Search Header Logo
  1. Resource Library
  2. Math
  3. Algebra
  4. Solving Radical Equations
  5. Solving Radical Equations
Solving Radical Equations

Solving Radical Equations

Assessment

Presentation

Mathematics

9th - 12th Grade

Practice Problem

Medium

CCSS
HSA.REI.A.2, RL.11-12.3, RL.6.3

+3

Standards-aligned

Created by

Natalie McIntosh

Used 116+ times

FREE Resource

12 Slides • 22 Questions

1

Solving Radical Equations

Have paper ready for notes.

Slide image

2

Poll

How are we feeling today?

3

Multiple Choice

Here is a radical equation to solve:

 x+1=4\sqrt{x+1}=4  
Which of the following values would make this a true statement when substituted in for x?

1

3

2

8

3

15

4

16

4

 x+1=4\sqrt{x+1}=4  

Substitute  x=15x=15  
 15+1=4\sqrt{15+1}=4  
 16=4\sqrt{16}=4 
  4=44=4  is a true statement.
Therefore,  x=15x=15  is a solution!

5

Multiple Choice

How about this one?

 5x+1=6\sqrt{5x+1}=6  
Which of the following values would make this a true statement when substituted in for x?

1

36

2

7

3

6

4

1

6

 5x+1=6\sqrt{5x+1}=6  

Substitute  x=7x=7  
 5(7)+1=6\sqrt{5\left(7\right)+1}=6  
 35+1=6\sqrt{35+1}=6  
 36=6\sqrt{36}=6 
  6=66=6  is a true statement.
Therefore,  x=7x=7  is a solution!


We won't always be given possible solutions to try. And so, we will need some steps!

7

Steps for Solving Radical Equations

  • Isolate the radical

  • Raise both sides to the same power as the index. (Clears the radical.)

  • Solve the equation

  • Check your answer

8

Multiple Choice

Remember our first equation:

 x+1=4\sqrt{x+1}=4  
Step One is to Isolate the square root. Is the radical isolated?

1

Yes. The Square Root is already isolated.

2

No. Subtract 1 from both sides.

3

No. Square both sides.

4

No. Take the square root of 4.

9

Multiple Choice

Step Two is to Square Both Sides which looks like this:

 (x+1)2=(4)2\left(\sqrt{x+1}\right)^2=\left(4\right)^2  
Which of the following would result from this step?

1

 x+1=16\sqrt{x+1}=16  

2

 x2+1=16\sqrt{x^2+1}=16  

3

 x+1=8x+1=8  

4

 x+1=16x+1=16  

10

Multiple Choice

So far, we have:

 x+1=4\sqrt{x+1}=4  
 (x+1)2=(4)2\left(\sqrt{x+1}\right)^2=\left(4\right)^2 
 x+1=16x+1=16   
What should we do next?

1

Nothing. x=16 is the solution.

2

Add 1 to both sides.

3

Subtract 1 from both sides.

4

Subtract 16 from both sides.

11

Completed Problem

 x+1=4\sqrt{x+1}=4  
 (x+1)2=(4)2\left(\sqrt{x+1}\right)^2=\left(4\right)^2  
 x+1=16x+1=16  
 x=15x=15  
Remember to check your answer!

12

Multiple Choice

Our second equation:

 5x+1=6\sqrt{5x+1}=6  
Is the radical isolated?

1

Yes. The radical is isolated.

2

No. Subtract 1 from both sides.

3

No. Divide both sides by 5.

4

No. Add 1 to both sides.

13

Multiple Choice

The radical is isolated.

 5x+1=6\sqrt{5x+1}=6  
What do I do next?

1

Subtract 1 from both sides

2

Raise both sides to the 2nd power.

3

Divide both sides by 5.

4

Add 1 to both sides.

14

Multiple Choice

So far we have:

 5x+1=6\sqrt{5x+1}=6  
 (5x+1)2=(6)2\left(\sqrt{5x+1}\right)^2=\left(6\right)^2  
What does my next step look like?

1

 5x+1=365x+1=36  

2

 5x+1=65x+1=6  

3

 25x+1=3625x+1=36  

4

 25x+1=625x+1=6  

15

Multiple Choice

 5x+1=6\sqrt{5x+1}=6   

 (5x+1)2=(6)2\left(\sqrt{5x+1}\right)^2=\left(6\right)^2  
 5x+1=365x+1=36  
What is the last thing to do?

1

First subtract 1 from both sides and then divide by 5.

2

First subtract 1 from both sides and then multiply by 5.

3

First divide by 5 and then subtract 1 from both sides.

4

First add 1 to both sides and then divide by 5.

16

Completed Problem

 5x+1=6\sqrt{5x+1}=6  
 (5x+1)2=(6)2\left(\sqrt{5x+1}\right)^2=\left(6\right)^2  
 5x+1=365x+1=36  
 5x=355x=35  
 x=7x=7  

Remember to check your answer!

17

Multiple Choice

Given the equation:
 3x2+2=6\sqrt{3x-2}+2=6  
What would you do first?

1

Square both sides

2

Divide both sides by 3

3

Add 2 to both sides

4

Subtract 2 from both sides

18

Multiple Choice

We isolate the radical by subtracting 2 from both sides:
 3x2+2=6\sqrt{3x-2}+2=6  
to get  3x2=4\sqrt{3x-2}=4  

Now what?

1

Square both sides

2

Divide both sides by 3

3

Add 2 to both sides

4

Subtract 2 from both sides

19

Multiple Choice

So far we have:
 3x2+2=6\sqrt{3x-2}+2=6  
 3x2=4\sqrt{3x-2}=4  

 (3x2)2=(4)2\left(\sqrt{3x-2}\right)^2=\left(4\right)^2  
 3x2=163x-2=16  
Now what?

1

Subtract 2 from both sides and then divide by 3

2

Divide both sides by 3 then add 2 to both sides

3

Add 2 to both sides and then divide by 3

4

Multiply 3 to both sides and then add 2 to both sides

20

Completed Problem

 3x2+2=6\sqrt{3x-2}+2=6  
 3x2=4\sqrt{3x-2}=4  
 (3x2)2=(4)2\left(\sqrt{3x-2}\right)^2=\left(4\right)^2  
 3x2=163x-2=16  
 3x=183x=18  
 x=6x=6  

Remember to check your answer!

21

Multiple Choice

Given the equation:

 3x+2=18-3\sqrt{x+2}=18  
What would the correct next step look like?

1

 x+2=6\sqrt{x+2}=-6  

2

 x+2=21\sqrt{x+2}=21  

3

 3x=16-3\sqrt{x}=16  

4

 3x=20-3\sqrt{x}=20  

22

Multiple Choice

So far we have:

 3x+2=18-3\sqrt{x+2}=18  
 x+2=6\sqrt{x+2}=-6  
What would the correct next step look like?

1

 x=8\sqrt{x}=-8  

2

 x=4\sqrt{x}=-4  

3

 (x+2)2=(6)2\left(\sqrt{x+2}\right)^2=\left(-6\right)^2  

4

 (x+2)3=(6)3\left(\sqrt{x+2}\right)^3=\left(-6\right)^3  

23

Multiple Choice

So far we have:

 3x+2=18-3\sqrt{x+2}=18  
 x+2=6\sqrt{x+2}=-6  
 (x+2)2=(6)2\left(\sqrt{x+2}\right)^2=\left(-6\right)^2  
 x+2=36x+2=36  
Solve and check your solution. What would the correct answer be?

1

No solution. The check failed.

2

 x=34x=34  

3

 x=38x=38  

4

 x=18x=18  

24

Remember to check your Solution!

Original Problem:  3x+2=18-3\sqrt{x+2}=18  

When we solved, we got x = 34.


 3(34)+2=18-3\sqrt{\left(34\right)+2}=18  
 336=18-3\sqrt{36}=18  
 3(6)=18-3\left(6\right)=18  
 18=18-18=18  which is not not a true statement.
Therefore, there is no solution to this problem.

25

Multiple Choice

What do we call a solution that does not work when you plug it back in to check?

1

Extraordinary Solution

2

Extraneous Solution

3

Involuntary Solution

4

Unnecessary Solution

26

We could have stopped earlier and said no solution:

Original Problem:  3x+2=18-3\sqrt{x+2}=18  

When we divided by -3, we got:

 x+2=18\sqrt{x+2}=-18  
It is impossible to take the square root of a number and get a negative number as a result.

Get ready for the next problem.

27

Multiple Choice

Question image

What would you do first?

1

Raise both sides to the 3rd power.

2

Raise both sides to the 2nd power.

3

Add 1 to both sides.

4

Divide both sides by 3.

28

Multiple Choice

Question image

We raise both sides to the 3rd power. What would it look like afterwards?

1

 3x1=43x-1=4  

2

 3x1=23x-1=2  

3

 3x1=83x-1=8  

4

 27x1=827x-1=8  

29

Multiple Choice

Question image

Solve and check.


What would the answer be?

1

 x=3x=3  

2

 x=73x=\frac{7}{3}  

3

 x=3x=-3  

4

 x=83x=\frac{8}{3}  

30

The Whole Solution:

And our check works!

Slide image

31

Poll

Question image

On your own, try to solve this problem using the steps.

Which of the following best describes your progress?

I was able to isolate the radical, but then got stuck.

I do not know where to start.

I think I got a solution using the steps.

I found a solution, but not using the steps.

I got stuck somewhere in the middle.

32

I shall work through the problem on the shared screen.

33

Poll

Question image

On your own, try to solve this problem using the steps.

Which of the following best describes your progress?

I was able to isolate the radical, but then got stuck.

I do not know where to start.

I think I got a solution using the steps.

I found a solution, but not using the steps.

I got stuck somewhere in the middle.

34

I shall work through the problem on the shared screen.

Solving Radical Equations

Have paper ready for notes.

Slide image

Show answer

Auto Play

Slide 1 / 34

SLIDE