x=6\). Since we’re taking an even root, we have to include both the. This is accomplished by multiplying the expression by a fraction having the value 1, in an appropriate form. Radical Form to Exponential Form Worksheets Exponential Form to Radical Form Worksheets Adding Subtracting Multiplying Radicals Worksheets Dividing Radicals Worksheets Algebra 1 Algebra 2 Square Roots Radical Expressions Introduction Topics: Simplifying radical expressions Simplifying radical expressions with variables Adding radical … In this example, we simplify 3√(500x³). Once a 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. In cases where you have help with math and in particular with Complex Fractions Online Calculator or lines come pay a visit to us at Solve-variable.com. 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. \({{\left( {-8} \right)}^{{\frac{2}{3}}}}={{\left( {\sqrt[3]{{-8}}} \right)}^{2}}={{\left( {-2} \right)}^{2}}=4\). We have to make sure our answers don’t produce any negative numbers under the square root; this looks good. It’s always easier to simply (for example. Some of the more complicated problems involve using Quadratics). In this case, the index is two because it is a square root, which means we need two of a kind. Radicals (which comes from the word “root” and means the same thing) means undoing the exponents, or finding out what numbers multiplied by themselves comes up with the number. Then we can put it all together, combining the radical. When radicals, it’s improper grammar to have a root on the bottom in a fraction – in the denominator. Here are the rules/properties with explanations and examples. Since we have to get \({{y}_{2}}\) by itself, we first have to take the square root of each side (and don’t forget to take the plus and the minus). To get rid of the radical on the left-hand side, we can cube both sides. Let’s check our answer:  \({{3}^{3}}-1=27-1=26\,\,\,\,\,\,\surd \), \(\displaystyle \begin{align}\sqrt[3]{{x+2}}&=3\\{{\left( {\sqrt[3]{{x+2}}} \right)}^{3}}&={{3}^{3}}\\x+2&=27\\x&=25\end{align}\). Also, since we squared both sides, let’s check our answer: \(\displaystyle 4\sqrt{{\frac{2}{{15}}}}=\sqrt{{\frac{{32}}{{15}}}}?\,\,\,\,\,\,\,\,\,\,\,\,\,4\sqrt{{\frac{2}{{15}}}}=\sqrt{{\left( {16} \right)\left( 2 \right)\frac{1}{{15}}}}\,\,?\,\,\,\,\,\,\,\,\,\,\,\,4\sqrt{{\frac{2}{{15}}}}=4\sqrt{{\frac{2}{{15}}}}\,\,\,\,\surd \), \(\displaystyle \begin{align}{{\left( {{{{\left( {x+2} \right)}}^{{\frac{4}{3}}}}} \right)}^{{\frac{3}{4}}}}&={{16}^{{\frac{3}{4}}}}\\x+2&=\pm {{2}^{3}}\\x&=\pm {{2}^{3}}-2\\x&=8-2=6\,\,\,\,\,\text{and}\\x&=-8-2=-10\end{align}\), \(\displaystyle \begin{array}{c}{{\left( {6+2} \right)}^{{\tfrac{4}{3}}}}+2={{\left( {\sqrt[3]{8}} \right)}^{4}}+2={{2}^{4}}+2=18\,\,\,\,\,\,\surd \\{{\left( {-10+2} \right)}^{{\tfrac{4}{3}}}}+2={{\left( {\sqrt[3]{{-8}}} \right)}^{4}}+2={{\left( {-2} \right)}^{4}}+2=18\,\,\,\,\,\,\surd \end{array}\), \(\begin{align}{{\left( {\sqrt{{2-x}}} \right)}^{2}}&={{\left( {\sqrt{{x-4}}} \right)}^{2}}\\\,2-x&=x-4\\\,2x&=6\\\,x&=3\end{align}\). 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). 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. We just have to work with variables as well as numbers. We could have also just put this one in the calculator (using parentheses around the fractional roots). Therefore, in this case, \(\sqrt{{{{a}^{3}}}}=\left| a \right|\sqrt{a}\). We have to make sure we square the, We correctly solved the equation but notice that when we plug in. Factor the expression completely (or find perfect squares). We can also use the MATH function to take the cube root (4, or scroll down) or nth root (5:). 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 \(a\)). Notice that when we moved the \(\pm \) to the other side, it’s still a \(\pm \). (You may have to do this a few times). 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. You can see that we have two points of intersections; therefore, we have two solutions. With MATH 5 (nth root), select the root first, then MATH 5, then what’s under the radical. Multiply fractions variables calculator, 21.75 decimal to hexadecimal, primary math poems, solving state equation using ode45. Show Instructions In general, you can skip the multiplication sign, so `5x` is equivalent to `5*x`. Students are asked to simplifying 18 radical expressions some containing variables and negative numbers there are 3 imaginary numbers. Note that this works when \(n\) is even too, if  \(x\ge 0\). Displaying top 8 worksheets found for - Simplifying Radicals With Fractions. \(\displaystyle {{x}^{2}}=16;\,\,\,\,\,\,\,x=\pm 4\), \({{\left( {xy} \right)}^{3}}={{x}^{3}}{{y}^{3}}\), \(\displaystyle {{\left( {\frac{x}{y}} \right)}^{m}}=\frac{{{{x}^{m}}}}{{{{y}^{m}}}}\), \(\displaystyle {{\left( {\frac{x}{y}} \right)}^{4}}=\frac{{{{x}^{4}}}}{{{{y}^{4}}}}\), \(\displaystyle  {{\left( {\frac{x}{y}} \right)}^{4}}=\frac{x}{y}\cdot \frac{x}{y}\cdot \frac{x}{y}\cdot \frac{x}{y}=\frac{{x\cdot x\cdot x\cdot x}}{{y\cdot y\cdot y\cdot y}}=\frac{{{{x}^{4}}}}{{{{y}^{4}}}}\), \({{x}^{m}}\cdot {{x}^{n}}={{x}^{{m+n}}}\), \({{x}^{4}}\cdot {{x}^{2}}={{x}^{{4+2}}}={{x}^{6}}\), \(\require{cancel} \displaystyle \frac{{{{x}^{5}}}}{{{{x}^{3}}}}=\frac{{x\cdot x\cdot x\cdot x\cdot x}}{{x\cdot x\cdot x}}=\frac{{x\cdot x\cdot \cancel{x}\cdot \cancel{x}\cdot \cancel{x}}}{{\cancel{x}\cdot \cancel{x}\cdot \cancel{x}}}=x\cdot x={{x}^{2}}\), \({{\left( {{{x}^{m}}} \right)}^{n}}={{x}^{{mn}}}\), \({{\left( {{{x}^{4}}} \right)}^{2}}={{x}^{{4\cdot 2}}}={{x}^{8}}\), \(\displaystyle 1=\frac{{{{x}^{5}}}}{{{{x}^{5}}}}={{x}^{{5-5}}}={{x}^{0}}\), \(\displaystyle \frac{1}{{{{x}^{m}}}}={{x}^{{-m}}}\), \(\displaystyle \frac{1}{{{{2}^{3}}}}=\frac{{{{2}^{0}}}}{{{{2}^{3}}}}={{2}^{{0-3}}}={{2}^{{-3}}}\), \(\displaystyle \sqrt[n]{x}={{x}^{{\frac{1}{n}}}}\). You move the base from the numerator to the denominator (or denominator to numerator) and make it positive! In these examples, we are taking the cube root of \({{8}^{2}}\). We can’t take the even root of a negative number and get a real number. If we don’t assume variables under the radicals are non-negative, we have to be careful with the signs and include absolute values for even radicals. The \(n\)th root of a base can be written as that base raised to the reciprocal of \(n\), or \(\displaystyle \frac{1}{n}\). eval(ez_write_tag([[320,50],'shelovesmath_com-box-3','ezslot_5',114,'0','0']));This section covers: We briefly talked about exponents in the Powers, Exponents, Radicals (Roots) and Scientific Notation section, but we need to go a little bit further in depth and talk about how to do algebra with them. 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 we’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. ], Convert Decimal To Fraction [ Def: A number that names a part of a whole or a part of a group. We can get an “imaginary number”, which we’ll see later. \(\displaystyle \begin{align}\sqrt[4]{{\frac{{{{x}^{6}}{{y}^{4}}}}{{162{{z}^{5}}}}}}&=\frac{{\sqrt[4]{{{{x}^{6}}{{y}^{4}}}}}}{{\sqrt[4]{{\left( {81} \right)\left( 2 \right){{z}^{5}}}}}}=\frac{{xy\sqrt[4]{{{{x}^{2}}}}}}{{3z\sqrt[4]{{2z}}}}\\&=\frac{{xy\sqrt[4]{{{{x}^{2}}}}}}{{3z\sqrt[4]{{2z}}}}\cdot \frac{{\sqrt[4]{{{{{\left( {2z} \right)}}^{3}}}}}}{{\sqrt[4]{{{{{\left( {2z} \right)}}^{3}}}}}}\\&=\frac{{xy\sqrt[4]{{{{x}^{2}}}}\sqrt[4]{{8{{z}^{3}}}}}}{{3z\sqrt[4]{{{{{\left( {2z} \right)}}^{4}}}}}}=\frac{{xy\sqrt[4]{{8{{x}^{2}}{{z}^{3}}}}}}{{3z\left( {2z} \right)}}\\&=\frac{{xy\sqrt[4]{{8{{x}^{2}}{{z}^{3}}}}}}{{6{{z}^{2}}}}\end{align}\). Be careful to make sure you cube all the numbers (and anything else on that side) too. Now let’s put it altogether. 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. Separate the numbers and variables. Put it all together, combining the radical. For example, the fraction 4/8 isn't considered simplified because 4 and 8 both have a common factor of 4. Then, combine like terms, where you need to have the same root and variables. A root “undoes” raising a number to that exponent. This calculator will simplify fractions, polynomial, rational, radical, exponential, logarithmic, trigonometric, and hyperbolic expressions. Simplifying Radical Expressions with Variables - Concept - Solved Questions. We have a tremendous amount of good reference information on matters ranging from mathematics i to precalculus i Simplify the roots (both numbers and variables) by taking out squares. To get rid of the square roots, we square each side, and we can leave the inequality signs the same since we’re multiplying by positive numbers. 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 \(2\cdot 2\cdot 2={{2}^{3}}=8\), \(\displaystyle {{x}^{{\frac{m}{n}}}}={{\left( {\sqrt[n]{x}} \right)}^{m}}=\,\sqrt[n]{{{{x}^{m}}}}\), \(\displaystyle {{x}^{{\frac{2}{3}}}}=\,\sqrt[3]{{{{8}^{2}}}}={{\left( {\sqrt[3]{8}} \right)}^{2}}={{2}^{2}}=4\). \(\displaystyle \sqrt[n]{{\frac{x}{y}}}=\frac{{\sqrt[n]{x}}}{{\sqrt[n]{y}}}\), \(\displaystyle \sqrt[3]{{\frac{{27}}{8}}}=\frac{{\sqrt[3]{{27}}}}{{\sqrt[3]{8}}}=\frac{3}{2}\), \(\displaystyle \begin{array}{c}\sqrt[{}]{{{{{\left( {-4} \right)}}^{2}}}}=\sqrt{{16}}=4\\\sqrt[{}]{{{{{\left( 4 \right)}}^{2}}}}=\sqrt{{16}}=4\end{array}\), \(\displaystyle \begin{align}\frac{x}{{\sqrt{y}}}&=\frac{x}{{\sqrt{y}}}\cdot \frac{{\sqrt{y}}}{{\sqrt{y}}}\\&=\frac{{x\sqrt{y}}}{y}\end{align}\), \(\displaystyle \begin{align}\frac{4}{{\sqrt{2}}}&=\frac{4}{{\sqrt{2}}}\cdot \frac{{\sqrt{2}}}{{\sqrt{2}}}\\&=\frac{{{}^{2}\cancel{4}\sqrt{2}}}{{{}^{1}\cancel{2}}}=2\sqrt{2}\end{align}\), \(\displaystyle \begin{align}\frac{x}{{x+\sqrt{y}}}&=\frac{x}{{x+\sqrt{y}}}\cdot \frac{{x-\sqrt{y}}}{{x-\sqrt{y}}}\\&=\frac{{x\left( {x-\sqrt{y}} \right)}}{{{{x}^{2}}-y}}\\\frac{x}{{x-\sqrt{y}}}&=\frac{x}{{x-\sqrt{y}}}\cdot \frac{{x+\sqrt{y}}}{{x+\sqrt{y}}}\\&=\frac{{x\left( {x+\sqrt{y}} \right)}}{{{{x}^{2}}-y}}\end{align}\), \(\displaystyle \begin{align}\frac{{\sqrt{3}}}{{1-\sqrt{3}}}&=\frac{{\sqrt{3}}}{{1-\sqrt{3}}}\cdot \frac{{1+\sqrt{3}}}{{1+\sqrt{3}}}\\&=\frac{{\sqrt{3}\left( {1+\sqrt{3}} \right)}}{{\left( {1-\sqrt{3}} \right)\left( {1+\sqrt{3}} \right)}}\\&=\frac{{\sqrt{3}+\sqrt{3}\cdot \sqrt{3}}}{{{{1}^{2}}-{{{\left( {\sqrt{3}} \right)}}^{2}}}}=\frac{{\sqrt{3}+3}}{{-2}}\end{align}\), 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, \({{\left( {9{{x}^{3}}y} \right)}^{2}}={{9}^{2}}{{x}^{6}}{{y}^{2}}=81{{x}^{6}}{{y}^{2}}\). 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