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Re: 3x^2+5 or \sqrt9x^4 [#permalink]
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Quote:
One way to do this is just to plug in x=0. The LHS is 5, and the RHS is sqrt (25) = +/- 5. So the answer cannot be determined.

More rigorously, we can factor the RHS, and get it as sqrt ((3x^2+5)^2) = +/- (3x^2+5), which gets us to the same spot.


Be careful with the +/- with the answer.

A general rule for the GRE is the following:

- If the radical appears in the question, only consider the positive root

- If you perform a square root during algebra, say to turn \(x^2\) to \(x\), then you should consider +/-

Since the former is true in this case, we only consider the positive root, so the answer is C.
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Re: 3x^2+5 or \sqrt9x^4 [#permalink]
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here quant B is the \((a+b)^2\) expansion of quant A

i.e, \((3x^2 +5)^2\)=\(9x^4+30x^2+25\)

so \sqrt{quant B} =quant A

option C is correct
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Re: 3x^2+5 or \sqrt9x^4 [#permalink]
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Re: 3x^2+5 or \sqrt9x^4 [#permalink]
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