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In our work with simplifying square root expressions, we noted that
$$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$$
Since this is an equation, we may write it as
$$\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$$
To divide two square root expressions, we use the division property of square roots.
As we can see by observing the right side of the equation governing the division of square roots, the process may produce a fraction in the radicand. This means, of course, that the square root expression is not in simplified form. It is sometimes more useful to rationalize the denominator of a square root expression before actually performing the division.
Simplify the square root expressions.
$\sqrt{\frac{3}{7}}.$
This radical expression is not in simplified form since there is a fraction under the radical sign. We can eliminate this problem using the division property of square roots.
$$\sqrt{\frac{3}{7}}=\frac{\sqrt{3}}{\sqrt{7}}=\frac{\sqrt{3}}{\sqrt{7}}\xb7\frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{3}\sqrt{7}}{7}=\frac{\sqrt{21}}{7}$$
$\frac{\sqrt{5}}{\sqrt{3}}.$
A direct application of the rule produces
$\sqrt{\frac{5}{3}},$ which must be simplified. Let us rationalize the denominator before we perform the division.
$$\frac{\sqrt{5}}{\sqrt{3}}=\frac{\sqrt{5}}{\sqrt{3}}\xb7\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{5}\sqrt{3}}{3}=\frac{\sqrt{15}}{3}$$
$\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{\frac{21}{7}}=\sqrt{3}.$
The rule produces the quotient quickly. We could also rationalize the denominator first and produce the same result.
$$\frac{\sqrt{21}}{\sqrt{7}}=\frac{\sqrt{21}}{7}\xb7\frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{21\xb77}}{7}=\frac{\sqrt{3\xb77\xb77}}{7}=\frac{\sqrt{3\xb7{7}^{2}}}{7}=\frac{7\sqrt{3}}{7}=\sqrt{3}$$
$$\frac{\sqrt{80{x}^{9}}}{\sqrt{5{x}^{4}}}=\sqrt{\frac{80{x}^{9}}{5{x}^{4}}}=\sqrt{16{x}^{5}}=\sqrt{16}\sqrt{{x}^{4}x}=4{x}^{2}\sqrt{x}$$
$$\frac{\sqrt{50{a}^{3}{b}^{7}}}{\sqrt{5a{b}^{5}}}=\sqrt{\frac{50{a}^{3}{b}^{7}}{5a{b}^{5}}}=\sqrt{10{a}^{2}{b}^{2}}=ab\sqrt{10}$$
$\frac{\sqrt{5a}}{\sqrt{b}}.$
Some observation shows that a direct division of the radicands will produce a fraction. This suggests that we rationalize the denominator first.
$$\frac{\sqrt{5a}}{\sqrt{b}}=\frac{\sqrt{5a}}{\sqrt{b}}\xb7\frac{\sqrt{b}}{\sqrt{b}}=\frac{\sqrt{5a}\sqrt{b}}{b}=\frac{\sqrt{5ab}}{b}$$
$$\frac{\sqrt{m-6}}{\sqrt{m+2}}=\frac{\sqrt{m-6}}{\sqrt{m+2}}\xb7\frac{\sqrt{m+2}}{\sqrt{m+2}}=\frac{\sqrt{{m}^{2}-4m-12}}{m+2}$$
$$\frac{\sqrt{{y}^{2}-y-12}}{\sqrt{y+3}}=\sqrt{\frac{{y}^{2}-y-12}{y+3}}=\sqrt{\frac{\left(y+3\right)\left(y-4\right)}{\left(y+3\right)}}=\sqrt{\frac{\overline{)\left(y+3\right)}\left(y-4\right)}{\overline{)\left(y+3\right)}}}=\sqrt{y-4}$$
Simplify the square root expressions.
$\frac{\sqrt{80{m}^{5}{n}^{8}}}{\sqrt{5{m}^{2}n}}$
$4m{n}^{3}\sqrt{mn}$
$\frac{\sqrt{196{\left(x+7\right)}^{8}}}{\sqrt{2{\left(x+7\right)}^{3}}}$
$7{\left(x+7\right)}^{2}\sqrt{2\left(x+7\right)}$
$\frac{\sqrt{n+4}}{\sqrt{n-5}}$
$\frac{\sqrt{{n}^{2}-n-20}}{n-5}$
To perform a division that contains a binomial in the denominator, such as $\frac{3}{4+\sqrt{6}},$ we multiply the numerator and denominator by a conjugate of the denominator.
Notice that when the conjugates
$a+b$ and
$a-b$ are multiplied together, they produce a difference of two squares.
$\left(a+b\right)\left(a-b\right)={a}^{2}-ab+ab-{b}^{2}={a}^{2}-{b}^{2}$
This principle helps us eliminate square root radicals, as shown in these examples that illustrate finding the product of conjugates.
$\begin{array}{lll}\left(5+\sqrt{2}\right)\left(5-\sqrt{2}\right)\hfill & =\hfill & {5}^{2}-{\left(\sqrt{2}\right)}^{2}\hfill \\ \hfill & =\hfill & 25-2\hfill \\ \hfill & =\hfill & 23\hfill \end{array}$
$\begin{array}{lll}\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{6}+\sqrt{7}\right)\hfill & =\hfill & {\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{7}\right)}^{2}\hfill \\ \hfill & =\hfill & 6-7\hfill \\ \hfill & =\hfill & -1\hfill \end{array}$
Simplify the following expressions.
$\frac{3}{4+\sqrt{6}}$ .
The conjugate of the denominator is
$4-\sqrt{6.}$ Multiply the fraction by 1 in the form of
$\frac{4-\sqrt{6}}{4-\sqrt{6}}$ .
$$\begin{array}{lll}\frac{3}{4+\sqrt{6}}\xb7\frac{4-\sqrt{6}}{4-\sqrt{6}}\hfill & =\hfill & \frac{3\left(4-\sqrt{6}\right)}{{4}^{2}-{\left(\sqrt{6}\right)}^{2}}\hfill \\ \hfill & =\hfill & \frac{12-3\sqrt{6}}{16-6}\hfill \\ \hfill & =\hfill & \frac{12-3\sqrt{6}}{10}\hfill \end{array}$$
$\frac{\sqrt{2x}}{\sqrt{3}-\sqrt{5x}}.$
The conjugate of the denominator is
$\sqrt{3}+\sqrt{5x.}$ Multiply the fraction by 1 in the form of
$\frac{\sqrt{3}+\sqrt{5x}}{\sqrt{3}+\sqrt{5x}}.$
$$\begin{array}{lll}\frac{\sqrt{2x}}{\sqrt{3}-\sqrt{5x}}\xb7\frac{\sqrt{3}+\sqrt{5x}}{\sqrt{3}+\sqrt{5x}}\hfill & =\hfill & \frac{\sqrt{2x}(\sqrt{3}+\sqrt{5x})}{{(\sqrt{3})}^{2}-{(\sqrt{5x})}^{2}}\hfill \\ \hfill & =\hfill & \frac{\sqrt{2x}\sqrt{3}+\sqrt{2x}\sqrt{5x}}{3-5x}\hfill \\ \hfill & =\hfill & \frac{\sqrt{6x}+\sqrt{10{x}^{2}}}{3-5x}\hfill \\ \hfill & =\hfill & \frac{\sqrt{6x}+x\sqrt{10}}{3-5x}\hfill \end{array}$$
Simplify the following expressions.
$\frac{\sqrt{8}}{\sqrt{3x}+\sqrt{2x}}$
$\frac{2\sqrt{6x}-4\sqrt{x}}{x}$
$\frac{\sqrt{2m}}{m-\sqrt{3m}}$
$\frac{\sqrt{2m}+\sqrt{6}}{m-3}$
For the following problems, simplify each expressions.
$\frac{\sqrt{200}}{\sqrt{10}}$
$\frac{\sqrt{96}}{\sqrt{24}}$
$\frac{\sqrt{336}}{\sqrt{21}}$
$\sqrt{\frac{25}{9}}$
$\sqrt{\frac{225}{16}}$
$\sqrt{\frac{3}{5}}$
$\sqrt{\frac{1}{2}}$
$\sqrt{\frac{11}{25}}$
$\sqrt{\frac{5}{16}}$
$\sqrt{\frac{32}{49}}$
$\frac{\sqrt{125{x}^{5}}}{\sqrt{5{x}^{3}}}$
$\frac{\sqrt{162{a}^{11}}}{\sqrt{2{a}^{5}}}$
$\frac{\sqrt{48{x}^{9}}}{\sqrt{3{x}^{2}}}$
$\frac{\sqrt{125{a}^{14}}}{\sqrt{5{a}^{5}}}$
$5{a}^{4}\sqrt{a}$
$\frac{\sqrt{27{a}^{10}}}{\sqrt{3{a}^{5}}}$
$\frac{\sqrt{108{x}^{21}}}{\sqrt{3{x}^{4}}}$
$6{x}^{8}\sqrt{x}$
$\frac{\sqrt{48{x}^{6}{y}^{7}}}{\sqrt{3xy}}$
$\frac{\sqrt{45{a}^{3}{b}^{8}{c}^{2}}}{\sqrt{5a{b}^{2}c}}$
$3a{b}^{3}\sqrt{c}$
$\frac{\sqrt{66{m}^{12}{n}^{15}}}{\sqrt{11m{n}^{8}}}$
$\frac{\sqrt{30{p}^{5}{q}^{14}}}{\sqrt{5{q}^{7}}}$
${p}^{2}{q}^{3}\sqrt{6pq}$
$\frac{\sqrt{b}}{\sqrt{5}}$
$\frac{\sqrt{2{a}^{3}b}}{\sqrt{14a}}$
$\frac{\sqrt{3{m}^{4}{n}^{3}}}{\sqrt{6m{n}^{5}}}$
$\frac{m\sqrt{2m}}{2n}$
$\frac{\sqrt{5{\left(p-q\right)}^{6}{\left(r+s\right)}^{4}}}{\sqrt{25{\left(r+s\right)}^{3}}}$
$\frac{\sqrt{r+1}}{\sqrt{r-1}}$
$\frac{\sqrt{s+3}}{\sqrt{s-3}}$
$\frac{\sqrt{{s}^{2}-9}}{s-3}$
$\frac{\sqrt{{a}^{2}+3a+2}}{\sqrt{a+1}}$
$\frac{\sqrt{{x}^{2}-2x-8}}{\sqrt{x+2}}$
$\frac{\sqrt{2{x}^{2}-x-1}}{\sqrt{x-1}}$
$\frac{1}{1+\sqrt{x}}$
$\frac{2}{1-\sqrt{a}}$
$\frac{2\left(1+\sqrt{a}\right)}{1-a}$
$\frac{-6}{\sqrt{5}-1}$
$\frac{3}{\sqrt{3}-\sqrt{2}}$
$\frac{\sqrt{5}}{\sqrt{8}-\sqrt{6}}$
$\frac{\sqrt{7x}}{2-\sqrt{5x}}$
$\frac{\sqrt{6y}}{1+\sqrt{3y}}$
$\frac{\sqrt{6y}-3y\sqrt{2}}{1-3y}$
$\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}$
$\frac{a-\sqrt{ab}}{a-b}$
$\frac{\sqrt{{8}^{3}{b}^{5}}}{4-\sqrt{2ab}}$
$\frac{\sqrt{7x}}{\sqrt{5x}+\sqrt{x}}$
$\frac{\sqrt{35}-\sqrt{7}}{4}$
$\frac{\sqrt{3y}}{\sqrt{2y}-\sqrt{y}}$
( [link] ) Simplify ${x}^{8}{y}^{7}\left(\frac{{x}^{4}{y}^{8}}{{x}^{3}{y}^{4}}\right).$
${x}^{9}{y}^{11}$
( [link] ) Solve the compound inequality $-8\le 7-5x\le -23.$
( [link] ) The symbol $\sqrt{x}$ represents which square root of the number $x,\text{\hspace{0.17em}}x\ge 0$ ?
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