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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?
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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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.^^;
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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