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Square root

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Square root

by ajmoney09 » Wed Jan 21, 2009 10:09 pm
P^2 = (Q^4)(R^6)

I had a problem like this, the solution said to take the "positive square root of this" which in turn would give me P= (Q^2)(R^3)

Why does Rs exponent go down to 3? why not 4?
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by mrsmarthi » Wed Jan 21, 2009 10:12 pm
Taking the positive square root means, dividing the exponent by 2.

p ^ 2 = (q ^ 4) (r ^ 6)

Taking the sqrt on both side, it is

p ^ (2 / 2) = (q ^ (4/2)) (r ^ (6/2))

==> p = (q ^ 2) (r ^ 3).

Hope you got it.
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by ajmoney09 » Wed Jan 21, 2009 10:21 pm
mrsmarthi wrote:Taking the positive square root means, dividing the exponent by 2.

p ^ 2 = (q ^ 4) (r ^ 6)

Taking the sqrt on both side, it is

p ^ (2 / 2) = (q ^ (4/2)) (r ^ (6/2))

==> p = (q ^ 2) (r ^ 3).

Hope you got it.
Thanks, that was simple enough....

if i was taking the cube root, would i divide by 3?
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by aroon7 » Wed Jan 21, 2009 10:33 pm
ajmoney09 wrote:
if i was taking the cube root, would i divide by 3?
You are right ajmoney...
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