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#1 2005-08-02 00:05:00

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Solving Radical equations

Is this solution correct?

1.
2[SQRT(x+4)]=3
2[SQRT(x+4)]^2=3^2
4x+16=9
4x=9-16
4x=-7
x=-7/4

2.
[SQRT(x+3)]=2[SQRT(x)]
[SQRT(x+3)]^2=2[SQRT(x)]^2
x+9=4x I'm not sure what to do here?

#2 2005-08-02 00:36:54

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,711

Re: Solving Radical equations

1. Looks right!

2.
√(x+3)=2 √x
√(x+3)²=2² √x²
x+3 = 4x    [Note: the square root and square cancel each other out, just leaving (x+3)]

Now, subtract 4x from both sides:  x+3 - 4x = 4x - 4x
Simplify:  -3x + 3 = 0
So, x must be 1

(Or, you could be formal, and follow the steps:
Subtract 3 from both sides: -3x = -3
divide by -3: x = 1)


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#3 2005-08-02 22:36:40

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Re: Solving Radical equations

Thanks to both of you! I wasn't sure how to check for valid solutions, but now I do!:) I'll register once I come up with a username. I'll need all the math help I can get.^^;

#4 2005-08-03 01:11:45

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,711

Re: Solving Radical equations

We look forward to more "radical questions" ...

Don't forget that a square root is just a half power:

√x = x^(1/2)


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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