Solving RadicalEquatio

ns

Module 14 Topic 4

Table of Contents

Slides 3-4: How to Solve Radical Equations Slides 5-20: Examples and Practice Problems Slides 21-22: TI Instructions

Audio/Video and Interactive Sites

Slide 23: Video/Interactive

A radical equation is an equation where there is a variable in the radicand.

Radicand: Number or expression under the radical symbol.

Examples of Radical Equations:

NOT Radical Equations:

103 x

215)2( x

25)8( 3

2

x

13 x

x9

7162 x

How do I solve a radical equation?

To solve a radical equation…1.Isolate the radical to one side of the equation 2.Then raise both sides of the equation to the same

power.3.Simplify Example:

1. Isolate the Radical (subtract 4 from both sides)

2. Raise both Sides to the same power (square both sides)

3. Simplify

104 x

6x

22 )6()( x

36x

Equations with Rational Exponents

7x

Recall: The square and square root are inverses ( cube and cube root are inverses, and so on.)

To solve this equation, you must use the inverse and square both sides.

22 )7()( x

49x

77

749

Check your answer.

843 x

20

603

6443

)8()43( 22

x

x

x

x

0315 q

2

105

915

)3()15(

315

0315

22

q

q

q

q

q

q

Did you check your answers? If so, you seen that in the second problem, q =2 does not work!!!!! Therefore, 2 is an extraneous solution and the solution is “No Solution”

No SolutionNo Solution

5

05 0

)5(0

50

4

444

)2()4(

2

2

2

22

x

xx

xx

xx

xxx

xxx

xx

24 xx

Did you check your answers? If so, you seen that in the second problem, x= -5 does not work!!!!! Therefore, -5 is an extraneous solution and the solution is x=0.

Solution Set { 0 }

1942 mmm

4

82

192

1294

1294

194

22

22

2

m

m

m

mm

mmmm

mmm

Solution Set { 4 }

5

1

5

1

)108()62( xx

5

5

15

5

1

)108()62(

xx

xxx 10862

n mn

m

aa that Recall

1412 x

6

7x

Equations in the form = k can be solved by raising

each side of the equation to the power since . Remember to check for extraneous solutions.

n

m

a

m

n1)( m

n

n

m

a

5

1

5

1

5

1

5

1

5

1

5

1

6

14

6

14

6

708

6

14

)6

7(1086)

6

7(2

Check:

3

13

2

12 ww 33 212 ww

Solution Set { ¼ , 1 }

1 4

1

14

01 014

ww

w

ww

33 2 144 www

333

3 2 144 www

www 144 2

0154 2 ww0)1)(14( ww

3837

4

x

No SolutionNo Solution

7

7

)3(837

4

x

218783 4 x

44 4 218783 x

The reason is shown below:

You can not get a real answer by taking the 4th root of a negative number.

You can not get a real answer by taking the 4th root of a negative number.

9

1

9

1

3264 xx

x

x

xx

xx

7

4

74

344

3264

Find the nth root of a if n = 2 and a = 81.

9

81

81

2

2

x

x

x

axn

Find the nth root of a if n = 5 and a = -1024.

4

1024

1024

55 5

5

x

x

x

axn

If there are two radicals, isolate both radicals by moving one to the other side of the equal sign.

Example: 0)105()23( xx

)105()23( xx

22)105()23( xx

10523 xx

1022 x

x212

x6 Divide both sides by 2

Isolate the radicals

Square both sides

Simplify

Subtract 3x from both sides

Add 10 to both sides

I squared both sides, but now I have an x2?!

Don’t panic! Continue to solve as a quadratic equation by either using the Quadratic Formula or by Factoring…but

you must check for extraneous solutions.

When solving radical equations, extra solutions may come up when you raise both sides to an even power. These extra solutions are called extraneous solutions.

Recall, to check for extraneous solutions by plugging in the values you found back into the original problem. If the left side does not equal the right side then you have an extraneous solution.

Extraneous Solution!

If you have a root other than a square root, simply raise both sides to the same power as the root.

So, if you have a cubic root, raise both sides to the third power, for a fourth root, raise both sides to the fourth power, etc.

Algebraic Rule for all Radical Equations:Algebraic Rule for all Radical Equations:

. , nnnnn kxandkxthenkxIf

16807 ,

.7 7 ,7 :x 55555

xtherefore

xandxthenxE

Isolate the radical

Cube both sides

SimplifySubtract 3 from both sides

843 x 0315 q

22843 x

20

603

6443

x

x

x

315 q

22315 q

2

105

915

q

q

qCheck:

88

864

8460

84)20(3

843

x

Check:

09

033

039

03110

031)2(5

0315

q

This solution does not work, therefore “No Solution”

24 xx

2224 xx

444 2 xxx

xx 50 2

)5(0 xx

x0

x

x

5

50 This solution, -5, does not work, therefore

the solution set is { 0 }

24 xx

Check:

33

39

2)5()5(4

:5

x

22

24

2004

:0

x

Using the TI to solveSimply graph both sides of the equation. The x-values of the intersection point(s) are your solution(s).

xx 2303

Solving Algebraically:

22 230)3( xx

xxx 230962

02142 xx

0)3)(7( xx

3 7 xorx

Why does the TI only show x = 7?

Because 3 is an extraneous solution!

Why does the TI only show x = 7?

Because 3 is an extraneous solution!

44

164

14304

)7(23037

3 7 xorx

620

240

6300

)3(23033

7 7 : orxSolution

Practice Problems

Practice Problems and Answers

Examples, Answers, and Videos