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Senior | Next Rank: 100 Posts
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by smushkas » Tue Mar 11, 2008 2:31 pm
Hey guys,

I was surprised that there were roots of 3 and 4 like in this question. Should we really have to know them or there is some kind of shortcut or trick how to compute them quickly?

Thanks in advance!
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by Stuart@KaplanGMAT » Tue Mar 11, 2008 2:43 pm
You don't actually have to know the values, you just have to apply a bit of common sense.

We know that sqrt(4) = 2.

We also know that both the cube root and quardic root of ANY number greater than 1 will themselves be greater than 1.

So, we have:

2 + (more than 1) + (more than 1) = more than 4: choose (e).
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by smushkas » Tue Mar 11, 2008 2:58 pm
Thanks Stuart!
Actually, I did answer by the same logic as you described.
Thanks again!
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by musicdaemon » Mon Mar 17, 2008 10:13 am
Stuart Kovinsky wrote:You don't actually have to know the values, you just have to apply a bit of common sense.

We know that sqrt(4) = 2.

We also know that both the cube root and quardic root of ANY number greater than 1 will themselves be greater than 1.

So, we have:

2 + (more than 1) + (more than 1) = more than 4: choose (e).
Dear Stuart,
What is your take at -> sqrt(4) = +2 or -2

if sqrt(4)=-2 then the answer is ambiguous


what do you think?
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by smushkas » Mon Mar 17, 2008 3:57 pm
Musicdaemon,
As Amitava wrote it "Actually musicdaemon, sqrt(x) is always positive and it can not be negative."
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