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square root question 13th ed.

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square root question 13th ed.

by hutch27 » Thu Feb 21, 2013 7:44 am
the square root of this entire thing. sqrt (16)(20)+(8)(32)

why cant you factor out a 4... for example

sqrt 4((4)(5)+(2)(8))
sqrt 4((20)+(16))
sqrt 4((36)

sqrt (4) x sqrt (36) = 2 x 6 = 12....

whats wrong with that math?

the oa is 24

and it's Q# 35 in the 13th edition, just as a reference.
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by Brent@GMATPrepNow » Thu Feb 21, 2013 8:11 am
hutch27 wrote:the square root of this entire thing. sqrt (16)(20)+(8)(32)

why cant you factor out a 4... for example

sqrt 4((4)(5)+(2)(8))
sqrt 4((20)+(16))
sqrt 4((36)

sqrt (4) x sqrt (36) = 2 x 6 = 12....

whats wrong with that math?

the oa is 24

and it's Q# 35 in the 13th edition, just as a reference.

You are suggesting that (16)(20)+(8)(32) = 4[(4)(5)+(2)(8)]
You have basically "double factored" here.
Notice that, if we multiply 4(4)(5), we get (16)(5), not (16)(20)

So, for example, you can factor a 4 out of the 16 or the 20, but not both.

Cheers,
Brent
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by Brent@GMATPrepNow » Thu Feb 21, 2013 8:20 am
hutch27 wrote: sqrt[(16)(20)+(8)(32)]
One option here is to evaluate (16)(20)+(8)(32), and then find the square root of the result. That's a bit of work.

We can also apply a technique called "Multiplying by Doubling and Halving"
We have a free video on this: https://www.gmatprepnow.com/module/gener ... es?id=1113

In the first part, (16)(20), I notice that 16 is a perfect square. Nice.
In the second part, (8)(32), I notice that we have no perfect squares. However, using the doubling and halving technique, we can see that (8)(32) = (16)(16)

So, sqrt[(16)(20)+(8)(32)] = sqrt[(16)(20)+(16)(16)]
= sqrt[16(20 + 16)] {I factored out the 16}
= sqrt[(16)(36)]

At this point, we can apply a useful rule: sqrt(xy) = [sqrt(x)][sqrt(y)]

sqrt[(16)(36)] = [sqrt(16)][sqrt(36)]
= (4)(6)
= 24

Cheers,
Brent
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