Roots and Radicals

76 Divide Square Roots

Learning Objectives

By the end of this section, you will be able to:

  • Divide square roots
  • Rationalize a one-term denominator
  • Rationalize a two-term denominator

Before you get started, take this readiness quiz.

  1. Find a fraction equivalent to \frac{5}{8} with denominator 48.
    If you missed this problem, review (Figure).
  2. Simplify: {\left(\sqrt{5}\right)}^{2}.
    If you missed this problem, review (Figure).
  3. Multiply: \left(7+3x\right)\left(7-3x\right).
    If you missed this problem, review (Figure).

Divide Square Roots

We know that we simplify fractions by removing factors common to the numerator and the denominator. When we have a fraction with a square root in the numerator, we first simplify the square root. Then we can look for common factors.

This figure shows two columns. The first is labeled “Common Factors” and has 3 times the square root of 2 over 3 times 5 beneath it. Both number threes are red. The second column is labeled “No common factors” and has 2 times the square root of 3 over 3 times 5.

Simplify: \frac{\sqrt{54}}{6}.

Solution

\begin{array}{cccc}& & & \frac{\sqrt{54}}{6}\hfill \\ \text{Simplify the radical.}\hfill & & & \frac{\sqrt{9}·\sqrt{6}}{6}\hfill \\ \text{Simplify.}\hfill & & & \frac{3\sqrt{6}}{6}\hfill \\ \text{Remove the common factors.}\hfill & & & \frac{\overline{)3}\sqrt{6}}{\overline{)3}·2}\hfill \\ \text{Simplify.}\hfill & & & \frac{\sqrt{6}}{2}\hfill \end{array}

Simplify: \frac{\sqrt{32}}{8}.

\frac{\sqrt{2}}{2}

Simplify: \frac{\sqrt{75}}{15}.

\frac{\sqrt{3}}{3}

Simplify: \frac{6-\sqrt{24}}{12}.

Solution

\begin{array}{cccc}& & & \frac{6-\sqrt{24}}{12}\hfill \\ \text{Simplify the radical.}\hfill & & & \frac{6-\sqrt{4}·\sqrt{6}}{12}\hfill \\ \text{Simplify.}\hfill & & & \frac{6-2\sqrt{6}}{12}\hfill \\ \begin{array}{c}\text{Factor the common factor from the}\hfill \\ \text{numerator.}\hfill \end{array}\hfill & & & \frac{2\left(3-\sqrt{6}\right)}{2·6}\hfill \\ \text{Remove the common factors.}\hfill & & & \frac{\overline{)2}\left(3-\sqrt{6}\right)}{\overline{)2}·6}\hfill \\ \text{Simplify.}\hfill & & & \frac{3-\sqrt{6}}{6}\hfill \end{array}

Simplify: \frac{8-\sqrt{40}}{10}.

\frac{4-\sqrt{10}}{5}

Simplify: \frac{10-\sqrt{75}}{20}.

\frac{5-\sqrt{3}}{4}

We have used the Quotient Property of Square Roots to simplify square roots of fractions. The Quotient Property of Square Roots says

\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}},b\ne 0

Sometimes we will need to use the Quotient Property of Square Roots ‘in reverse’ to simplify a fraction with square roots.

\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}},b\ne 0

We will rewrite the Quotient Property of Square Roots so we see both ways together. Remember: we assume all variables are greater than or equal to zero so that their square roots are real numbers.

Quotient Property of Square Roots

If a, b are non-negative real numbers and b\ne 0, then

\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}

We will use the Quotient Property of Square Roots ‘in reverse’ when the fraction we start with is the quotient of two square roots, and neither radicand is a perfect square. When we write the fraction in a single square root, we may find common factors in the numerator and denominator.

Simplify: \frac{\sqrt{27}}{\sqrt{75}}.

Solution

\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\frac{\sqrt{27}}{\sqrt{75}}\hfill \\ \\ \\ \begin{array}{c}\text{Neither radicand is a perfect square, so}\hfill \\ \text{rewrite using the quotient property of}\hfill \\ \text{square roots.}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}\sqrt{\frac{27}{75}}\hfill \\ \\ \\ \begin{array}{c}\text{Remove common factors in the numerator}\hfill \\ \text{and denominator.}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}\sqrt{\frac{\overline{)3}·9}{\overline{)3}·25}}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\sqrt{\frac{9}{25}}\hfill \\ & & & \phantom{\rule{4em}{0ex}}\frac{3}{5}\hfill \end{array}

Simplify: \frac{\sqrt{48}}{\sqrt{108}}.

\frac{2}{3}

Simplify: \frac{\sqrt{96}}{\sqrt{54}}.

\frac{4}{3}

We will use the Quotient Property for Exponents, \frac{{a}^{m}}{{a}^{n}}={a}^{m-n}, when we have variables with exponents in the radicands.

Simplify: \frac{\sqrt{6{y}^{5}}}{\sqrt{2y}}.

Solution

\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\frac{\sqrt{6{y}^{5}}}{\sqrt{2y}}\hfill \\ \\ \\ \begin{array}{c}\text{Neither radicand is a perfect square, so}\hfill \\ \text{rewrite using the quotient property of}\hfill \\ \text{square roots.}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}\sqrt{\frac{6{y}^{5}}{2y}}\hfill \\ \\ \\ \begin{array}{c}\text{Remove common factors in the numerator}\hfill \\ \text{and denominator.}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}\sqrt{\frac{\overline{)2}·3·{y}^{4}·\overline{)y}}{\overline{)2}·\overline{)y}}}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\sqrt{3{y}^{4}}\hfill \\ \\ \\ \text{Simplify the radical.}\hfill & & & \phantom{\rule{4em}{0ex}}{y}^{2}\sqrt{3}\hfill \end{array}

Simplify: \frac{\sqrt{12{r}^{3}}}{\sqrt{6r}}.

n\sqrt{2}

Simplify: \frac{\sqrt{14{p}^{9}}}{\sqrt{2{p}^{5}}}.

{p}^{2}\sqrt{7}

Simplify: \frac{\sqrt{72{x}^{3}}}{\sqrt{162x}}.

Solution

\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\frac{\sqrt{72{x}^{3}}}{\sqrt{162x}}\hfill \\ \\ \\ \begin{array}{c}\text{Rewrite using the quotient property of}\hfill \\ \text{square roots.}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}\sqrt{\frac{72{x}^{3}}{162x}}\hfill \\ \\ \\ \text{Remove common factors.}\hfill & & & \phantom{\rule{4em}{0ex}}\sqrt{\frac{\overline{)18}·4·{x}^{2}·\overline{)x}}{\overline{)18}·9·\overline{)x}}}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\sqrt{\frac{4{x}^{2}}{9}}\hfill \\ \\ \\ \text{Simplify the radical.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{2x}{3}\hfill \end{array}

Simplify: \frac{\sqrt{50{s}^{3}}}{\sqrt{128s}}.

\frac{5s}{8}

Simplify: \frac{\sqrt{75{q}^{5}}}{\sqrt{108q}}.

\frac{5{q}^{2}}{6}

Simplify: \frac{\sqrt{147a{b}^{8}}}{\sqrt{3{a}^{3}{b}^{4}}}.

Solution

\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\frac{\sqrt{147a{b}^{8}}}{\sqrt{3{a}^{3}{b}^{4}}}\hfill \\ \\ \\ \begin{array}{c}\text{Rewrite using the quotient property of}\hfill \\ \text{square roots.}\hfill \end{array}\hfill & & & \phantom{\rule{4em}{0ex}}\sqrt{\frac{147a{b}^{8}}{3{a}^{3}{b}^{4}}}\hfill \\ \\ \\ \text{Remove common factors.}\hfill & & & \phantom{\rule{4em}{0ex}}\sqrt{\frac{49{b}^{4}}{{a}^{2}}}\hfill \\ \\ \\ \text{Simplify the radical.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{7{b}^{2}}{a}\hfill \end{array}

Simplify: \frac{\sqrt{162{x}^{10}{y}^{2}}}{\sqrt{2{x}^{6}{y}^{6}}}.

\frac{9{x}^{2}}{{y}^{2}}

Simplify: \frac{\sqrt{300{m}^{3}{n}^{7}}}{\sqrt{3{m}^{5}n}}.

\frac{10{n}^{3}}{m}

Rationalize a One Term Denominator

Before the calculator became a tool of everyday life, tables of square roots were used to find approximate values of square roots. (Figure) shows a portion of a table of squares and square roots. Square roots are approximated to five decimal places in this table.

A table of square roots was used to find approximate values of square roots before there were calculators.

This table has three solumn and eleven rows. The columns are labeled, “n,” “n squared,” and “the square root of n.” Under the column labeled “n” are the following numbers: 200; 201; 202; 203; 204; 205; 206; 207; 208; 209; and 210. Under the column labeled, “n squared” are the following numbers: 40,000; 40,401; 40,804; 41,209; 41,616; 42,025; 42,436; 42,849; 43,264; 43,681; 44,100. Under the column labeled, “the square root of n” are the following numbers: 14.14214; 14.17745; 14.21267; 14.24781; 14.28286; 14.31782; 14.35270; 14.38749; 14.42221; 14.45683; 14.49138.

If someone needed to approximate a fraction with a square root in the denominator, it meant doing long division with a five decimal-place divisor. This was a very cumbersome process.

For this reason, a process called rationalizing the denominator was developed. A fraction with a radical in the denominator is converted to an equivalent fraction whose denominator is an integer. This process is still used today and is useful in other areas of mathematics, too.

Rationalizing the Denominator

The process of converting a fraction with a radical in the denominator to an equivalent fraction whose denominator is an integer is called rationalizing the denominator.

Square roots of numbers that are not perfect squares are irrational numbers. When we rationalize the denominator, we write an equivalent fraction with a rational number in the denominator.

Let’s look at a numerical example.

\begin{array}{cccc}\text{Suppose we need an approximate value for the fraction.}\hfill & & & \frac{1}{\sqrt{2}}\hfill \\ \text{A five decimal place approximation to}\phantom{\rule{0.2em}{0ex}}\sqrt{2}\phantom{\rule{0.2em}{0ex}}\text{is}\phantom{\rule{0.2em}{0ex}}1.41421.\hfill & & & \frac{1}{1.41421}\hfill \\ \text{Without a calculator, would you want to do this division?}\hfill & & & 1.41421\overline{)1.0}\hfill \end{array}

But we can find a fraction equivalent to \frac{1}{\sqrt{2}} by multiplying the numerator and denominator by \sqrt{2}.

This figure shows three fractions. The first fraction is 1 over the square root of 2. The second is 1 times the square root of 2 over the square root of 2 times the square root of 2. The third shows the square root of 2 over 2.

Now if we need an approximate value, we divide 2\overline{)1.41421}. This is much easier.

Even though we have calculators available nearly everywhere, a fraction with a radical in the denominator still must be rationalized. It is not considered simplified if the denominator contains a square root.

Similarly, a square root is not considered simplified if the radicand contains a fraction.

Simplified Square Roots

A square root is considered simplified if there are

  • no perfect-square factors in the radicand
  • no fractions in the radicand
  • no square roots in the denominator of a fraction

To rationalize a denominator, we use the property that {\left(\sqrt{a}\right)}^{2}=a. If we square an irrational square root, we get a rational number.

We will use this property to rationalize the denominator in the next example.

Simplify: \frac{4}{\sqrt{3}}.

Solution

To rationalize a denominator, we can multiply a square root by itself. To keep the fraction equivalent, we multiply both the numerator and denominator by the same factor.

\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\frac{4}{\sqrt{3}}\hfill \\ \\ \\ \text{Multiply both the numerator and denominator by}\phantom{\rule{0.2em}{0ex}}\sqrt{3}.\hfill & & & \phantom{\rule{4em}{0ex}}\frac{4·\sqrt{3}}{\sqrt{3}·\sqrt{3}}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & \phantom{\rule{4em}{0ex}}\frac{4\sqrt{3}}{3}\hfill \end{array}

Simplify: \frac{5}{\sqrt{3}}.

\frac{5\sqrt{3}}{3}

Simplify: \frac{6}{\sqrt{5}}.

\frac{6\sqrt{5}}{5}

Simplify: -\frac{8}{3\sqrt{6}}.

Solution

To remove the square root from the denominator, we multiply it by itself. To keep the fractions equivalent, we multiply both the numerator and denominator by \sqrt{6}.

.
Multiply both the numerator and the denominator by \sqrt{6}. .
Simplify. .
Remove common factors. .
Simplify. .

Simplify: \frac{5}{2\sqrt{5}}.

\frac{\sqrt{5}}{2}

Simplify: -\frac{9}{4\sqrt{3}}.

-\frac{3\sqrt{3}}{4}

Always simplify the radical in the denominator first, before you rationalize it. This way the numbers stay smaller and easier to work with.

Simplify: \sqrt{\frac{5}{12}}.

Solution
.
The fraction is not a perfect square, so rewrite using the
Quotient Property.
.
Simplify the denominator .
Rationalize the denominator. .
Simplify. .
Simplify. .

Simplify: \sqrt{\frac{7}{18}}.

\frac{\sqrt{14}}{6}

Simplify: \sqrt{\frac{3}{32}}.

\frac{\sqrt{6}}{8}

Simplify: \sqrt{\frac{11}{28}}.

Solution
.
Rewrite using the Quotient Property. .
Simplify the denominator. .
Rationalize the denominator. .
Simplify. .
Simplify. .

Simplify: \sqrt{\frac{3}{27}}.

\frac{1}{3}

Simplify: \sqrt{\frac{10}{50}}.

\frac{\sqrt{5}}{5}

Rationalize a Two-Term Denominator

When the denominator of a fraction is a sum or difference with square roots, we use the Product of Conjugates pattern to rationalize the denominator.

\begin{array}{cccc}\hfill \left(a-b\right)\left(a+b\right)\hfill & & & \hfill \phantom{\rule{2em}{0ex}}\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\hfill \\ \hfill {a}^{2}-{b}^{2}\hfill & & & \hfill \phantom{\rule{2em}{0ex}}{2}^{2}-{\left(\sqrt{5}\right)}^{2}\hfill \\ & & & \hfill \phantom{\rule{2em}{0ex}}4-5\hfill \\ & & & \hfill \phantom{\rule{2em}{0ex}}-1\hfill \end{array}

When we multiply a binomial that includes a square root by its conjugate, the product has no square roots.

Simplify: \frac{4}{4+\sqrt{2}}.

Solution
.
Multiply the numerator and denominator by the conjugate of the denominator. .
Multiply the conjugates in the denominator. .
Simplify the denominator. .
Simplify the denominator. .
Remove common factors from the numerator and denominator. .
We leave the numerator in factored form to make it easier to look for common factors after we have simplified the denominator.

Simplify: \frac{2}{2+\sqrt{3}}.

\frac{2\left(2-\sqrt{3}\right)}{1}

Simplify: \frac{5}{5+\sqrt{3}}.

\frac{5\left(5-\sqrt{3}\right)}{22}

Simplify: \frac{5}{2-\sqrt{3}}.

Solution
.
Multiply the numerator and denominator by the conjugate of the denominator. .
Multiply the conjugates in the denominator. .
Simplify the denominator. .
Simplify the denominator. .
Simplify. .

Simplify: \frac{3}{1-\sqrt{5}}.

-\frac{3\left(1+\sqrt{5}\right)}{4}

Simplify: \frac{2}{4-\sqrt{6}}.

\frac{4+\sqrt{6}}{5}

Simplify: \frac{\sqrt{3}}{\sqrt{u}-\sqrt{6}}.

Solution
.
Multiply the numerator and denominator by the conjugate of the denominator. .
Multiply the conjugates in the denominator. .
Simplify the denominator. .

Simplify: \frac{\sqrt{5}}{\sqrt{x}+\sqrt{2}}.

\frac{\sqrt{5}\left(\sqrt{x}-\sqrt{2}\right)}{x-2}

Simplify: \frac{\sqrt{10}}{\sqrt{y}-\sqrt{3}}.

\frac{\sqrt{10}\left(\sqrt{y}+\sqrt{3}\right)}{y-3}

Simplify: \frac{\sqrt{x}+\sqrt{7}}{\sqrt{x}-\sqrt{7}}.

Solution
.
Multiply the numerator and denominator by the conjugate of the denominator. .
Multiply the conjugates in the denominator. .
Simplify the denominator. .
We do not square the numerator. In factored form, we can see there are no common factors to remove from the numerator and denominator.

Simplify: \frac{\sqrt{p}+\sqrt{2}}{\sqrt{p}-\sqrt{2}}.

{\frac{\left(\sqrt{p}+\sqrt{2}\right)}{p-2}}^{2}

Simplify: \frac{\sqrt{q}-\sqrt{10}}{\sqrt{q}+\sqrt{10}}.

{\frac{\left(\sqrt{q}-\sqrt{10}\right)}{q-10}}^{2}

Access this online resource for additional instruction and practice with dividing and rationalizing.

Key Concepts

  • Quotient Property of Square Roots
    • If a, b are non-negative real numbers and b\ne 0, then
      \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}
  • Simplified Square Roots
    A square root is considered simplified if there are
    • no perfect square factors in the radicand
    • no fractions in the radicand
    • no square roots in the denominator of a fraction

Practice Makes Perfect

Divide Square Roots

In the following exercises, simplify.

\frac{\sqrt{27}}{6}

\frac{\sqrt{3}}{2}

\frac{\sqrt{50}}{10}

\frac{\sqrt{72}}{9}

\frac{2\sqrt{2}}{3}

\frac{\sqrt{243}}{6}

\frac{2-\sqrt{32}}{8}

\frac{1-2\sqrt{2}}{4}

\frac{3+\sqrt{27}}{9}

\frac{6+\sqrt{45}}{6}

\frac{2+\sqrt{5}}{2}

\frac{10-\sqrt{200}}{20}

\frac{\sqrt{80}}{\sqrt{125}}

\frac{4}{5}

\frac{\sqrt{72}}{\sqrt{200}}

\frac{\sqrt{128}}{\sqrt{72}}

\frac{4}{3}

\frac{\sqrt{48}}{\sqrt{75}}

\frac{\sqrt{8{x}^{6}}}{\sqrt{2{x}^{2}}}\frac{\sqrt{200{m}^{5}}}{\sqrt{98m}}

2{x}^{2}\frac{10{m}^{2}}{7}

\frac{\sqrt{10{y}^{3}}}{\sqrt{5y}}\frac{\sqrt{108{n}^{7}}}{\sqrt{243{n}^{3}}}

\frac{\sqrt{75{r}^{3}}}{\sqrt{108r}}

\frac{5r}{6}

\frac{\sqrt{196{q}^{5}}}{\sqrt{484q}}

\frac{\sqrt{108{p}^{5}{q}^{2}}}{\sqrt{34{p}^{3}{q}^{6}}}

\frac{3p\sqrt{102}}{17{q}^{2}}

\frac{\sqrt{98r{s}^{10}}}{\sqrt{2{r}^{3}{s}^{4}}}

\frac{\sqrt{320m{n}^{5}}}{\sqrt{45{m}^{7}{n}^{3}}}

\frac{8n}{3{m}^{3}}

\frac{\sqrt{810{c}^{3}{d}^{7}}}{\sqrt{1000{c}^{5}d}}

\frac{\sqrt{98}}{14}

\frac{\sqrt{2}}{2}

\frac{\sqrt{72}}{18}

\frac{5+\sqrt{125}}{15}

\frac{1+\sqrt{3}}{3}

\frac{6-\sqrt{45}}{12}

\frac{\sqrt{96}}{\sqrt{150}}

\frac{4}{5}

\frac{\sqrt{28}}{\sqrt{63}}

\frac{\sqrt{26{y}^{7}}}{\sqrt{2y}}

{y}^{3}\sqrt{13}

\frac{\sqrt{15{x}^{3}}}{\sqrt{3x}}

Rationalize a One-Term Denominator

In the following exercises, simplify and rationalize the denominator.

\frac{10}{\sqrt{6}}

\frac{5\sqrt{6}}{3}

\frac{8}{\sqrt{3}}

\frac{6}{\sqrt{7}}

\frac{6\sqrt{7}}{7}

\frac{4}{\sqrt{5}}

\frac{3}{\sqrt{13}}

\frac{3\sqrt{13}}{13}

\frac{10}{\sqrt{11}}

\frac{10}{3\sqrt{10}}

\frac{\sqrt{10}}{3}

\frac{2}{5\sqrt{2}}

\frac{4}{9\sqrt{5}}

\frac{4\sqrt{5}}{45}

\frac{9}{2\sqrt{7}}

-\frac{9}{2\sqrt{3}}

-\frac{3\sqrt{3}}{2}

-\frac{8}{3\sqrt{6}}

\sqrt{\frac{3}{20}}

\frac{\sqrt{15}}{10}

\sqrt{\frac{4}{27}}

\sqrt{\frac{7}{40}}

\frac{\sqrt{70}}{20}

\sqrt{\frac{8}{45}}

\sqrt{\frac{19}{175}}

\frac{\sqrt{133}}{35}

\sqrt{\frac{17}{192}}

Rationalize a Two-Term Denominator

In the following exercises, simplify by rationalizing the denominator.

\frac{3}{3+\sqrt{11}}\frac{8}{1-\sqrt{5}}

\frac{3\left(3-\sqrt{11}\right)}{-2}-2\left(1+\sqrt{5}\right)

\frac{4}{4+\sqrt{7}}\frac{7}{2-\sqrt{6}}

\frac{5}{5+\sqrt{6}}\frac{6}{3-\sqrt{7}}

\frac{5\left(5-\sqrt{6}\right)}{19}3\left(3+\sqrt{7}\right)

\frac{6}{6+\sqrt{5}}\frac{5}{4-\sqrt{11}}

\frac{\sqrt{3}}{\sqrt{m}-\sqrt{5}}

\frac{\sqrt{3}\left(\sqrt{m}+\sqrt{5}\right)}{m-5}

\frac{\sqrt{5}}{\sqrt{n}-\sqrt{7}}

\frac{\sqrt{2}}{\sqrt{x}-\sqrt{6}}

\frac{\sqrt{2}\left(\sqrt{x}+\sqrt{3}\right)}{x-6}

\frac{\sqrt{7}}{\sqrt{y}+\sqrt{3}}

\frac{\sqrt{r}+\sqrt{5}}{\sqrt{r}-\sqrt{5}}

{\frac{\left(\sqrt{r}+\sqrt{5}\right)}{r-5}}^{2}

\frac{\sqrt{s}-\sqrt{6}}{\sqrt{s}+\sqrt{6}}

\frac{\sqrt{150{x}^{2}{y}^{6}}}{\sqrt{6{x}^{4}{y}^{2}}}

\frac{5{y}^{2}}{x}

\frac{\sqrt{80{p}^{3}q}}{\sqrt{5p{q}^{5}}}

\frac{15}{\sqrt{5}}

3\sqrt{5}

\frac{3}{5\sqrt{8}}

\sqrt{\frac{8}{54}}

\frac{2\sqrt{3}}{9}

\sqrt{\frac{12}{20}}

\frac{3}{5+\sqrt{5}}

\frac{3\left(5-\sqrt{5}\right)}{20}

\frac{20}{4-\sqrt{3}}

\frac{\sqrt{2}}{\sqrt{x}-\sqrt{3}}

\frac{\sqrt{2}\left(\sqrt{x}+\sqrt{3}\right)}{x-3}

\frac{\sqrt{5}}{\sqrt{y}-\sqrt{7}}

\frac{\sqrt{x}+\sqrt{8}}{\sqrt{x}-\sqrt{8}}

{\frac{\left(\sqrt{x}+2\sqrt{2}\right)}{x-8}}^{2}

\frac{\sqrt{m}-\sqrt{3}}{\sqrt{m}+\sqrt{3}}

Everyday Math

A supply kit is dropped from an airplane flying at an altitude of 250 feet. Simplify \sqrt{\frac{250}{16}} to determine how many seconds it takes for the supply kit to reach the ground.

\frac{5\sqrt{10}}{4}\phantom{\rule{0.2em}{0ex}}\text{seconds}

A flare is dropped into the ocean from an airplane flying at an altitude of 1,200 feet. Simplify \sqrt{\frac{1200}{16}} to determine how many seconds it takes for the flare to reach the ocean.

Writing Exercises


Simplify \sqrt{\frac{27}{3}} and explain all your steps.
Simplify \sqrt{\frac{27}{5}} and explain all your steps.
Why are the two methods of simplifying square roots different?

Answers will vary.

  1. Approximate \frac{1}{\sqrt{2}} by dividing \frac{1}{1.414} using long division without a calculator.
  2. Rationalizing the denominator of \frac{1}{\sqrt{2}} gives \frac{\sqrt{2}}{2}. Approximate \frac{\sqrt{2}}{2} by dividing \frac{1.414}{2} using long division without a calculator.
  3. Do you agree that rationalizing the denominator makes calculations easier? Why or why not?

Self Check

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and four rows. The columns are labeled, “I can…,” “confidently.,” “with some help.,” and “no – I don’t get it!” The rows under the column “I can…” read, “divide square roots,” “rationalize a one term denominator.,” and “rationalize a two term denominator.” All the other rows under the columns are empty.

After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Glossary

rationalizing the denominator
The process of converting a fraction with a radical in the denominator to an equivalent fraction whose denominator is an integer is called rationalizing the denominator.

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Elementary Algebra by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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