When we solve equations with radicals, we turn to our squaring property of equality. When we use this property of equality, we must guard against extraneous solutions or solutions that do not work in the original equation. To do this, we simply check all proposed solutions of the transformed equation in the original equation. Any solution that does not work in the original is rejected.

Test Objectives
• Demonstrate the ability to remove radicals from an equation
• Demonstrate the ability to solve an equation with radicals
• Demonstrate the ability to find any extraneous solutions for a radical equation

#1:

Instructions: solve each equation.

$$a)\hspace{.2em}x=\sqrt{5x + 50}$$

$$b)\hspace{.2em}\sqrt{-x - 2}=\sqrt{x + 10}$$

#2:

Instructions: solve each equation.

$$a)\hspace{.2em}\sqrt{x - 8}=\sqrt{-2x + 43}$$

$$b)\hspace{.2em}\sqrt{2x - 6}=\sqrt{-2x + 22}$$

#3:

Instructions: solve each equation.

$$a)\hspace{.2em}\sqrt{3x - 13}=\sqrt{7 - x}$$

$$b)\hspace{.2em}x + 2=\sqrt{2x + 67}$$

#4:

Instructions: solve each equation.

$$a)\hspace{.2em}\sqrt{-2 - 3x}=1 - \sqrt{-5x - 1}$$

$$b)\hspace{.2em}-3 - \sqrt{-x - 2}=\sqrt{-4x + 1}$$

#5:

Instructions: solve each equation.

$$a)\hspace{.2em}\sqrt{2x - 1}- \sqrt{10x - 1}=-2$$

$$b)\hspace{.2em}\sqrt{8 - x}- 1=\sqrt{1 - x}$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}x=10$$

$$b)\hspace{.2em}x=-6$$

#2:

Solutions:

$$a)\hspace{.2em}x=17$$

$$b)\hspace{.2em}x=7$$

#3:

Solutions:

$$a)\hspace{.2em}x=5$$

$$b)\hspace{.2em}x=7$$

#4:

Solutions:

$$a)\hspace{.2em}No \hspace{.2em}Solution$$

$$b)\hspace{.2em}No \hspace{.2em}Solution$$

#5:

Solutions:

$$a)\hspace{.2em}x=\frac{1}{2}, 1$$

$$b)\hspace{.2em}x=-8$$