Lesson 8.5: Divide Radical Expressions

New Concept

Master dividing radical expressions by using the Quotient Property, $\frac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\frac{a}{b}}$, to combine terms. Then, learn to rationalize denominators with one or two terms, a key step for writing expressions in simplest form.

What’s next

Get ready to apply the Quotient Property. You’ll work through guided examples and practice cards to master dividing and simplifying radical expressions.

Property

If $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers, $b \neq 0$, and for any integer $n \geq 2$ then,

We use the form $\frac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\frac{a}{b}}$ when the fraction we start with is the quotient of two radicals. When we write the fraction in a single radical, we may find common factors in the numerator and denominator.

Examples

• Simplify $\dfrac{\sqrt{48x^5}}{\sqrt{3x}}$. We combine them to get $\sqrt{\dfrac{48x^5}{3x}} = \sqrt{16x^4}$. This simplifies to $4x^2$.
• Simplify $\dfrac{\sqrt[3]{54y^2}}{\sqrt[3]{2y^5}}$. We combine them to get $\sqrt[3]{\dfrac{54y^2}{2y^5}} = \sqrt[3]{\dfrac{27}{y^3}}$. This simplifies to $\dfrac{3}{y}$.
• Simplify $\dfrac{\sqrt{75a^7b^5}}{\sqrt{3a^2b}}$. This becomes $\sqrt{\dfrac{75a^7b^5}{3a^2b}} = \sqrt{25a^5b^4} = \sqrt{25a^4b^4 \cdot a} = 5a^2b^2\sqrt{a}$.

Explanation

Think of this property as a grouping rule. You can either handle two separate radicals or combine them under one radical roof. Combining them often makes simplifying easier by letting you cancel out common factors before taking the root.

Property

A radical expression is considered simplified if there are

• no factors in the radicand have perfect powers of the index
• no fractions in the radicand
• no radicals in the denominator of a fraction

Examples

• Is $\sqrt{50}$ simplified? No, because $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$. It contained a perfect square factor.
• Is $\sqrt{\dfrac{5}{9}}$ simplified? No, because it has a fraction in the radicand. It simplifies to $\dfrac{\sqrt{5}}{\sqrt{9}} = \dfrac{\sqrt{5}}{3}$.
• Is $\dfrac{7}{\sqrt{3}}$ simplified? No, because there is a radical in the denominator. It must be rationalized to become $\dfrac{7\sqrt{3}}{3}$.

Explanation

A radical is fully simplified when it's completely tidy. This means no perfect squares (or cubes, etc.) are left inside, no fractions are under the radical sign, and no radicals are hiding in the denominator of a fraction.

Property

Rationalizing the denominator is the process of converting a fraction with a radical in the denominator to an equivalent fraction whose denominator is an integer. To rationalize a denominator with a higher index radical, we multiply the numerator and denominator by a radical that would give us a radicand that is a perfect power of the index.

Examples

• To simplify $\dfrac{7}{\sqrt{5}}$, multiply by $\dfrac{\sqrt{5}}{\sqrt{5}}$ to get $\dfrac{7\sqrt{5}}{(\sqrt{5})^2} = \dfrac{7\sqrt{5}}{5}$.
• To simplify $\dfrac{2}{\sqrt[3]{9}}$, rewrite as $\dfrac{2}{\sqrt[3]{3^2}}$. Multiply by $\dfrac{\sqrt[3]{3}}{\sqrt[3]{3}}$ to get $\dfrac{2\sqrt[3]{3}}{\sqrt[3]{3^3}} = \dfrac{2\sqrt[3]{3}}{3}$.
• To simplify $\dfrac{5}{\sqrt{8x}}$, first simplify to $\dfrac{5}{2\sqrt{2x}}$. Then multiply by $\dfrac{\sqrt{2x}}{\sqrt{2x}}$ to get $\dfrac{5\sqrt{2x}}{2(2x)} = \dfrac{5\sqrt{2x}}{4x}$.

Explanation

Radicals in the denominator are considered un-simplified. To fix this, we multiply the numerator and denominator by a value that makes the denominator's radicand a perfect power of the index, which removes the radical and makes the denominator rational.

Property

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.

Examples

• To simplify $\dfrac{4}{3 - \sqrt{2}}$, multiply by the conjugate $(3 + \sqrt{2})$. This gives $\dfrac{4(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})} = \dfrac{4(3 + \sqrt{2})}{9 - 2} = \dfrac{4(3 + \sqrt{2})}{7}$.
• To simplify $\dfrac{\sqrt{2}}{\sqrt{x} + \sqrt{5}}$, multiply by the conjugate $(\sqrt{x} - \sqrt{5})$. This gives $\dfrac{\sqrt{2}(\sqrt{x} - \sqrt{5})}{(\sqrt{x})^2 - (\sqrt{5})^2} = \dfrac{\sqrt{2x} - \sqrt{10}}{x - 5}$.
• To simplify $\dfrac{\sqrt{y} + \sqrt{3}}{\sqrt{y} - \sqrt{3}}$, multiply by the conjugate $(\sqrt{y} + \sqrt{3})$. This gives $\dfrac{(\sqrt{y} + \sqrt{3})^2}{(\sqrt{y})^2 - (\sqrt{3})^2} = \dfrac{(\sqrt{y} + \sqrt{3})^2}{y - 3}$.

Explanation

To remove a two-term radical denominator, multiply the numerator and denominator by its conjugate. The conjugate has the opposite sign in the middle. This trick uses the difference of squares formula to eliminate the radicals from the denominator.