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Re: GRE Math Challenge #17 - what is the value of 2x? [#permalink]
I would have checked x=-1 first since its easier to calculate.
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Re: GRE Math Challenge #17 [#permalink]
GreenlightTestPrep wrote:
soumya1989 wrote:
If \(\sqrt{8x^2 +17} = 3x-2\), what is the value of 2x? _________________________


x = -1
√[(8)((-1)²) + 17] = 3(-1) - 2
STOP
When we evaluate the RIGHT side, we get: √[(8)((-1)²) + 17] = -5
The square root of a value cannot equal -5
So, the solution x = -1 is not valid



Probably, it's me that I am tired but I cannot understand how sqrt((8)((-1)^2)+17)) could be negative. -1^2 = 1 so we get sqrt(25). Am I wrong?
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Re: GRE Math Challenge #17 - what is the value of 2x? [#permalink]
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Re: GRE Math Challenge #17 - what is the value of 2x? [#permalink]
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