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Square roots

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by Morgoth » Wed Oct 01, 2008 10:41 am
sqrt [2sqrt63 + 2/(8+3sqrt7)]

sqrt [{2sqrt63*(8+3sqrt7) + 2}/(8+3sqrt7)]

sqrt [{48sqrt7+126 + 2}/(8+3sqrt7)]

sqrt [{48sqrt7+128}/(8+3sqrt7)]


sqrt [16{3sqrt7+8}/(8+3sqrt7)]

you cancel the numerator and denominator part

sqrt 16 = 4.

Hope this helps.
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by cramya » Wed Oct 01, 2008 2:06 pm
Nicely done Morgoth!

I got up to sqrt [{48sqrt7+128}/(8+3sqrt7)] and the light bulb dint click to take out the common factor 16!
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by beater » Wed Oct 01, 2008 2:12 pm
Same here. I was stuck and didnt know how to proceed from that point forward
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by cramya » Wed Oct 01, 2008 2:23 pm
Hoping all bulbs in reserve now click on the Gday for both you and me Beater!!

Good luck guys!
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by Morgoth » Wed Oct 01, 2008 2:51 pm
Here is a tip for such question types.

You will have a whole number answer on these types 99.9% of times.

What I did in this question when I reached the last part was either take 64 common or 16 common. Because srt16 = 4 and sqrt64 = 8, which are two of the answer choices.
You obviously cannot take 64 common because the term is 48.

Thus, sqrt16 = 4.

All the best!
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by beater » Wed Oct 01, 2008 9:10 pm
Great tip - thanks for sharing Morgoth!
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by mental » Thu Oct 02, 2008 5:09 am
SLIGHTLY DIFF APPROACH

sqrt [2sqrt63 + 2/(8+3sqrt7)]

TAKING PORTION 2/(8+3sqrt7)
(3sqrt7 = sqrt9*7 = sqrt63)
multiply and divide by (8-3sqrt7)
2/(8+3sqrt7)] = 2(8-3sqrt7) /(8+3sqrt7)(8-3sqrt7)
= 2(8-3sqrt7) /(64 - 63) = 2(8-3sqrt7)

Back to question
sqrt [2sqrt63 + 2/(8+3sqrt7)]

we replace: 2/(8+3sqrt7) by 2(8-3sqrt7)
we get
sqrt [2sqrt63 + 2(8-3sqrt7)]
= sqrt [2sqrt63 + 16 -2*3sqrt7)]
= sqrt [2sqrt63 + 16 -2sqrt63)]
= sqrt [16] = 4

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