Rules of Roots
What is the nth root?
The number that must be multiplied itself n times to equal a given value. The nth root of x is written $ \displaystyle \sqrt[n]{x}$ or $ \displaystyle {{x}^{{\frac{1}{n}}}}$.
The rules below are a subset of the rules of exponents, because roots are the inverse operations of exponentiation.
Definitions
1. $ \displaystyle b=\sqrt[n]{a}$ if both $ \displaystyle b\ge 0$ and $ \displaystyle {{b}^{n}}=a$
$ \displaystyle \sqrt[4]{{16}}=2$ because $ \displaystyle {{2}^{4}}=16$
2. If n is odd then $ \displaystyle \sqrt[n]{{{{a}^{n}}}}=a$
$ \displaystyle \sqrt[5]{{{{{(3)}}^{5}}}}=3$
3. If n is even then $ \displaystyle \sqrt[n]{{{{a}^{n}}}}=\left| a \right|$
$ \displaystyle \sqrt[6]{{{{{(-3)}}^{6}}}}=\left| {-3} \right|=3$
4. If $ \displaystyle a\ge 0$ then $ \displaystyle \sqrt[n]{{{{a}^{n}}}}=a$
$ \displaystyle \sqrt[3]{{{{{({{{\log }}_{5}}7)}}^{3}}}}={{\log }_{5}}7$
For all the above rules for n even, $ \displaystyle a\ge 0$, for n odd $ \displaystyle a\in R$,$ \displaystyle b\in R$.
Distributing Rules
First Rule
$ \displaystyle \sqrt[n]{{a\cdot b}}=\sqrt[n]{a}\cdot \sqrt[n]{b}$
Second Rule
$ \displaystyle \sqrt[n]{{\frac{a}{b}}}=\frac{{\sqrt[n]{a}}}{{\sqrt[n]{b}}}$
Third Rule
$ \displaystyle \sqrt[n]{{{{a}^{m}}}}={{(\sqrt[n]{a})}^{m}}$ for $ \displaystyle m\ge 0$
Fourth Rule
$ \displaystyle \sqrt[n]{a}\cdot \sqrt[n]{a}\cdot \sqrt[n]{a}\cdot \cdot \cdot \cdot \cdot \cdot \sqrt[n]{a}=a$
$ \displaystyle \sqrt[n]{a}$ multiplied n times by itself equal a.
Derived Rules
$ \displaystyle \sqrt[n]{a}\cdot \sqrt[m]{b}=\sqrt[{mn}]{{{{a}^{m}}{{b}^{n}}}}$
$ \displaystyle \frac{{\sqrt[n]{a}}}{{\sqrt[m]{b}}}=\frac{{\sqrt[{mn}]{{{{a}^{m}}}}}}{{\sqrt[{mn}]{{{{b}^{m}}}}}}=\sqrt[{mn}]{{\frac{{{{a}^{m}}}}{{{{b}^{n}}}}}}$
$ \displaystyle {{(\sqrt[n]{{{{a}^{m}}}})}^{x}}=\sqrt[n]{{{{a}^{{mx}}}}}$
$ \displaystyle (\sqrt[n]{{{{a}^{m}}}})=\sqrt[{np}]{{{{a}^{{mp}}}}}$
$ \displaystyle \sqrt[m]{{\sqrt[n]{a}}}=\sqrt[{mn}]{a}$
$ \displaystyle \frac{x}{{\sqrt[n]{a}}}=\frac{{x(\sqrt[n]{{{{a}^{{n-1}}}}})}}{a}$
Be careful!
$ \displaystyle \sqrt[n]{{a+b}}\ne \sqrt[n]{a}+\sqrt[n]{b}$
$ \displaystyle \sqrt[n]{{a-b}}\ne \sqrt[n]{a}-\sqrt[n]{b}$
$ \displaystyle \sqrt[n]{{{{a}^{n}}+{{b}^{n}}}}\ne a+b$
What is square root?
A nonnegative number that needs to multiplies with itself to equal a given number. The square root of x is written $ \displaystyle \sqrt{x}$ or $ \displaystyle {{x}^{{\frac{1}{2}}}}$
Definitions
$ \displaystyle \sqrt{{{{a}^{2}}}}=\left| a \right|=\left\{ \begin{array}{l}a~~~~for~~a\ge 0~~\\-a~for~~~a\ge 0~~~~\end{array} \right.$
$ \displaystyle \sqrt{{{{{(-6)}}^{2}}}}=\left| {-6} \right|=6$
$ \displaystyle \sqrt{{{{{(7)}}^{2}}}}=7$
Distributing Rules
First Rule
$ \displaystyle \sqrt{{a\cdot b}}=\sqrt{a}\cdot \sqrt{b}$
Second Rule
$ \displaystyle \sqrt{{\frac{a}{b}}}=\frac{{\sqrt{a}}}{{\sqrt{b}}}$ $ \displaystyle (b\ne 0)$
Third Rule
$ \displaystyle \sqrt{a}\sqrt{a}=a$
Fourth Rule
$ \displaystyle \sqrt{{{{a}^{n}}}}={{(\sqrt{a})}^{n}}$
Be careful!
$ \displaystyle \sqrt{{a+b}}\ne \sqrt{a}+\sqrt{b}$
$ \displaystyle \sqrt{{a-b}}\ne \sqrt{a}-\sqrt{b}$
$\displaystyle \sqrt{{{{a}^{2}}+{{b}^{2}}}}\ne a+b$
Example 1: Simplify by using the rules of roots and exponents.
a) $ \displaystyle \sqrt[5]{{\frac{{{{{15}}^{{10}}}-{{{10}}^{{10}}}}}{{{{3}^{{10}}}-{{2}^{{10}}}}}}}$
$ \displaystyle \sqrt[5]{{\frac{{{{{15}}^{{10}}}-{{{10}}^{{10}}}}}{{{{3}^{{10}}}-{{2}^{{10}}}}}}}=$
$ \displaystyle \sqrt[5]{{\frac{{{{{(5\cdot 3)}}^{{10}}}-{{{(2\cdot 5)}}^{{10}}}}}{{{{3}^{{10}}}-{{2}^{{10}}}}}}}=$
$ \displaystyle \sqrt[5]{{\frac{{{{5}^{{10}}}({{3}^{{10}}}-{{2}^{{10}}})}}{{{{3}^{{10}}}-{{2}^{{10}}}}}}}=$
$ \displaystyle \sqrt[5]{{\frac{{{{5}^{{10}}}\cancel{{({{3}^{{10}}}-{{2}^{{10}}})}}}}{{\cancel{{{{3}^{{10}}}-{{2}^{{10}}}}}}}}}=$
$ \displaystyle \sqrt[5]{{{{5}^{{10}}}}}=$
$ \displaystyle {{5}^{{\frac{{10}}{5}}}}=$
$ \displaystyle {{5}^{2}}=25$
b) $ \displaystyle \frac{{3+\sqrt{{18}}+\sqrt{5}+\sqrt{{10}}}}{{3+\sqrt{5}}}-\sqrt{2}$
$ \displaystyle \frac{{3+\sqrt{{18}}+\sqrt{5}+\sqrt{{10}}}}{{3+\sqrt{5}}}-\sqrt{2}=$
$ \displaystyle \frac{{\cancel{{3+\sqrt{5}}}}}{{\cancel{{3+\sqrt{5}}}}}+\frac{{\sqrt{{18}}+\sqrt{{10}}}}{{3+\sqrt{5}}}-\sqrt{2}=$
$ \displaystyle 1+\frac{{\sqrt{{2\cdot 9}}+\sqrt{{2\cdot 5}}}}{{3+\sqrt{5}}}-\sqrt{2}=$
$ \displaystyle 1+\frac{{\sqrt{2}\sqrt{9}+\sqrt{2}\sqrt{5}}}{{3+\sqrt{5}}}-\sqrt{2}=$
$ \displaystyle 1+\frac{{\sqrt{2}\cancel{{(3+\sqrt{5})}}}}{{\cancel{{3+\sqrt{5}}}}}-\sqrt{2}$
$\displaystyle 1+\cancel{{\sqrt{2}}}-\cancel{{\sqrt{2}}}=$
$\displaystyle 1+0=1$
c) $ \displaystyle \sqrt{{{{{\left( {\sqrt{3}-2} \right)}}^{{-2}}}}}$
$ \displaystyle \sqrt{{{{{\left( {\sqrt{3}-2} \right)}}^{{-2}}}}}=$$ \displaystyle {{\left( {\sqrt{3}-2} \right)}^{{-2\cdot \frac{1}{2}}}}=$$ \displaystyle {{\left( {\sqrt{3}-2} \right)}^{{-1}}}=$
$\displaystyle \frac{1}{{\sqrt{3}-2}}=$$\displaystyle \frac{{(\sqrt{3}+2)}}{{(\sqrt{3}-2)(\sqrt{3}+2)}}=$$ \displaystyle \frac{{\sqrt{3}+2}}{{3-4}}=$
$ \displaystyle \frac{{\sqrt{3}+2}}{{-1}}=$$ \displaystyle -2-\sqrt{3}$
d) Simplify $\displaystyle \sqrt{{22+\sqrt{{12-\sqrt[3]{{22+\sqrt{{25}}}}}}}}$
Solution
$ \displaystyle \sqrt{{22+\sqrt{{12-\sqrt[3]{{22+\sqrt{{25}}}}}}}}=$ $ \displaystyle \sqrt{{22+\sqrt{{12-\sqrt[3]{{22+5}}}}}}=$ $ \displaystyle \sqrt{{22+\sqrt{{12-\sqrt[3]{{27}}}}}}=$ $ \displaystyle \sqrt{{22+\sqrt{{12-3}}}}=$ $ \displaystyle \sqrt{{22+\sqrt{9}}}=$ $ \displaystyle \sqrt{{22+3}}=$ $ \displaystyle \sqrt{{25}}=5$
Example 2: Find x
a) $ \displaystyle \sqrt[{2x-1}]{{{{{16}}^{{-1}}}}}=\sqrt[3]{{0.125}}$
$ \displaystyle \sqrt[{2x-1}]{{{{2}^{{-4}}}}}=\sqrt[3]{{\frac{{125}}{{1000}}}}$
$ \displaystyle \sqrt[{2x-1}]{{{{2}^{{-4}}}}}=\sqrt[3]{{\frac{1}{8}}}$
$ \displaystyle \sqrt[{2x-1}]{{{{2}^{{-4}}}}}=\sqrt[3]{{{{2}^{{-3}}}}}$
$ \displaystyle {{2}^{{\frac{{-4}}{{2x-1}}}}}={{2}^{{\frac{{-3}}{3}}}}$
$ \displaystyle -\frac{4}{{2x-1}}=-1$
$ \displaystyle -4=-1(2x-1)$
$ \displaystyle -4=-2x+1$
$ \displaystyle 2x=1+4$
$ \displaystyle x=\frac{5}{2}$
b) $ \displaystyle \frac{{\sqrt{{{{3}^{x}}+{{3}^{x}}+{{3}^{x}}}}}}{{\sqrt[3]{{{{3}^{x}}+{{3}^{x}}+{{3}^{x}}}}}}=\frac{1}{3}$
$ \displaystyle \frac{{\sqrt{{3\cdot {{3}^{x}}}}}}{{\sqrt[3]{{3\cdot {{3}^{x}}}}}}=\frac{1}{3}$
$ \displaystyle \frac{{\sqrt{{{{3}^{{1+x}}}}}}}{{\sqrt[3]{{{{3}^{{1+x}}}}}}}={{3}^{{-1}}}$
$ \displaystyle \frac{{{{3}^{{\frac{{x+1}}{2}}}}}}{{{{3}^{{\frac{{x+1}}{3}}}}}}={{3}^{{-1}}}$
$ \displaystyle \frac{{x+1}}{2}-\frac{{x+1}}{3}=-1$
$ \displaystyle \frac{{3(x+1)}}{6}-\frac{{2(x+1)}}{6}=-1$
$ \displaystyle 3x+3-2x-2=-6$
$ \displaystyle x+1=-6$
$ \displaystyle x=-7$
c) $ \displaystyle \frac{{\sqrt{{{{3}^{{6x+4}}}}}}}{{\sqrt[3]{{{{9}^{{x-1}}}}}}}=243$
$ \displaystyle \frac{{{{3}^{{\frac{{6x+4}}{2}}}}}}{{{{9}^{{\frac{{x-1}}{3}}}}}}={{3}^{5}}$
$ \displaystyle \frac{{{{3}^{{\frac{{6x+4}}{2}}}}}}{{{{3}^{{2(\frac{{x-1}}{3})}}}}}={{3}^{5}}$
$ \displaystyle \frac{{{{3}^{{\frac{{6x+4}}{2}}}}}}{{{{3}^{{\frac{{2x-2}}{3}}}}}}={{3}^{5}}$
$ \displaystyle \frac{{6x+4}}{2}-\frac{{2x-2}}{3}=5$
$ \displaystyle \frac{{3(6x+4)}}{6}-\frac{{2(2x-2)}}{6}=5$
$ \displaystyle 3(6x+4)-2(2x-2)=30$
$ \displaystyle 18x+12-4x+4=30$
$ \displaystyle 14x+16=30$
$ \displaystyle 14x=14$
$ \displaystyle x=1$
d) Simplify $ \displaystyle \sqrt[x]{{\frac{{{{4}^{{x+3}}}}}{{64}}}}$
Solution
$ \displaystyle \sqrt[x]{{\frac{{{{4}^{{x+3}}}}}{{64}}}}=$ $ \displaystyle \sqrt[x]{{\frac{{{{4}^{{x+3}}}}}{{{{4}^{3}}}}}}=$ $\displaystyle \sqrt[x]{{{{4}^{{x+3-3}}}}}=$ $ \displaystyle \sqrt[x]{{{{4}^{x}}}}=$ $ \displaystyle {{4}^{{\frac{x}{x}}}}=4$
Example 3: We have $\displaystyle a=\sqrt{3}-1$ and $ \displaystyle b=\sqrt{3}+1$. Find $\displaystyle {{\left( {\frac{1}{a}+\frac{1}{b}} \right)}^{{-\frac{1}{3}}}}$
Solution: $ \displaystyle \left( {\frac{1}{a}+\frac{1}{b}} \right)=\frac{1}{{\sqrt{3}-1}}+\frac{1}{{\sqrt{3}+1}}=$
$ \displaystyle \frac{{\sqrt{3}+1+\sqrt{3}-1}}{{(\sqrt{3}-1)(\sqrt{3}+1)}}=$ $\displaystyle \frac{{2\sqrt{3}}}{{{{{\left( {\sqrt{3}} \right)}}^{2}}-1}}=\frac{{2\sqrt{3}}}{{3-1}}=$ $ \displaystyle \frac{{2\sqrt{3}}}{2}=\sqrt{3}={{3}^{{\frac{1}{2}}}}$
$\displaystyle {{\left( {\frac{1}{a}+\frac{1}{b}} \right)}^{{-\frac{1}{3}}}}={{\left( {{{3}^{{\frac{1}{2}}}}} \right)}^{{-\frac{1}{3}}}}$ $ \displaystyle {{3}^{{^{{\frac{1}{2}\cdot (-\frac{1}{3})}}}}}={{3}^{{-\frac{1}{6}}}}=\frac{1}{{\sqrt[6]{3}}}$
