solving radical equations module 14 topic 4. table of contents slides 3-4: how to solve radical...
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Solving RadicalEquatio
ns
Module 14 Topic 4
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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
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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
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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
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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.
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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
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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 }
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1942 mmm
4
82
192
1294
1294
194
22
22
2
m
m
m
mm
mmmm
mmm
Solution Set { 4 }
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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
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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:
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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
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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
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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
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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
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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.
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Extraneous Solution!
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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
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Isolate the radical
Cube both sides
SimplifySubtract 3 from both sides
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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”
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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
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Using the TI to solveSimply graph both sides of the equation. The x-values of the intersection point(s) are your solution(s).
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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
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Practice Problems
Practice Problems and Answers
Examples, Answers, and Videos