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Course: Algebra 2 > Unit 10

Lesson 2: Square-root equations

Intro to solving square-root equations

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Practice some problems before going into the exercise.

Introduction

In this article, we will practice the ways we can solve equations with square-roots in them.

Practice question 1

Solve the following equation for $x$ ‍.

$\sqrt{2 x - 5} = 7$ ‍

x =

Step 1: Solving the equation by taking the square

To solve this equation, we can simply square both sides to eliminate the radical.

$\begin{aligned} \sqrt{2 x - 5} & = 7 \\ {(\sqrt{2 x - 5})}^{2} & = 7^{2} \\ 2 x - 5 & = 49 \\ 2 x & = 54 \\ x & = 27 \end{aligned}$ ‍

Step 2: Checking for extraneous solutions

In square-root equations, we should always check for extraneous solutions! This is done by plugging the solution we found back into the equation.

The original equation is

\sqrt{2 x - 5} = 7

. Let's plug

x = 27

into it.

$\begin{aligned} \sqrt{2 x - 5} & = 7 \\ \sqrt{2 (27) - 5} & \overset{?}{=} 7 \\ \sqrt{54 - 5} & \overset{?}{=} 7 \\ \sqrt{49} & \overset{?}{=} 7 \\ 7 & = 7 \end{aligned}$ ‍

x = 27

is indeed a solution.

Practice question 2

Solve the following equation for $x$ ‍.

$\sqrt{10 - 3 x} = \sqrt{2 x + 5}$ ‍

x =

Step 1: Solving the equation by taking the square

To solve this equation, we can simply square both sides to eliminate the radical.

$\begin{aligned} \sqrt{10 - 3 x} & = \sqrt{2 x + 5} \\ {(\sqrt{10 - 3 x})}^{2} & = {(\sqrt{2 x + 5})}^{2} \\ 10 - 3 x & = 2 x + 5 \\ 10 - 5 & = 2 x + 3 x \\ 5 & = 5 x \\ 1 & = x \end{aligned}$ ‍

Step 2: Checking for extraneous solutions

In square-root equations, we should always check for extraneous solutions! This is done by plugging the solution we found back into the equation.

The original equation is

\sqrt{10 - 3 x} = \sqrt{2 x + 5}

. Let's plug

x = 1

into it.

$\begin{aligned} \sqrt{10 - 3 x} & = \sqrt{2 x + 5} \\ \sqrt{10 - 3 (1)} & \overset{?}{=} \sqrt{2 (1) + 5} \\ \sqrt{7} & = \sqrt{7} \end{aligned}$ ‍

x = 1

is indeed a solution.

Practice question 3

Solve the following equation for $x$ ‍.

$\sqrt{5 x - 4} = x - 2$ ‍

x =

Step 1: Solving the equation by taking the square

To solve this equation, we can simply square both sides to eliminate the radical.

$\begin{aligned} \sqrt{5 x - 4} & = x - 2 \\ {(\sqrt{5 x - 4})}^{2} & = (x - 2)^{2} \\ 5 x - 4 & = x^{2} - 4 x + 4 \\ 0 & = x^{2} - 9 x + 8 \\ 0 & = (x - 8) (x - 1) \end{aligned}$ ‍

We obtained the solutions

x = 1

and

x = 8

Step 2: Checking for extraneous solutions

In square-root equations, we should always check for extraneous solutions! This is done by plugging the solution we found back into the equation.

The original equation is

\sqrt{5 x - 4} = x - 2

When

x = 8

, we get:

$\begin{aligned} \sqrt{5 x - 4} & = x - 2 \\ \sqrt{5 (8) - 4} & \overset{?}{=} 8 - 2 \\ \sqrt{36} & \overset{?}{=} 6 \\ 6 & = 6 \end{aligned}$ ‍

x = 8

is a solution.

When

x = 1

, we get:

$\begin{aligned} \sqrt{5 x - 4} & = x - 2 \\ \sqrt{5 (1) - 4} & \overset{?}{=} 1 - 2 \\ \sqrt{1} & \overset{?}{=} - 1 \\ 1 & \neq - 1 \end{aligned}$ ‍

x = 1

is not a solution.

Conclusion

We found that the equation has a single solution,

x = 8

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