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square, square, square roots!

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square, square, square roots!

by Bnow » Sun Dec 19, 2010 5:45 pm
Tricky question for me to translate into this text format, but here it goes -- I hope the equation makes sense!

For which of the following values of x is
_________
√(1-√(2-√x) )

defined as a real #?

1
2
3
4
5

Answer is 5, but I'm not sure I totally see how. If I plug in, and square 5, I get 25, then minus 2 to get square root 23, then square that to get 529 minus 1, to get 528, and then square that...? Argh. Feel like I should get this right away![/img]
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by VivianKerr » Sun Dec 19, 2010 7:21 pm
Hey Bnow,

I think I understand the concept being tested here.

"Real" numbers includes positives, negatives, integers, non-integer rational numbers, square roots, cube roots , pi, etc. Basically "real" numbers include rational AND irrational numbers.

An "imaginary" number is a square root of a non-positive real number. Basically, you can't take the square root of a negative number. You may have heard of the imaginary number "i".

i = the square root of -1

Back to your question, we are plugging the answer choices in for x. As you plug in, make sure that there is never a negative value underneath the square root. If there is, then the answer will not be a real number.

Hope this helps! :)
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by Rahul@gurome » Sun Dec 19, 2010 9:10 pm
Bnow wrote:Answer is 5, but I'm not sure I totally see how. If I plug in, and square 5, I get 25, then minus 2 to get square root 23, then square that to get 529 minus 1, to get 528, and then square that...? Argh. Feel like I should get this right away
If you plug in the answer choices, you must do it in place of x. Why are you squaring 5 and subtracting 2 from it! If you plug 5 for x the expression will look like: √(1 - √(2 - √5)), which is not real as √5 > 2, and thus √(2 - √5) is an imaginary number. I believe the posted question is not the original one (otherwise 5 shouldn't be the answer and other fours will be). I've made the necessary changes in the question quoted below.
For which of the following values of x, √(1 - √(2 - √x)) is NOT defined as a real number?
  • 1
    2
    3
    4
    5
Though plugging the options is a very effective way to solve this problem, I want to provide some insight.

An expression containing square roots will be a real number only if there are no negative numbers under the square root. The expression given has three square roots in it.

For √(1 - √(2 - √x)) to be real (1 - √(2 - √x)) shouldn't be negative, i.e. must be greater than or equal to zero.

i.e. (1 - √(2 - √x)) ≥ 0
=> 1 ≥ √(2 - √x)

Now for √(2 - √x) to be real (2 - √x) shouldn't be negative, i.e. must be greater than or equal to zero. Combining this new condition with the above one we get,
=> 0 ≤ (2 - √x) ≤ 1
=> 1 ≤ √x ≤ 2
=> 1 ≤ x ≤ 4

Thus for the expression to be NOT real x must be less than 1 or greater than 4. Only such option is 5.

The correct answer is E.
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by gmat7202011 » Tue Dec 21, 2010 9:30 am
Thanks Rahul,

"Thus for the expression to be NOT real x must be less than 1 or greater than 4"

I think this expression will be real for a value of x =0, isn't it. Am i missing something

Thanks again
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by Rahul@gurome » Tue Dec 21, 2010 9:36 am
gmat7202011 wrote:I think this expression will be real for a value of x =0, isn't it. Am i missing something

Thanks again
Put x = 0 in the expression: √(1 - √(2 - √0)) = √(1 - √2) = √(1 - 1.414) = √(-0.414) = NOT real
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by gmat7202011 » Tue Dec 21, 2010 9:38 am
Ah.. the most important mistake type i need to avoid. Careless...

Thanks
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by Bnow » Tue Dec 21, 2010 2:42 pm
Yes -- I had a typo in the question, which you totally caught! Attached is the original question. And thank you for all your help!
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