Solving Equations Involving Radical Expressions

In this lesson we will continue to explore solving various types of equations including quadratic equations and radical equations. We will continue to explore the real-world context of agriculture. Some of these contexts include farming and cattle ranching.

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For now, let’s review some topics from previous modules.

Prior Learning – Evaluating Functions and Real-World Models

Recall that in Module 2, we evaluated algebraic functions and models for real-world contexts. We also used models to solve real-world problems. We will review some of those topics here.

Practice:

1. Given $LaTeX: f\left(x\right)=\sqrt{2x-4}$ , Evaluate and Simplify your answer for A & B

a. $LaTeX: f\left(10\right)$

$LaTeX: f\left(x\right)=\sqrt{2x-4}$

$LaTeX: f\left(10\right)=\sqrt{2\left(10\right)-4}$

$LaTeX: =\sqrt{20-4}$

$LaTeX: =\sqrt{16}$

LaTeX: =4

b. $LaTeX: f\left(18\right)$

$LaTeX: f\left(x\right)=\sqrt{2x-4}$

$LaTeX: f\left(18\right)=\sqrt{2\left(18\right)-4}$

$LaTeX: =\sqrt{36-4}$

$LaTeX: =\sqrt{32}$

$LaTeX: =4\sqrt{2}$

2. Suppose a rectangle has a length that is twice the width. We want to explore a function or model for the area of the rectangle.

Recall that the area of a rectangle is length times width

Area = Length X Width

Since the length of the rectangle is twice the width, we have

Area = (2 X Width) X Width

If x = width of the of the rectangle, the function or model for the area is given by

$LaTeX: f\left(x\right)=2x^2$

Evaluate the following and explain the meaning of the value of the expression:

$LaTeX: f\left(4\right)$

Where 4 is measured in inches.

$LaTeX: f\left(4\right)=2\left(4\right)^2=2\left(16\right)=32$

The area of the rectangle is 32 square inches.

Solving Equations Involving Radical Expressions

Recall that radical functions are functions of the form

In this lesson, we will learn to solve equations involving radical expressions using algebraic techniques. In the table below, we see examples of radical equations.

1	$LaTeX: 5=\sqrt{x+6}$
2	$LaTeX: 2+\sqrt{x+6}=10$
3	$LaTeX: \sqrt{7x-4}=\sqrt{3+5x}$
4	$LaTeX: \sqrt{3x}-8=12$

Let’s consider how to solve each of these equations.

Problem 1
Problem 2
Problem 3
Problem 4

Solve for x:

$LaTeX: 5=\sqrt{x+6}$

In this equation, we say that the radical is “isolated” because it is all by itself on one side of the equation. This is important because when we are solving radical equations, we will want to isolate the radical and then square both sides of the equation to eliminate the radical.

Squaring both sides of the equations yields:

$LaTeX: \left(5\right)^2=\left(\sqrt{x+6}\right)^2$

LaTeX: 25=x+6

LaTeX: 19=x

We can check our answer by substituting the value for x into the equation.

$LaTeX: 5=\sqrt{19+6}$

$LaTeX: 5=\sqrt{25}$

Solve for x:

$LaTeX: 2+\sqrt{x+6}=10$

In this equation, we can see that the radical is not isolated. We can isolate the radical by subtracting 2 from both sides of the equation.

$LaTeX: \sqrt{x+6}=10-2$

$LaTeX: \sqrt{x+6}=8$

Now that the radical is isolated, we can proceed as we did in example 1, squaring both sides of the equation.

$LaTeX: \left(\sqrt{x+6}\right)^2=8^2$

LaTeX: x+6=64

LaTeX: x=58

Let’s check our answer.

$LaTeX: 2+\sqrt{58+6}=10$

$LaTeX: 2+\sqrt{64}=10$

LaTeX: 2+8=10

Solve for x:

$LaTeX: \sqrt{7x-4}=\sqrt{3+5x}$

In this equation, we have two radical expressions. We can square both sides to eliminate the radical as before.

$LaTeX: \left(\sqrt{7x-4}\right)^2=\left(\sqrt{3+5x}\right)^2$

LaTeX: 7x-4=3+5x

LaTeX: 2x=7

$LaTeX: x=\frac{7}{2}=3.5$

Solve for x:

$LaTeX: \sqrt{3x}-8=12$

$LaTeX: \sqrt{3x}=20$

$LaTeX: \left(\sqrt{3x}\right)^2=20^2$

LaTeX: 3x=400

$LaTeX: x=\frac{400}{3}\approx133.33$

Checking our solution you get:

$LaTeX: \sqrt{3}\left(\frac{400}{3}\right)-8=12$

$LaTeX: \sqrt{400}-8=12$

LaTeX: 20-8=12

Watch this video which demonstrates how to solve more examples of equations involving radical expressions. Stop at 5:50

Extraneous Solutions

Extraneous solutions are values that we get when solving an equation that are not really solutions to the equation. Below is an example that illustrates an extraneous solution.

EXAMPLE: Solve the following equation for x. Explain your solution method and check your solution.

$LaTeX: \sqrt{x+4}=x-2$

1. Square both sides of the equation: $LaTeX: x+4=\left(x-2\right)^2$

2. Expand the squared binomial: LaTeX: x+4=x^2-4x+4

3. Rearrange into a quadratic equation: LaTeX: x^2-5x=0

4. Factor out LaTeX: x : $LaTeX: x\left(x-5\right)=0$

So, the solutions are LaTeX: x=0 and LaTeX: x=5 .

However, we need to check the solutions back into the original equation to see if they are valid.

Checking LaTeX: x=0 :

$LaTeX: \sqrt{0+4}=0-2$

$LaTeX: \sqrt{4}=-2$

This gives us $LaTeX: \sqrt{4}=-2$ , which is not true. So, LaTeX: x=0 is an extraneous solution.

Checking LaTeX: x=5 :

$LaTeX: \sqrt{5+4}=5-2$

$LaTeX: \sqrt{9}=3$

This gives us $LaTeX: \sqrt{9}=3$ , which is true. So, LaTeX: x=5 is a valid solution.

Extraneous solutions can occur when solving equations involving radical expressions. Therefore, you should check your solutions when solving equations involving radicals.

In the next part of this lesson, we will see an example of a function with a radical expression in a real-world farming context.

Real-World Problem Involving A Radical Expression : The Broken Down Tractor Problem

In the photo below, we see a long straight road that runs along a field.

Narrative: Tractor is working in the field. It breaks down and needs a “rescue” vehicle to either bring it a part or to tow it out of the field. - nee help with this.

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We know that the rescue vehicle can travel faster on the road than on the field. Along the road, the vehicle can travel at most 45 mph. Since that is the speed limit. Along the field, the vehicle can only travel 20 mph.

We are given a model for the total time it will take the rescue vehicle to reach the tractor as a function of where the vehicle turns off the road onto the field.

A model for the total time it takes for the rescue vehicle to reach the tractor is shown below:

$LaTeX: T\left(x\right)=\frac{6-x}{45}+\frac{\sqrt{\left(0.8\right)^2+x^2}}{10}$

The distance from the barn is to where the tractor is broken down (straight on the road) is 6 miles and the distance across field from tractor to the road is 0.8 miles:

where x is the distance marked on the diagram above.

We can use Desmos to graph this function as shown below. Notice the scale on the x- and y-axes. This will be important for our work with this function later in this lesson.

Later in this lesson, we will use this model to solve problems.

When you have completed the above lesson, click the next button in Canvas to progress.