Divide Square Roots

By the end of this section, you will be able to: divide square roots, rationalize a one-term denominator, and rationalize a two-term denominator.

Throughout this section we assume all variables are greater than or equal to zero, so that every square root is a real number.

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. Simplify $\tfrac{\sqrt{54}}{6}$ .

Simplify the radical, then remove the common factors:

\begin{array}{lrcl} \text{Simplify the radical.} & \tfrac{\sqrt{54}}{6} &=& \tfrac{\sqrt{9} \cdot \sqrt{6}}{6} \\[10pt] \text{Simplify.} && & \tfrac{3\sqrt{6}}{6} \\[10pt] \text{Remove the common factors.} && & \tfrac{\sqrt{6}}{2} \end{array}

Simplify: $\tfrac{\sqrt{32}}{8}$ .

Simplify: $\tfrac{\sqrt{75}}{15}$ .

When there is a sum or difference in the numerator, we simplify the radical first and then factor out the common factor before removing it.

Example. Simplify $\tfrac{6 - \sqrt{24}}{12}$ .

Simplify the radical, factor the common factor from the numerator, and then remove the common factors:

\begin{array}{lrcl} \text{Simplify the radical.} & \tfrac{6 - \sqrt{24}}{12} &=& \tfrac{6 - \sqrt{4} \cdot \sqrt{6}}{12} \\[10pt] \text{Simplify.} && & \tfrac{6 - 2\sqrt{6}}{12} \\[10pt] \text{Factor the common factor from the numerator.} && & \tfrac{2(3 - \sqrt{6})}{2 \cdot 6} \\[10pt] \text{Remove the common factors.} && & \tfrac{3 - \sqrt{6}}{6} \end{array}

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

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

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

\sqrt{\tfrac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}, \quad b \neq 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{\tfrac{a}{b}}, \quad b \neq 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 $b \neq 0$ , then

\sqrt{\tfrac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \quad \text{and} \quad \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\tfrac{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.

Example. Simplify $\tfrac{\sqrt{27}}{\sqrt{75}}$ .

Neither radicand is a perfect square, so rewrite using the Quotient Property, remove common factors, and simplify:

\begin{array}{lrcl} \text{Rewrite using the Quotient Property.} & \tfrac{\sqrt{27}}{\sqrt{75}} &=& \sqrt{\tfrac{27}{75}} \\[10pt] \text{Remove common factors.} && & \sqrt{\tfrac{9}{25}} \\[10pt] \text{Simplify.} && & \tfrac{3}{5} \end{array}

Simplify: $\tfrac{\sqrt{48}}{\sqrt{108}}$ .

Simplify: $\tfrac{\sqrt{96}}{\sqrt{54}}$ .

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

Example. Simplify $\tfrac{\sqrt{6y^5}}{\sqrt{2y}}$ .

Rewrite using the Quotient Property, remove common factors, and then simplify the radical:

\begin{array}{lrcl} \text{Rewrite using the Quotient Property.} & \tfrac{\sqrt{6y^5}}{\sqrt{2y}} &=& \sqrt{\tfrac{6y^5}{2y}} \\[10pt] \text{Remove common factors.} && & \sqrt{3y^4} \\[10pt] \text{Simplify the radical.} && & y^2\sqrt{3} \end{array}

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

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

Example. Simplify $\tfrac{\sqrt{72x^3}}{\sqrt{162x}}$ .

Rewrite using the Quotient Property, remove common factors, and then simplify the radical:

\begin{array}{lrcl} \text{Rewrite using the Quotient Property.} & \tfrac{\sqrt{72x^3}}{\sqrt{162x}} &=& \sqrt{\tfrac{72x^3}{162x}} \\[10pt] \text{Remove common factors.} && & \sqrt{\tfrac{4x^2}{9}} \\[10pt] \text{Simplify the radical.} && & \tfrac{2x}{3} \end{array}

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

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

Example. Simplify $\tfrac{\sqrt{147ab^8}}{\sqrt{3a^3 b^4}}$ .

Rewrite using the Quotient Property, remove common factors, and then simplify the radical:

\begin{array}{lrcl} \text{Rewrite using the Quotient Property.} & \tfrac{\sqrt{147ab^8}}{\sqrt{3a^3 b^4}} &=& \sqrt{\tfrac{147ab^8}{3a^3 b^4}} \\[10pt] \text{Remove common factors.} && & \sqrt{\tfrac{49b^4}{a^2}} \\[10pt] \text{Simplify the radical.} && & \tfrac{7b^2}{a} \end{array}

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

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

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. 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.

Consider the fraction $\tfrac{1}{\sqrt{2}}$ . A five-decimal-place approximation to $\sqrt{2}$ is $1.41421$ , so without a calculator we would have to divide $1$ by $1.41421$ — an awkward computation. But we can find an equivalent fraction by multiplying the numerator and denominator by $\sqrt{2}$ :

\frac{1}{\sqrt{2}} = \frac{1 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{\sqrt{2}}{2}.

Now if we need an approximate value, we divide $\sqrt{2} \approx 1.41421$ by $2$ . This is much easier.

Even though calculators are 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.

Example. Simplify $\tfrac{4}{\sqrt{3}}$ .

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}{lrcl} \text{Multiply by } \tfrac{\sqrt{3}}{\sqrt{3}}. & \tfrac{4}{\sqrt{3}} &=& \tfrac{4 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} \\[10pt] \text{Simplify.} && & \tfrac{4\sqrt{3}}{3} \end{array}

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

Simplify: $\tfrac{6}{\sqrt{5}}$ .

Rationalize a one-term denominator. To rationalize a denominator with a single square root, multiply the numerator and denominator by

\sqrt{b}

, where

\sqrt{b}

is the square root in the denominator. This makes the denominator

\sqrt{b} \cdot \sqrt{b} = b

, a rational number.

Example. Simplify $-\tfrac{8}{3\sqrt{6}}$ .

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

\begin{array}{lrcl} \text{Multiply by } \tfrac{\sqrt{6}}{\sqrt{6}}. & -\tfrac{8}{3\sqrt{6}} &=& -\tfrac{8 \cdot \sqrt{6}}{3\sqrt{6} \cdot \sqrt{6}} \\[10pt] \text{Simplify.} && & -\tfrac{8\sqrt{6}}{3 \cdot 6} \\[10pt] \text{Remove common factors.} && & -\tfrac{4 \cdot 2\sqrt{6}}{3 \cdot 2 \cdot 3} \\[10pt] \text{Simplify.} && & -\tfrac{4\sqrt{6}}{9} \end{array}

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

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

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

Example. Simplify $\sqrt{\tfrac{5}{12}}$ .

The fraction is not a perfect square, so rewrite it using the Quotient Property, simplify the denominator, and then rationalize:

\begin{array}{lrcl} \text{Rewrite using the Quotient Property.} & \sqrt{\tfrac{5}{12}} &=& \tfrac{\sqrt{5}}{\sqrt{12}} \\[10pt] \text{Simplify the denominator.} && & \tfrac{\sqrt{5}}{2\sqrt{3}} \\[10pt] \text{Rationalize the denominator.} && & \tfrac{\sqrt{5} \cdot \sqrt{3}}{2\sqrt{3} \cdot \sqrt{3}} \\[10pt] \text{Simplify.} && & \tfrac{\sqrt{15}}{2 \cdot 3} \\[10pt] \text{Simplify.} && & \tfrac{\sqrt{15}}{6} \end{array}

Simplify: $\sqrt{\tfrac{7}{18}}$ .

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

Example. Simplify $\sqrt{\tfrac{11}{28}}$ .

Rewrite using the Quotient Property, simplify the denominator, rationalize, and then simplify:

\begin{array}{lrcl} \text{Rewrite using the Quotient Property.} & \sqrt{\tfrac{11}{28}} &=& \tfrac{\sqrt{11}}{\sqrt{28}} \\[10pt] \text{Simplify the denominator.} && & \tfrac{\sqrt{11}}{2\sqrt{7}} \\[10pt] \text{Rationalize the denominator.} && & \tfrac{\sqrt{11} \cdot \sqrt{7}}{2\sqrt{7} \cdot \sqrt{7}} \\[10pt] \text{Simplify.} && & \tfrac{\sqrt{77}}{2 \cdot 7} \\[10pt] \text{Simplify.} && & \tfrac{\sqrt{77}}{14} \end{array}

Simplify: $\sqrt{\tfrac{3}{27}}$ .

Simplify: $\sqrt{\tfrac{10}{50}}$ .

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. Recall that the conjugate of $a + b$ is $a - b$ , and the product of conjugates is a difference of squares:

(a - b)(a + b) = a^2 - b^2.

For example, multiplying $2 - \sqrt{5}$ by its conjugate $2 + \sqrt{5}$ gives

\left(2 - \sqrt{5}\right)\left(2 + \sqrt{5}\right) = 2^2 - \left(\sqrt{5}\right)^2 = 4 - 5 = -1.

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

Example. Simplify $\tfrac{4}{4 + \sqrt{2}}$ .

Multiply the numerator and denominator by the conjugate $4 - \sqrt{2}$ , then simplify the denominator:

\begin{array}{lrcl} \text{Multiply by the conjugate.} & \tfrac{4}{4 + \sqrt{2}} &=& \tfrac{4\left(4 - \sqrt{2}\right)}{\left(4 + \sqrt{2}\right)\left(4 - \sqrt{2}\right)} \\[10pt] \text{Multiply the conjugates in the denominator.} && & \tfrac{4\left(4 - \sqrt{2}\right)}{4^2 - \left(\sqrt{2}\right)^2} \\[10pt] \text{Simplify the denominator.} && & \tfrac{4\left(4 - \sqrt{2}\right)}{16 - 2} \\[10pt] \text{Simplify the denominator.} && & \tfrac{4\left(4 - \sqrt{2}\right)}{14} \\[10pt] \text{Remove common factors.} && & \tfrac{2\left(4 - \sqrt{2}\right)}{7} \end{array}

We leave the numerator in factored form to make it easier to look for common factors after we have simplified the denominator.

Simplify: $\tfrac{2}{2 + \sqrt{3}}$ .

Simplify: $\tfrac{5}{5 + \sqrt{3}}$ .

Example. Simplify $\tfrac{5}{2 - \sqrt{3}}$ .

Multiply the numerator and denominator by the conjugate $2 + \sqrt{3}$ , then simplify the denominator:

\begin{array}{lrcl} \text{Multiply by the conjugate.} & \tfrac{5}{2 - \sqrt{3}} &=& \tfrac{5\left(2 + \sqrt{3}\right)}{\left(2 - \sqrt{3}\right)\left(2 + \sqrt{3}\right)} \\[10pt] \text{Multiply the conjugates in the denominator.} && & \tfrac{5\left(2 + \sqrt{3}\right)}{2^2 - \left(\sqrt{3}\right)^2} \\[10pt] \text{Simplify the denominator.} && & \tfrac{5\left(2 + \sqrt{3}\right)}{4 - 3} \\[10pt] \text{Simplify the denominator.} && & \tfrac{5\left(2 + \sqrt{3}\right)}{1} \\[10pt] \text{Simplify.} && & 5\left(2 + \sqrt{3}\right) \end{array}

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

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

Example. Simplify $\tfrac{\sqrt{3}}{\sqrt{u} - \sqrt{6}}$ .

Multiply the numerator and denominator by the conjugate $\sqrt{u} + \sqrt{6}$ , then simplify the denominator:

\begin{array}{lrcl} \text{Multiply by the conjugate.} & \tfrac{\sqrt{3}}{\sqrt{u} - \sqrt{6}} &=& \tfrac{\sqrt{3}\left(\sqrt{u} + \sqrt{6}\right)}{\left(\sqrt{u} - \sqrt{6}\right)\left(\sqrt{u} + \sqrt{6}\right)} \\[10pt] \text{Multiply the conjugates in the denominator.} && & \tfrac{\sqrt{3}\left(\sqrt{u} + \sqrt{6}\right)}{\left(\sqrt{u}\right)^2 - \left(\sqrt{6}\right)^2} \\[10pt] \text{Simplify the denominator.} && & \tfrac{\sqrt{3}\left(\sqrt{u} + \sqrt{6}\right)}{u - 6} \end{array}

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

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

When both terms of the fraction contain square roots, we do not square the numerator — leaving it in factored form lets us check for common factors to remove.

Example. Simplify $\tfrac{\sqrt{x} + \sqrt{7}}{\sqrt{x} - \sqrt{7}}$ .

Multiply the numerator and denominator by the conjugate $\sqrt{x} + \sqrt{7}$ , then simplify the denominator:

\begin{array}{lrcl} \text{Multiply by the conjugate.} & \tfrac{\sqrt{x} + \sqrt{7}}{\sqrt{x} - \sqrt{7}} &=& \tfrac{\left(\sqrt{x} + \sqrt{7}\right)\left(\sqrt{x} + \sqrt{7}\right)}{\left(\sqrt{x} - \sqrt{7}\right)\left(\sqrt{x} + \sqrt{7}\right)} \\[10pt] \text{Multiply the conjugates in the denominator.} && & \tfrac{\left(\sqrt{x} + \sqrt{7}\right)\left(\sqrt{x} + \sqrt{7}\right)}{\left(\sqrt{x}\right)^2 - \left(\sqrt{7}\right)^2} \\[10pt] \text{Simplify the denominator.} && & \tfrac{\left(\sqrt{x} + \sqrt{7}\right)^2}{x - 7} \end{array}

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: $\tfrac{\sqrt{p} + \sqrt{2}}{\sqrt{p} - \sqrt{2}}$ .

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

Key terms

Quotient Property of Square Roots — for non-negative real numbers $a$ and $b$ with $b \neq 0$ , $\sqrt{\tfrac{a}{b}} = \tfrac{\sqrt{a}}{\sqrt{b}}$ (and its reverse), used to simplify a quotient of square roots. rationalizing the denominator — converting a fraction with a radical in the denominator to an equivalent fraction whose denominator is an integer. conjugate — for a two-term expression $a + b$ , the expression $a - b$ ; multiplying a binomial containing a square root by its conjugate gives a product with no square roots. simplified square root — a square root with no perfect-square factors in the radicand, no fractions in the radicand, and no square roots in a denominator.

This section is adapted from Elementary Algebra 2e, 9.5 Divide Square Roots by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the square-root table as prose, recast the worked examples as aligned step tables, and converted the practice problems (“Try Its”) into interactive exercises with instant feedback; omitted the Be Prepared quiz, media links, and end-of-section exercises.

Multiply Square Roots Solve Equations with Square Roots