9.5 Divide Square Roots

Learning objectives

Learning objectives

• Divide square roots

• Rationalize a one-term denominator

• Rationalize a two-term denominator

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.

Example 9.60

Solution

Example 9.61

Solution

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

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

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.

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.

Example 9.62

Solution

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.

Example 9.63

Solution

Example 9.64

Solution

Example 9.65

Solution

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 9.3 shows a portion of a table of squares and square roots. Square roots are approximated to five decimal places in this table.

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.

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.

$\begin{array}{lccl} \mathrm{Suppose we need an approximate value for the fraction.} & & & \frac{1}{\sqrt{2}} \\ \mathrm{A five decimal place approximation to}\,\sqrt{2}\,\mathrm{is}\,1.41421. & & & \frac{1}{1.41421} \\ \mathrm{Without a calculator, would you want to do this division?} & & & 1.414211.0 \\ \end{array}$

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

Now if we need an approximate value, we divide $21.41421$. 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.

To rationalize a denominator, we use the property that ${(\sqrt{a})}^{2}=a$. 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.

Example 9.66

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.

Example 9.67

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 $\sqrt{6}$.

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

Example 9.68

Solution

Example 9.69

Solution

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.

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

Example 9.70

Solution

Example 9.71

Solution

Example 9.72

Solution

Example 9.73

Solution

Section 9.5 Exercises

Section 9.5 Exercises

Divide Square Roots

In the following exercises, simplify.

Rationalize a One-Term Denominator

In the following exercises, simplify and rationalize the denominator.

Rationalize a Two-Term Denominator

In the following exercises, simplify by rationalizing the denominator.

Everyday Math

Writing Exercises

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

4. $\frac{\sqrt{27}}{6}$

5. $\frac{\sqrt{72}}{9}$

1. $\frac{2-\sqrt{32}}{8}$

2. $\frac{6+\sqrt{45}}{6}$

3. $\frac{\sqrt{80}}{\sqrt{125}}$

4. $\frac{\sqrt{128}}{\sqrt{72}}$

5. ⓐ $\frac{\sqrt{8{x}^{6}}}{\sqrt{2{x}^{2}}}$ ⓑ $\frac{\sqrt{200{m}^{5}}}{\sqrt{98m}}$

1. $\frac{\sqrt{75{r}^{3}}}{\sqrt{108r}}$

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

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

4. $\frac{\sqrt{98}}{14}$

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

1. $\frac{\sqrt{96}}{\sqrt{150}}$

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

3. $\frac{10}{\sqrt{6}}$

4. $\frac{6}{\sqrt{7}}$

5. $\frac{3}{\sqrt{13}}$

1. $\frac{10}{3\sqrt{10}}$

2. $\frac{4}{9\sqrt{5}}$

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

4. $\sqrt{\frac{3}{20}}$

5. $\sqrt{\frac{7}{40}}$

1. $\sqrt{\frac{19}{175}}$

2. ⓐ $\frac{3}{3+\sqrt{11}}$ ⓑ $\frac{8}{1-\sqrt{5}}$

3. ⓐ $\frac{5}{5+\sqrt{6}}$ ⓑ $\frac{6}{3-\sqrt{7}}$

4. $\frac{\sqrt{3}}{\sqrt{m}-\sqrt{5}}$

5. $\frac{\sqrt{2}}{\sqrt{x}-\sqrt{6}}$

1. $\frac{\sqrt{r}+\sqrt{5}}{\sqrt{r}-\sqrt{5}}$

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

3. $\frac{15}{\sqrt{5}}$

4. $\sqrt{\frac{8}{54}}$

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

1. $\frac{\sqrt{2}}{\sqrt{x}-\sqrt{3}}$

2. $\frac{\sqrt{x}+\sqrt{8}}{\sqrt{x}-\sqrt{8}}$

3. 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.

4. 1. ⓐ Simplify $\sqrt{\frac{27}{3}}$ and explain all your steps.

2. ⓑ Simplify $\sqrt{\frac{27}{5}}$ and explain all your steps.

3. ⓒ Why are the two methods of simplifying square roots different?

5. 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?