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Solve a Radical Equation in One Variable

Solve a Radical Equation in One Variable

Assessment

Presentation

•

Mathematics

•

9th - 12th Grade

•

Hard

Created by

Joseph Anderson

FREE Resource

15 Slides • 31 Questions

1

Poll

How are we feeling today?

2

2.6 Radical Equations in One Variable

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  • ​ A radical equation is an equation that has at least one radical expression containing a variable, while any nonradical expressions are polynomial terms.

  • In math, an extraneous solution is a solution that emerges during the process of solving a problem. It does not work when you plug it back in to check.

4

Multiple Choice

Examine the following radical expression. What is the index?

1085\sqrt[5]{108}  

1

5

2

108

3

36

5

Multiple Choice

Examine the following radical expression. What is the radicand?

1085\sqrt[5]{108}

1

5

2

108

3

36

6

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Rational Exponents

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8

Multiple Choice

Write the index/root: 13 1413\ ^{\frac{1}{4}}

1

1313

2

44

3

11

9

Multiple Choice

Write the exponent: 13 1413\ ^{\frac{1}{4}}

1

1313

2

44

3

11

10

Multiple Choice

Write the radicand: 13 1413\ ^{\frac{1}{4}}

1

1313

2

44

3

11

11

Multiple Choice

Write as a radical expression: 13 1413\ ^{\frac{1}{4}}

1

413\sqrt[13]{4}

2

131\sqrt[1]{13}

3

134\sqrt[4]{13}

12

Multiple Choice

Write the following in rational exponent (exponential) form:

x73\sqrt[3]{x^7}

1

x73x^{\frac{7}{3}}

2

3x73x^7

3

x37x^{\frac{3}{7}}

4

7x37x^3

13

Multiple Choice

Write the following in rational exponent (exponential) form:

26\sqrt[6]{2}

1

262^6

2

2x62x^6

3

2162^{\frac{1}{6}}

4

6x26x^2

14

Multiple Choice

Write the following in rational exponent (exponential) form:

7\sqrt[]{7}

1

727^2

2

7x27x^2

3

7127^{\frac{1}{2}}

4

272^7

15

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

16

17

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

18

19

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

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20

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.

21

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  

22

Multiple Choice

So far, we have:

(x+1)2=(4)2\left(\sqrt{x+1}\right)^2=\left(4\right)^2  
x+1=4\sqrt{x+1}=4
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.

23

Completed Problem

24

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.

25

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.

26

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  

27

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.

28

Completed Problem

29

Multiple Choice

Given the equation:
3x−2+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

30

Multiple Choice

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

31

Multiple Choice

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

3x−2=163x-2=16  
(3x−2)2=(4)2\left(\sqrt{3x-2}\right)^2=\left(4\right)^2  
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

32

Completed Problem

33

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  

34

Multiple Choice

So far we have:

−3x+2=18-3\sqrt{x+2}=18  
−3x+2=18-3\sqrt{x+2}=18  
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  

35

Multiple Choice

So far we have:

−3x+2=18-3\sqrt{x+2}=18  
−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  
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  

36

Remember to check your Solution!

37

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

38

We could have stopped earlier and said no solution:

39

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.

40

Multiple Choice

Question image

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

1

3x−1=43x-1=4  

2

3x−1=23x-1=2  

3

3x−1=83x-1=8  

4

27x−1=827x-1=8  

41

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}  

42

The Whole Solution:

And our check works!

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43

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.

44

I shall work through the problem on the board.

45

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.

46

I shall work through the problem on the board.

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