Exponents and Radicals in Algebra

We briefly talked about exponents in the Powers, Exponents, Radicals (Roots) and Scientific Notation section, but now we’ll work with them using algebra. Note that we’ll see more radicals in the Solving Radical Equations and Inequalities section, and we’ll talk about Factoring and Solving with Exponents here in the Advanced Factoring section. Exponential Functions are discussed in the Exponential Functions section. Also, imaginary (non-real) roots are discussed in the Imaginary (Non-Real) and Complex Numbers section.

Introducing Exponents and Radicals (Roots) with Variables

Here are some basics of Exponents and Radicals, as we start to use them in Algebra:

  • An exponent, or “raising” a number to a power, is just the number of times that a base is multiplied by itself. In this example, the exponent is 3 and the base is 5:  $ {{5}^{3}}=5\times 5\times 5=125$.
  • 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. For example, $ \sqrt[3]{{125}}=5$, since $ 5\times 5\times5=125$.
  • When we take the square root, there’s an invisible 2 in the radical, like this: $ \sqrt[2]{x}$).
  • When taking the square root (or any even root), we always take the positive value (just memorize this). For example, $ \sqrt{4}=2$. But when solving for an even root in an equation, we have to take plus and minus; for example: $ {{x}^{2}}=4;\,\,x=\pm 2$.
  • $ \sqrt{{{x}^{2}}}=\left| x \right|$ since $ x$ can be negative; try it: $ \sqrt{{{{{\left( {-3} \right)}}^{2}}}}=\sqrt{9}=3=\left| {-3} \right|$. More generally, $ \sqrt[n]{{{{x}^{n}}}}=\left| x \right|$, if $ n$ is even. Note that $ \left| {} \right|$ signifies the Absolute Value of a number, which is the positive value only.
  • Taking a root of a number is the same as raising it to $ \displaystyle \frac{1}{{\text{that root}}}$; for example, $ \displaystyle {{x}^{\frac{1}{2}}}=\sqrt{x}$.
  • A negative exponent has nothing to do with negative numbers as we know them! Move the base from the numerator to the denominator (or denominator to numerator) and make the exponent positive! If the negative exponent is on the outside of the parentheses of a fraction, take the reciprocal of the fraction (base) and make the exponent positive. 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}}$.
  • The two basic ways to write radicals (roots) are via a radical expression (such as $ \sqrt[3]{x}$), and a Rational expression (such as $ {{x}^{{\frac{1}{3}}}}$; “rational” means fractional). I remember this since the $ \sqrt{{}}$ is a radical sign, and rational sounds like fractional. As another example, the rational expression $ \displaystyle {{x}^{{\frac{3}{5}}}}$ can be written in radical form as $ \displaystyle \sqrt[5]{{{{x}^{3}}}}$ or $ \displaystyle {{\left( {\sqrt[5]{x}} \right)}^{3}}$; note that the exponent can be either inside or outside the radical.
  • What’s under the radical sign is called the radicand, and the index is the actual root. In the example $ \sqrt{x}$, $ x$ is the radicand and (“invisible”) 2 is the index.

Note that in these discussions, 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.

Properties of Exponents and Radicals

To summarize, here are some basic rules:

Exponent and Radical Rules

Example

Notes

$ {{x}^{m}}=x\cdot x\cdot x\cdot x….. (m\, \text{times})$

 $ {{2}^{3}}=2\cdot 2\cdot 2=8$

$ x$ is the base, $ m$ is the exponent.
 $ \displaystyle \sqrt[{m\text{ }}]{x}=y$  means  $ \displaystyle {{y}^{m}}=x$  $ \sqrt[3]{8}=2$,  since $ 2\cdot 2\cdot 2={{2}^{3}}=8$ $ x$ is the radicand, $ m$ is the index (root).
 $ \sqrt[{}]{x}$  means  $ \sqrt[2]{x}$

$ \sqrt{4}=2$  means  $ \sqrt[2]{4}=2$

The default root is 2 (square root).
$ \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$ If a root is raised to a fraction (rational), the numerator of the exponent is the power and the denominator is the root. When raising a radical to an exponent, the exponent can be on the “inside” or “outside”.

$ \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}}$

$ \displaystyle \begin{align}{{2}^{{-2}}}=\frac{1}{{{{2}^{2}}}}=\frac{1}{4}\\\frac{1}{{{{2}^{{-2}}}}}={{2}^{2}}=4\\{{\left( {\frac{2}{3}} \right)}^{{-2}}}={{\left( {\frac{3}{2}} \right)}^{2}}=\frac{9}{4}\end{align}$ Raising a base to a negative exponent means taking the reciprocal and making the exponent positive.

$ a\sqrt[{}]{x}\times b\sqrt[{}]{y}=ab\sqrt[{}]{{xy}}$

(Doesn’t work for imaginary numbers under radicals!)

$ 2\sqrt{3}\times \,4\sqrt{5}\,=\,8\sqrt{{15}}$

When you multiply two radical terms, you can multiply what’s on the outside (coefficient), and also what’s in the inside (radicand). You can only do this if the roots (indices) are the same (like square root, cube root).

$ \sqrt[{\text{even} }]{{\text{negative number}}}$   exists for imaginary numbers, but not for real numbers.

$ \sqrt[4]{{-16}}=$ no real solution We can’t take the even root of a negative number and get a real number. We can get an “imaginary number”, which we’ll see later.
$ \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[3]{{{{{\left( {-2} \right)}}^{3}}}}=\sqrt[3]{{-8}}=-2\\\sqrt{{{{{\left( {-2} \right)}}^{2}}}}=\sqrt{4}=2\end{array}$ A root “undoes” raising a number to that exponent. For an even root, we only take positive value, even if original was negative. For example, we squared  –2 under the square root, but our answer is 2, which is $ \left| {-2} \right|$ (the absolute value of 2).

For $ \displaystyle y={{x}^{{\text{even}}}},\,\,\,\,\,x=\pm \sqrt[{\text{even} }]{y}$

$ \displaystyle {{x}^{2}}=16;\,\,\,\,\,\,\,x=\pm 4$ Since we’re taking an even root, we have to include both the positive and negative solutions in an equation with an even exponent. Remember that the square root sign only gives you the positive solutions.

This is because both the positive root and negative roots work, when raised to that even power.

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 the rules/properties with explanations and examples. 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.

Unless otherwise indicated, assume numbers under radicals with even roots are positive, and numbers in denominators are nonzero.

Exponential/Radical Property

Example

“Proof”/Explanation

$ {{(xy)}^{m}}={{x}^{m}}\cdot {{y}^{m}}$

 $ {{\left( {xy} \right)}^{3}}={{x}^{3}}{{y}^{3}}$

$ {{(xy)}^{3}}=xy\cdot xy\cdot xy=\left( {x\cdot x\cdot x} \right)\cdot \left( {y\cdot y\cdot y} \right)={{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}}$

$ {{x}^{4}}\cdot {{x}^{2}}=(x\cdot x\cdot x\cdot x)\cdot (x\cdot x)=x\cdot x\cdot x\cdot x\cdot x\cdot x={{x}^{6}}$

$ \displaystyle \frac{{{{x}^{m}}}}{{{{x}^{n}}}}={{x}^{{m-n}}}$

$ \displaystyle \frac{{{{x}^{5}}}}{{{{x}^{3}}}}={{x}^{{5-3}}}={{x}^{2}}$

 $ \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 {{({{x}^{4}})}^{2}}={{x}^{4}}\cdot {{x}^{4}}=\left( {x\cdot x\cdot x\cdot x} \right)\cdot \left( {x\cdot x\cdot x\cdot x} \right)={{x}^{8}}$

 $ {{x}^{1}}=x$

$ {{\left( {473,837,843} \right)}^{1}}=473,837,843$

$ x$ isn’t multiplied by anything, so it’s just $ x$.

 $ {{x}^{0}}=1$

$ 473,837,{{843}^{0}}=1$

 $ \displaystyle 1=\frac{{{{x}^{5}}}}{{{{x}^{5}}}}={{x}^{{5-5}}}={{x}^{0}}$
 $ \displaystyle \frac{1}{{{{x}^{m}}}}={{x}^{{-m}}}$

$ \displaystyle {{\left( {\frac{x}{y}} \right)}^{{-m}}}=\,{{\left( {\frac{y}{x}} \right)}^{m}}$

$ \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 \frac{1}{{{{2}^{3}}}}=\frac{{{{2}^{0}}}}{{{{2}^{3}}}}={{2}^{{0-3}}}={{2}^{{-3}}}$

$ \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[n]{x}={{x}^{{\frac{1}{n}}}}$

$ \displaystyle \sqrt[3]{8}={{8}^{{\frac{1}{3}}}}=2$

The $ n$th root of a base can be written as that base raised to the reciprocal of $ n$, or $ \displaystyle \frac{1}{n}$.

$ \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.)

 $ \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 \sqrt[3]{{\frac{{{{x}^{3}}}}{{{{y}^{3}}}}}}=\sqrt[3]{{\frac{{x\cdot x\cdot x}}{{y\cdot y\cdot y}}}}=\sqrt[3]{{\frac{x}{y}}}\cdot \sqrt[3]{{\frac{x}{y}}}\cdot \sqrt[3]{{\frac{x}{y}}}={{\left( {\sqrt[3]{{\frac{x}{y}}}} \right)}^{3}}=\frac{x}{y}=\frac{{\sqrt[3]{{{{x}^{3}}}}}}{{\sqrt[3]{{{{y}^{3}}}}}}$

$ \displaystyle {{\left( {\sqrt[n]{x}} \right)}^{m}}=\,\sqrt[n]{{{{x}^{m}}}}={{x}^{{\frac{m}{n}}}}$

(if $ n$ is even, $ x\ge 0$)

$ \displaystyle {{8}^{{\frac{2}{3}}}}=\sqrt[3]{{{{8}^{2}}}}={{\left( {\sqrt[3]{8}} \right)}^{2}}=\,\,{{2}^{2}}\,\,\,=4$

You can take an expression in radical form (with a radical sign) and turn it into rational form (fractional form) by using a fractional exponent with the original exponent on top and root on bottom (remember: the root is in a “cave” so it needs to go on the bottom).

 $ n$ is odd:

$ \displaystyle {{\left( {\sqrt[n]{x}} \right)}^{n}}=\sqrt[n]{{{{x}^{n}}}}=\,\,\,x$

$ \displaystyle \begin{array}{c}{{\left( {\sqrt[3]{{-2}}} \right)}^{3}}=\sqrt[3]{{{{{\left( {-2} \right)}}^{3}}}}\\=\sqrt[3]{{-8}}=-2\end{array}$

$ \displaystyle {{\left( {\sqrt[5]{x}} \right)}^{5}}=\sqrt[5]{{{{x}^{5}}}}\,\,={{x}^{{\frac{5}{5}}}}={{x}^{1}}=x$

Note that this works when $ n$ is even also, but only if $ x\ge 0$.

$ n$ is even:

$ \displaystyle \sqrt[n]{{{{x}^{n}}}}=\,\left| x \right|$

 $ \displaystyle \begin{array}{c}\sqrt[{}]{{{{{\left( {-4} \right)}}^{2}}}}=\sqrt{{16}}=4\\\sqrt[{}]{{{{{\left( 4 \right)}}^{2}}}}=\sqrt{{16}}=4\end{array}$

For when $ x$ is negative:

$ \displaystyle \sqrt[4]{{{{{\left( {\text{neg number }x} \right)}}^{4}}}}=\sqrt[4]{{\text{pos number }{{x}^{4}}}}=\text{positive }x=\left| x \right|$

(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!!)

$ \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}$ This is called “rationalizing” the denominator (getting rid of the radical in the denominator) and is considered better “grammar” in math.
$ \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 conjugate of the denominator, which you can get by changing the signs between the terms.

I know this seems like a lot to know, but after a lot of practice, they become second nature. You will have to learn the basic properties, but after that, the rest of it will fall in place!

Exponents and Radicals in the Calculator

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. Be careful though, because if there’s not a perfect square root, the calculator will give you a long decimal number that’s not the “exact value”. The “exact value” would be the answer with the root sign in it. You need to know your calculator!

Here are some exponent and radical calculator examples (TI 83/84 Graphing Calculator):

Keystrokes Screen Notes
To put a radical in the calculator, type “8 ^ (2/3)”. In the calculator “CLASSIC mode”, put parentheses around the $ \displaystyle \frac{2}{3}$ because the calculator naturally follows the PEMDAS order of operations and would otherwise perform the 8 squared first. (This is not necessary in the newer calculators’ MATHPRINT mode).

 

When there is negative on the outside of the 8, the calculator performs the radicals first (cube root of 8, and then squared) and then puts the negative in front of it.

Take the square root by using the 2nd button, and then the $ {{x}^{2}}$ button.

We can also use the MATH function to take the cube root (4:, or scroll down) or $ n$th root (5:). With MATH 5 ($ n$th root), select the root first, then MATH 5, then what’s under the radical.

 

In these examples, we are taking the cube root of $ {{8}^{2}}$.

Remember that you can always take the $ n$th of any number by raising the number to $ \displaystyle \frac{1}{n}$.

 

Again, we don’t need the parentheses around the exponent in the newer calculator operating systems (but it won’t hurt to have them).

Using a TI30 Scientific Calculator, here are the steps:

Keystrokes Screen Notes
To put a radical in the calculator, type “8 ^ (2/3)”. Some calculators need the parentheses around the $ \frac{2}{3}$ as the calculator follows the PEMDAS order of operations and may square 8 and then divide by 3. On this calculator, this is not required.

 

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.

Take the square root by using the 2nd button, and then the $ {{x}^{2}}$ button.

 

Take the $ n$th roots with the 2nd button, and the ^ button, followed by the radicand; example is the cube root of $ {{8}^{2}}$.

Take the $ n$th root of any number by raising the number to $ \displaystyle \frac{1}{n}$.

 

Parentheses may be optional around exponents.

Rationalizing Radicals

In math, sometimes we have to worry about “proper grammar”. With radicals, it’s improper grammar to have a root in the denominator of a fraction, and thus we have to “rationalize” it. To rationalize, multiply by a fraction by 1 such that the denominator “cancels” out the radical. If two terms are in the denominator, multiply the top and bottom by a conjugate. Here are some examples:

Math

Notes

$ \displaystyle \color{#800000}{{\frac{1}{{\sqrt{2}}}}}\color{#222222}{=\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, get rid of it by multiplying by 1, or $ \displaystyle \frac{{\sqrt{2}}}{{\sqrt{2}}}$.
$ \require{cancel} \displaystyle \color{#800000}{{\frac{4}{{2\sqrt{3}}}}}\color{#222222}{=\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, multiply by 1, or $ \displaystyle \frac{{\sqrt{3}}}{{\sqrt{3}}}$. Note that we didn’t need to multiply by $ 2\sqrt{3}$, only by the radical.
$ \displaystyle \color{#800000}{{\frac{5}{{2\sqrt[4]{3}}}}}\color{#222222}{=\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 \,\,\,\,\,\,\,\,\,\,\,\,=\frac{{5{{{(\sqrt[4]{3})}}^{3}}}}{{2{{{(\sqrt[4]{3})}}^{4}}}}=\frac{{5{{{(\sqrt[4]{3})}}^{3}}}}{{2\cdot 3}}=\frac{{5{{{(\sqrt[4]{3})}}^{3}}}}{6}$

Since we have the 4th root of 3 on the bottom ($ \displaystyle \sqrt[4]{3}$), multiply by 1, with the numerator and denominator being that radical cubed, to eliminate the 4th root.
$ \displaystyle \begin{align}\color{#800000}{{\frac{{6x}}{{x{{y}^{2}}\sqrt[5]{{4{{x}^{3}}{{y}^{2}}}}}}}}&=\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: multiply by 1, with the numerator and denominator a radical ($ \displaystyle \sqrt[5]{{8{{x}^{2}}{{y}^{3}}}}$ ) that gets rid of the radical in the denominator. See how we picked a radical where the exponents in the radicand add up to a multiple of 5, and the product of the constants in the radicand is a fifth power of 2 (32)? Tricky!
$ \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 conjugate of the denominator. The denominator will turn into a difference of squares.

Simplifying Exponential Expressions

There are some hints for simplifying exponents and radicals. For the purpose of the examples below, we are assuming that variables in radicals are non-negative, and denominators are nonzero.

  •  Get rid of parentheses (). When an exponential expression is raised to another exponent, multiply exponents. When an algebraic expression, for example, with coefficients and variables in parentheses, is raised to an exponent, remove parentheses and “push through” the exponent. Example: $ {{\left( {6{{x}^{3}}y} \right)}^{2}}=36{{x}^{6}}{{y}^{2}}$.
  • Combine bases to combine exponents. Add exponents of common bases if you are multiplying, and subtract exponents of common bases if you are dividing (you can subtract “up”, or subtract “down”, starting with the largest exponent, to get the positive exponent). Sometimes you have to match the bases first in order to combine exponents. Examples: $ \displaystyle {{a}^{2}}{{a}^{3}}={{a}^{5}};\,\,\,\frac{{{{a}^{5}}}}{{{{a}^{3}}}}={{a}^{{5-3}}}={{a}^{2}};\,\,\,\frac{{{{a}^{3}}}}{{{{a}^{5}}}}=\frac{1}{{{{a}^{{5-3}}}}}=\frac{1}{{{{a}^{2}}}}\,\,(\text{which is }{{a}^{{-2}}})$,  $ \displaystyle \frac{{{{9}^{3}}}}{{{{3}^{{-4}}}}}=\frac{{{{{\left( {{{3}^{2}}} \right)}}^{3}}}}{{{{3}^{{-4}}}}}=\frac{{{{3}^{6}}}}{{{{3}^{{-4}}}}}={{3}^{{6-\left( {-4} \right)}}}={{3}^{{10}}}$.
  •  Get rid of negative exponents. To get rid of negative exponents, simply move a negative exponent in the denominator to the numerator and make it positive, or vice versa. Examples: $ \displaystyle {{a}^{{-4}}}=\frac{1}{{{{a}^{4}}}};\,\,{{\left( {\frac{a}{b}} \right)}^{{-2}}}={{\left( {\frac{b}{a}} \right)}^{2}}=\frac{{{{b}^{2}}}}{{{{a}^{2}}}}$.
  •  Simplify any numbers (like $ \sqrt{4}=2$). Also, remember to simplify radicals by taking out any factors of perfect squares (under a square root), cubes (under a cube root), and so on. Example: $ \sqrt{{50{{x}^{2}}}}=\sqrt{{25\cdot 2\cdot {{x}^{2}}}}=\sqrt{{25}}\cdot \sqrt{2}\cdot \sqrt{{{{x}^{2}}}}=5x\sqrt{2}$.
  • For exponents inside root signs, divide exponents by their root index, and if it goes in exactly, move to the outside with as the exponent. Even if it doesn’t go in exactly, move highest factor to outside, and leave remainders under the root sign. For example, $ \sqrt[3]{{{{x}^{5}}{{y}^{{12}}}}}={{x}^{1}}{{y}^{4}}\sqrt[3]{{{{x}^{2}}}}=x{{y}^{4}}\sqrt[3]{{{{x}^{2}}}}$, since, for $ {{x}^{5}}$, 5 divided by 3 is 1, with 2 left over, and for $ {{y}^{{12}}}$, 12 divided by 3 is 4. You can also use rational (fractional) exponents to see this, which is sort of like turning improper fractions into mixed fractions: $ \sqrt[3]{{{{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[3]{{{{x}^{2}}}}$. Pretty cool!
  •  Combine any like terms. If you’re adding or subtracting terms with the same roots and/or variables, you can put these together. Almost think of a radical expression (like $ \sqrt{2}$) like another variable. Example: $ 4{{x}^{2}}\sqrt{2}-2{{x}^{2}}\sqrt{2}=2{{x}^{2}}\sqrt{2}$.
  • Rationalize radical fractions, which means getting rid of radicals in the denominator.

Now let’s put it altogether. Here are some (difficult) examples; just remember that you have to be really, really careful doing these! Make sure the answers have no negative exponents.

Expression

Simplification

Explanation

 $ {{\left( {9{{x}^{3}}y} \right)}^{2}}$  $ {{\left( {9{{x}^{3}}y} \right)}^{2}}={{9}^{2}}{{x}^{6}}{{y}^{2}}=81{{x}^{6}}{{y}^{2}}$ “Push through” the exponent when eliminating the parentheses. When raising an exponent to another exponent, multiply exponents to simplify.
 $ \displaystyle \left( {6{{a}^{{-2}}}b} \right){{\left( {\frac{{2a{{b}^{3}}}}{{4{{a}^{3}}}}} \right)}^{2}}$ $ \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}$ Eliminate the parentheses with the squared first. Then combine variables and add or subtract exponents. Remember that $ {{a}^{0}}=1$.

$ \displaystyle \frac{{34{{n}^{{2x+y}}}}}{{17{{n}^{{x-y}}}}}$

 $ \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}$ Divide 34 by 17 to get 2; subtract exponents with base of $ n$.
 $ \displaystyle \frac{{{{{\left( {4{{a}^{{-3}}}{{b}^{2}}} \right)}}^{{-2}}}{{{\left( {{{a}^{3}}{{b}^{{-1}}}} \right)}}^{3}}}}{{{{{\left( {-2{{a}^{{-3}}}} \right)}}^{2}}}}$ $ \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}$ Move what’s inside the negative exponent down first and make exponent positive. Then get rid of parentheses first, by pushing the exponents through.

 

Notice that, since we wanted to end up with positive exponents, we kept the positive exponents where they were in the fraction. Then subtract up or down (starting where the exponents are larger) to turn the negative exponents positive.

 $ {{\left( {-8} \right)}^{{\frac{2}{3}}}}$  $ {{\left( {-8} \right)}^{{\frac{2}{3}}}}={{\left( {\sqrt[3]{{-8}}} \right)}^{2}}={{\left( {-2} \right)}^{2}}=4$ The bottom of the fraction is what goes in the root, and we typically take the root first. We could also put this in our calculator! Note also that if the negative were on the outside, like $ -{{8}^{{\frac{2}{3}}}}$, the answer would be –4.
 $ \displaystyle {{\left( {\frac{{{{a}^{9}}}}{{27}}} \right)}^{{-\frac{2}{3}}}}$ $ \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[3]{{27}}} \right)}}^{2}}}}{{{{a}^{{\frac{{18}}{3}}}}}}\\&=\frac{{{{{\left( {\sqrt[3]{{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. “Carry through” the exponent to both the top and bottom of the fraction and remember that the cube root of 27 is 3.

$ \displaystyle {{\left( {\frac{{{{2}^{{-1}}}+{{2}^{{-2}}}}}{{{{2}^{{-4}}}}}} \right)}^{{-1}}}$

 $ \require{cancel} \displaystyle \begin{align}{{\left( {\frac{{{{2}^{{-1}}}+{{2}^{{-2}}}}}{{{{2}^{{-4}}}}}} \right)}^{{-1}}}&=\frac{{{{2}^{{-4}}}}}{{{{2}^{{-1}}}+{{2}^{{-2}}}}}=\frac{{\frac{1}{{{{2}^{4}}}}}}{{\frac{1}{2}+\frac{1}{4}}}\\&=\frac{{\frac{1}{{{{2}^{4}}}}}}{{\frac{2}{4}+\frac{1}{4}}}=\frac{{\frac{1}{{{{2}^{4}}}}}}{{\frac{3}{4}}}\\&=\frac{1}{{{}_{4}\cancel{{16}}}}\cdot \frac{{{{{\cancel{4}}}^{1}}}}{3}=\frac{1}{{12}}\end{align}$ We can’t push the –1 exponent through since we’re adding on the top and not multiplying.

 

Flip the fraction, and then do the math with each term separately.

$ \displaystyle {{\left( {\frac{{{{2}^{{-1}}}+{{2}^{{-2}}}}}{{{{2}^{{-4}}}}}} \right)}^{{-1}}}$ $ \displaystyle \begin{align}{{\left( {\frac{{{{2}^{{-1}}}+{{2}^{{-2}}}}}{{{{2}^{{-4}}}}}} \right)}^{{-1}}}&=\frac{{{{2}^{{-4}}}}}{{{{2}^{{-1}}}+{{2}^{{-2}}}}}\,\,\times \,\,\frac{{{{2}^{4}}}}{{{{2}^{4}}}}\\&=\frac{{\left( {{{2}^{{-4}}}} \right)\left( {{{2}^{4}}} \right)}}{{{{2}^{{-1}}}\left( {{{2}^{4}}} \right)+{{2}^{{-2}}}\left( {{{2}^{4}}} \right)}}\\&=\frac{1}{{{{2}^{3}}+{{2}^{2}}}}=\frac{1}{{12}}\end{align}$ Easier way to do problems like this: If the same base is on the top and bottom, multiply by 1, with the numerator and denominator being that base with the positive exponent that is the opposite of the smallest (largest negative) of all the exponents; in our case, multiply by $ \displaystyle \frac{{{{2}^{4}}}}{{{{2}^{4}}}}$, since –4 is the smallest exponent.

Here are even more examples. Assume variables under radicals are non-negative.

Expression Simplification Explanation

$ \sqrt[4]{{64{{a}^{7}}{{b}^{8}}}}$

$ \displaystyle \begin{align}\sqrt[4]{{64{{a}^{7}}{{b}^{8}}}}&=\left( {\sqrt[4]{{64}}} \right)\sqrt[4]{{{{a}^{7}}{{b}^{8}}}}\\&=\left( {\sqrt[4]{{16}}} \right)\left( {\sqrt[4]{4}} \right)\left( {\sqrt[4]{{{{a}^{7}}}}} \right)\sqrt[4]{{{{b}^{8}}}}\\&=2\left( {\sqrt[4]{4}} \right){{a}^{1}}\sqrt[4]{{{{a}^{3}}}}{{b}^{2}}\\&=2a{{b}^{2}}\sqrt[4]{{4{{a}^{3}}}}\end{align}$ METHOD 1:

Separate the coefficients (numbers) and variables. With $ \sqrt[4]{{64}}$, factor 64 into 16 and 4, since $ \displaystyle \sqrt[4]{{16}}=2$. The 4th root of $ {{a}^{7}}$ is  $ a\,\sqrt[4]{{{{a}^{3}}}}$, since 4 goes into 7 one time (take one $ a$ out), and there’s 3 left over (to get the $ {{a}^{3}}$). The 4th root of $ {{b}^{8}}$ is $ {{b}^{2}}$, since 4 goes into 8 exactly 2 times. Put it all together, combining the radicals.

 $ \sqrt[4]{{64{{a}^{7}}{{b}^{8}}}}$ $ \displaystyle \begin{align}\sqrt[4]{{64{{a}^{7}}{{b}^{8}}}}&={{\left( {64{{a}^{7}}{{b}^{8}}} \right)}^{{\frac{1}{4}}}}\\&={{\left( {64} \right)}^{{\frac{1}{4}}}}{{\left( {{{a}^{7}}} \right)}^{{\frac{1}{4}}}}{{\left( {{{b}^{8}}} \right)}^{{\frac{1}{4}}}}\\&={{\left( {16} \right)}^{{\frac{1}{4}}}}{{\left( 4 \right)}^{{\frac{1}{4}}}}{{a}^{{\frac{7}{4}}}}{{b}^{{\frac{8}{4}}}}\\&=2{{\left( 4 \right)}^{{\frac{1}{4}}}}{{a}^{{\frac{4}{4}}}}{{a}^{{\frac{3}{4}}}}{{b}^{2}}\\&=2a{{b}^{2}}{{\left( {4{{a}^{3}}} \right)}^{{\frac{1}{4}}}}=2a{{b}^{2}}\sqrt[4]{{4{{a}^{3}}}}\end{align}$ METHOD 2:

Turn the fourth root into a rational (fractional) exponent and “carry it through”.  When we end up with exponential “improper fractions” (numerator > denominator), separate the exponents (almost like “mixed fractions”) and the move the variables with integer exponents to the outside.

With $ {{64}^{{\frac{1}{4}}}}$, factor it into 16 and 4, since $ {{16}^{{\frac{1}{4}}}}$ is 2.

$ 6{{x}^{2}}\sqrt{{48{{y}^{2}}}}-4y\sqrt{{27{{x}^{4}}}}$  $ \begin{align}6{{x}^{2}}\sqrt{{48{{y}^{2}}}}-4y\sqrt{{27{{x}^{4}}}}\\=6{{x}^{2}}y\sqrt{{16\cdot 3}}-4{{x}^{2}}y\sqrt{{9\cdot 3}}\\=6\cdot 4\cdot {{x}^{2}}y\sqrt{3}-3\cdot 4{{x}^{2}}y\sqrt{3}\\=24\sqrt{3}{{x}^{2}}y-12\sqrt{3}{{x}^{2}}y\\=12\sqrt{3}{{x}^{2}}y\end{align}$ Simplify the roots (both coefficents and variables) by taking out squares. Then, combine like terms, where you need to have the same root and variables.

We could have turned the roots into fractional exponents and gotten the same answer – it’s a matter of preference.

$ \displaystyle \sqrt[4]{{\frac{{{{x}^{6}}{{y}^{4}}}}{{162{{z}^{5}}}}}}$

$ \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}$

This one’s pretty complicated since we have to simplify and rationalize. I like to separate the numerator and the denominator first, and then take out everything that can be raised to the 4th under the radicals. Remember the trick with variables, like with the $ x$’s on the top: 4 goes into one time, with 2 left over, so $ x$ is on the outside and $ {{x}^{2}}$ is on the inside.

 

To rationalize, since we have a 4th root, multiply by a radical that has the 3rd root on top and bottom.

$ {{32}^{{\tfrac{3}{5}}}}\cdot {{81}^{{\tfrac{1}{4}}}}\cdot {{27}^{{-\tfrac{1}{3}}}}$

$ \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}$ Each root had a “perfect” answer, so take roots first. Then apply the exponents, and multiply across.

We can also put this one in the calculator (using parentheses around the fractional roots, if necessary).

$ {{9}^{{x-2}}}\cdot {{3}^{{x-1}}}$

$ \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}$ We will visit Matching Exponential Bases here in the Exponential Functions section later, but to add the exponents, get a common base, which is 3.

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. Here’s an example:

Expression Simplification Explanation

$ \sqrt[{}]{{45{{a}^{8}}{{b}^{2}}}}$

 

($ a$ and $ b$ not necessarily positive)

$ \displaystyle \begin{align}\sqrt[{}]{{45{{a}^{8}}{{b}^{2}}}}&=\left( {\sqrt[{}]{{45}}} \right)\sqrt[{}]{{{{a}^{8}}{{b}^{2}}}}\\&=\left( {\sqrt[{}]{9}} \right)\left( {\sqrt[{}]{5}} \right)\left( {\sqrt[{}]{{{{a}^{8}}}}} \right)\left( {\sqrt[{}]{{{{b}^{2}}}}} \right)\\&=3\left( {\sqrt[{}]{5}} \right)\left( {\sqrt[{}]{{{{a}^{8}}}}} \right)\left( {\sqrt[{}]{{{{b}^{2}}}}} \right)\\&=3\left( {\sqrt[{}]{5}} \right){{\left| a \right|}^{4}}\cdot \left| b \right|\\&=3\sqrt{5}{{a}^{4}}\left| b \right|\end{align}$ Separate the coefficients and variables. With $ \sqrt[{}]{{45}}$, factor 45 into 9 and 5, since $ \displaystyle \sqrt{9}=3$.

 

If $ a$ is positive, the square root of $ {{a}^{8}}$ is $ {{a}^{4}}$, since 2 goes into 8 two times. At this point, it doesn’t matter if $ a$ is positive or negative; the result will be positive. If $ b$ is negative, however, when it is squared, it will be positive, and the square root is positive. Therefore, in this case, $ \displaystyle \sqrt{{{{b}^{2}}}}=\left| b \right|$.

 

Again, remember that 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!

For all these examples, see how we’re doing the same steps over and over again – just with different problems? If you don’t get them at first, don’t worry; just try to go over them again. You’ll get it!  And don’t forget that there are many ways to arrive at the same answers!

Solving Exponential and Radical Equations

Note: We’ll see more of these types of problems here in the Solving Radical Equations and Inequalities section. Some of the more complicated problems involve using Quadratics.

Now we’ll try solving equations involving exponents and roots. Generally, to get rid of the exponents, take radicals of both sides, and to get rid of radicals, raise both sides of the equation to that power.

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). 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. Also, all the answers we get may not work, since we can’t take the even roots of negative numbers. Thus, it’s a good idea to always check our answers when we solve for roots (especially even roots)!

Let’s first try some equations with odd exponents and roots, since these are a little more straightforward. Notice when we have fractional exponents, the radical is still odd when the numerator is odd.

Radical Equation Notes
$ \displaystyle {{x}^{3}}-1=26$

$ \displaystyle \begin{align}{{x}^{3}}&=27\\\,\sqrt[3]{{{{x}^{3}}}}&=\sqrt[3]{{27}}\\\,x&=3\end{align}$

Move all the constants (numbers) to the right. To get rid of the $ {{x}^{3}}$, take the cube root of each side; this is the same as raising each side to the $ \displaystyle \frac{1}{3}$ power. (Remember that when we cube a cube root, we end up with what’s under the root sign.)

Since the root is odd, we don’t have to worry about the signs.

Check our answer: $ {{3}^{3}}-1=27-1=26\,\,\,\,\,\,\surd $

$ \displaystyle 2\sqrt[3]{{x+2}}=6$

$ \displaystyle \begin{align}\sqrt[3]{{x+2}}&=3\\{{\left( {\sqrt[3]{{x+2}}} \right)}^{3}}&={{3}^{3}}\\x+2&=27\\x&=25\end{align}$

First, divide both sides by 2; always try to simplify before solving with radicals. To get rid of the radical on the left-hand side, cube both sides. Then solve for $ x$.

Check our answer: $ 2\sqrt[3]{{25+2}}=2(3)=6\,\,\,\,\,\,\surd $

$ {{\left( {y+2} \right)}^{{\frac{3}{2}}}}=8\,\,\,$

$ \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[3]{8}} \right)}^{2}}={{2}^{2}}\\y+2&=4\\y&=2\end{align}$

Raise both sides to the reciprocal of the exponent on the left, so it turns into a 1. Remember that when you raise an exponent to another exponent, you multiply exponents. To raise 8 to the $ \displaystyle \frac{2}{3}$, either use a calculator, or take the cube root of 8 and square it. Then solve for $ y$ by subtracting 2 from each side.

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 we have to make sure $ \left( {y+2} \right)$ 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. If the problem were $ {{\left( {y+2} \right)}^{{\frac{3}{2}}}}=-8$, for example, we would have no solution.

$ 4\sqrt[3]{x}=2\sqrt[3]{{x+7}}\,\,\,\,$

$ \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}$

Since we have the cube root on each side, cube each side. It’s always easier to simply (for example, divide both sides by 2) first, but you don’t have to. Be careful to make sure you cube all the numbers (and anything else on that side) too. Then, solve for $ x$.

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. 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. When we have fractional exponents, the radical is even when the numerator is even.

Also, when the original problem contains an even root sign, we need to check answers to make sure there are no negative numbers under the even root sign (no negative radicands). We also must check answers when we raise both sides to an even exponent (for example, square both sides).

The solutions that don’t work when you put them back in the original equation are called extraneous solutions. Again, we’ll see more of these types of problems in the here in the Solving Radical Equations and Inequalities section.

Radical Equation Notes
$ {{x}^{2}}+9=5$

$ \begin{array}{c}{{x}^{2}}=-4\\\emptyset \text{ or no solution}\end{array}$

Since we can never square any real number and end up with a negative number, there is no real solution for this equation. The solutions are imaginary, like we’ll learn in the Imaginary (Non-Real) and Complex Numbers section.

$ {{x}^{2}}-1=24$

$ \begin{array}{c}{{x}^{2}}=25\\x=\pm 5\end{array}$

Since we’re taking an even root, include both the positive and negative solutions. Remember that the square root sign only gives you the positive solutions.

Check our answers: $ {{\left( 5 \right)}^{2}}-1=24\,\,\,\,\surd \,\,\,\,\,\,\,\,{{\left( {-5} \right)}^{2}}-1=24\,\,\,\,\surd $

$ \sqrt[4]{{x+3}}=2$

$ \begin{array}{c}{{\left( {\sqrt[4]{{x+3}}} \right)}^{4}}={{2}^{4}}\\x+3=16\\x=13\end{array}$

“Undo” the fourth root by raising both sides to the 4th power. Don’t worry about plus and minuses since we’re not taking the root of a number.

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}$

Since we have square roots on both sides, square both sides (including the 4) to get rid of them. Then, solve for $ x$ to get $ \displaystyle \frac{{17}}{{15}}$.

Make sure our answers don’t produce any negative numbers under the square root; this looks good. Also, since we squared both sides, check our answer:

$ \displaystyle 4\sqrt{{\frac{{17}}{{15}}-1}}=\sqrt{{\frac{{17}}{{15}}+1}}\,\,?\,\,\,\,\,\,\,\,\,4\sqrt{{\frac{2}{{15}}}}=\sqrt{{\frac{{\left( {16} \right)\left( 2 \right)}}{{15}}}}\,\,?\,\,\,\,\,\,\,\,\,4\sqrt{{\frac{2}{{15}}}}=4\sqrt{{\frac{2}{{15}}}}\,\,\,\,\surd $

$ \displaystyle {{\left( {x+2} \right)}^{{\frac{4}{3}}}}+2=18$

$ \displaystyle \begin{align}{{\left( {{{{\left( {x+2} \right)}}^{{\frac{4}{3}}}}} \right)}^{{\frac{3}{4}}}}&={{16}^{{\frac{3}{4}}}}\\x+2&=\pm \left( {{{2}^{3}}} \right)\\x&=\pm {{2}^{3}}-2\\x&=8-2=6\,\,\,\,\,\text{and}\\x&=-8-2=-10\end{align}$

You can raise a term to the reciprocal of a fractional exponent to “get rid of it”, or turn it to 1. Be careful though; if an even number is in the numerator of the original fraction, take the positive and negative solutions on the other side. We also have to make sure the that we’re not trying to take an even root of a negative number. Check our answers:

$ \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}$

$ \sqrt{{2-x}}=\sqrt{{x-4}}$

$ \begin{align}{{\left( {\sqrt{{2-x}}} \right)}^{2}}&={{\left( {\sqrt{{x-4}}} \right)}^{2}}\\\,2-x&=x-4\\\,2x&=6\\\,x&=3\end{align}$

$ \emptyset \text{ or no solution}$

We correctly solved the equation but notice that when we plug 3 in the first radical (and the second one, too!), we get a negative number $ (2-3=-1)$. “Throw away” our answer; the correct answer is “no solution” or $ \emptyset $.

 

This shows us that we must plug in our answer when we’re dealing with even roots to identify extraneous roots!

And here’s one more where we’re solving for one variable in terms of the other variables:

Math

Notes

Solve for $ {{y}_{2}}$:

$ \begin{array}{c}\color{#800000}{{d=\sqrt{{{{{\left( {{{x}_{1}}-{{x}_{2}}} \right)}}^{2}}+{{{\left( {{{y}_{1}}-{{y}_{2}}} \right)}}^{2}}}}}}\\{{d}^{2}}={{\left( {{{x}_{1}}-{{x}_{2}}} \right)}^{2}}+{{\left( {{{y}_{1}}-{{y}_{2}}} \right)}^{2}}\\{{d}^{2}}-{{\left( {{{x}_{1}}-{{x}_{2}}} \right)}^{2}}=\,\,{{\left( {{{y}_{1}}-{{y}_{2}}} \right)}^{2}}\\\pm \,\sqrt{{{{d}^{2}}-{{{\left( {{{x}_{1}}-{{x}_{2}}} \right)}}^{2}}}}={{y}_{1}}-{{y}_{2}}\\{{y}_{2}}={{y}_{1}}\pm \,\sqrt{{{{d}^{2}}-{{{\left( {{{x}_{1}}-{{x}_{2}}} \right)}}^{2}}}}\end{array}$

Move variables around until $ {{y}_{2}}$ is on one side. Since we have to get $ {{y}_{2}}$ by itself, first square each side. Then, to get $ {{y}_{1}}-{{y}_{2}}$, take the square root of each side (and don’t forget to take the plus and the minus).

 

Then, solve for $ {{y}_{2}}$. Notice that when the $ \pm $ moved to the other side, it’s still a $ \pm $.

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}}=$”. Then, use the intersection feature to find the solution(s); the solution(s) will be what $ x$ is at that point. Here are those instructions, using an example from above:

Instructions

Screen

Push y= and enter the two equations in Y1=  and Y2=, respectively.

 

Push graph. You may need to hit “zoom 6” (ZStandard) and/or “zoom 0” (ZoomFit) to make sure you see the lines crossing in the graph. (You can also use the window button to change the minimum and maximum values of the $ x$- and $ y$-values.) I also used “zoom 3” (Zoom Out) enter to see the intersections a little better. (You may have to do this a few times).

 

You can see that we have two points of intersections; therefore, we have two solutions.

To get the first point of intersection, push “2nd trace” (calc), and then either push 5, or move the cursor down to intersect. You should see “First curve?” at the bottom. Then push enter, enter, enter. The first point of intersection that it found is $ x=6$.

 

To find the other point of intersection, move the cursor closer to that point by pressing “trace” and an arrow key (it should follow along one of the curves).

 

Repeat with “2nd trace” (calc), 5, enter, enter, enter. You should see the second solution is $ x=-10$. These are the answers we got above!

Solving Radical Inequalities

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.

Just like we had to solve linear inequalities, we also have to learn how to solve inequalities that involve exponents and radicals (roots). 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.

We also must make sure our answer takes into account what we call the domain restriction: we must make sure what’s under an even radical is 0 or positive, so we may have to create another inequality. To do this, we’ll set what’s under the even radical to greater than or equal to 0, solve for $ x$, Then we take the intersection of both solutions. The reason we take the intersection of the two solutions is because both must work. With odd roots, we don’t have to worry – we just raise each side that power, and solve!

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. It gets trickier when we don’t know the sign of one of the sides.

Radical Inequality Problem

Notes

$ \sqrt{{x+2}}\le 4$

$ \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}$

Solve two inequalities, since $ x$ must work in the original, but also make sure the radical in the original problem is $ \ge 0$. To get rid of the square roots, square each side, and leave the inequality signs the same since we’re multiplying by positive numbers.

 

Check our answer by trying random numbers in our solution (like $ x=2$) in the original inequality (which works). We should also try numbers outside our solution (like $ x=-6$ and $ x=20$) to see that they don’t work.

$ \displaystyle \sqrt{{5x-16}}<\sqrt{{2x-4}}$

$ \displaystyle \begin{align}{{\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{align}$

Solve three inequalities: one for the main problem, and one each for the even root radicands that have to be $ \ge 0$.

 

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 will give us $ \displaystyle \frac{{16}}{5}\le x<4$. (Try it yourself on a number line). Watch out for the hard and soft brackets.

 

Check our answer by trying $ x=3.5$ in the original inequality (which works) and $ x=3$ or $ x=5$ (which don’t work).

$ \sqrt{{x+6}}\le -2$

$ \{\}\text{ }\,\,\text{ or }\emptyset $

Before we even need to get started with this inequality, notice that the square root of anything can never be $ \displaystyle \le -2$  ($ <0$), by definition.

The answer is no solution, or {}, or $ \emptyset $.

(Note that for $ \sqrt{{x+6}}\ge -2$, the answer would be $ x\ge -6$, since we only have to worry about what’s under the square root sign being positive or 0).

$ \,\sqrt[3]{{x-3}}>4\,\,\,$

$ \begin{array}{c}{{\left( {\sqrt[3]{{x-3}}} \right)}^{3}}>{{4}^{3}}\,\,\,\,\\x-3>64\\x>67\\\{x:\,\,x>67\}\text{ or }\left( {67,\infty } \right)\end{array}$

With odd roots, we don’t have to worry about checking underneath the radical sign, since we can have positive or negative numbers as a radicand.

Learn these rules, and practice, practice, practice!


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On to Introduction to Multiplying Polynomials – you are ready!