Rationalizing the Denominator Test

When working with radicals, we always want to report a simplified result. One rule of simplification states that we can’t have a radical in the denominator. In order to deal with this problem, we follow a procedure known as rationalizing the denominator. This will give us a rational number in the denominator.

Test Objectives:

•Demonstrate the ability to simplify a square root, cube root, or higher level root

•Demonstrate the ability to rationalize a denominator that contains a square root

•Demonstrate the ability to rationalize a denominator that contains a cube root, or higher level root

Rationalizing the Denominator Test:

#1:

Instructions: Simplify each.

a) $$\frac{\sqrt{4}}{\sqrt{6}}$$

b) $$\frac{\sqrt{8}}{4\sqrt{6}}$$

c) $$\frac{\sqrt{5}}{6\sqrt{8}}$$

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#2:

Instructions: Simplify each.

a) $$\frac{3\sqrt{5}}{5\sqrt{7}}$$

b) $$\frac{\sqrt{2}}{\sqrt{7}}$$

c) $$\frac{\sqrt{49}}{3\sqrt{14}}$$

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#3:

Instructions: Simplify each.

a) $$\frac{\sqrt{32}}{\sqrt{20}}$$

b) $$\frac{\sqrt{3}}{4\sqrt{5}}$$

c) $$\frac{\sqrt{10}}{\sqrt{15}}$$

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#4:

Instructions: Simplify each.

a) $$\frac{8\sqrt{10} + \sqrt{5}}{13}$$

b) $$\frac{5 + 4\sqrt{3}}{10\sqrt{31}}$$

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#5:

Instructions: Simplify each.

a) $$\frac{3}{\sqrt[3]{3x}}$$

b) $$\frac{u}{\sqrt[4]{216}}$$

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Written Solutions:

#1:

Solution:

a) $$\frac{\sqrt{6}}{3}$$

b) $$\frac{\sqrt{3}}{6}$$

c) $$\frac{\sqrt{10}}{24}$$

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#2:

Solution:

a) $$\frac{3\sqrt{35}}{35}$$

b) $$\frac{\sqrt{14}}{7}$$

c) $$\frac{\sqrt{14}}{6}$$

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#3:

Solution:

a) $$\frac{2\sqrt{10}}{5}$$

b) $$\frac{\sqrt{15}}{20}$$

c) $$\frac{\sqrt{6}}{3}$$

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#4:

Solution:

a) $$\frac{8\sqrt{130} + \sqrt{65}}{13}$$

b) $$\frac{5\sqrt{31} + 4\sqrt{93}}{310}$$

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#5:

Solution:

a) $$\frac{\sqrt[3]{9x^2}}{x}$$

b) $$\frac{\sqrt[4]{6u^3}}{6}$$

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