For which of the following values of x is (1 - (2 - x^(1/2))^(1/2))

Question

For which of the following values of x is  \sqrt{1-\sqrt{2-\sqrt{x}}}  NOT defined as a real number?

A. 1
B. 2
C. 3
D. 4
E. 5

Bunuel · Accepted Answer

For which of the following values of x is  \sqrt{1-\sqrt{2-\sqrt{x}}}  NOT defined as a real number?

A. 1
B. 2
C. 3
D. 4
E. 5

Real Numbers are: Integers, Fractions and Irrational Numbers. Non-real numbers are even roots (such as square roots) of negative numbers.

We have  \sqrt{1-\sqrt{2-\sqrt{x}}} . For  {x=5}  expression becomes: \sqrt{1-\sqrt{2-\sqrt{5}}}  and  2-\sqrt{5}<0 , thus square root from this expression is not a real number.

Answer: E.

Hope it helps.

Bunuel · Answer

Consider another example: For which of the following values of x  \sqrt{1-\sqrt{4-\sqrt{x}}}  is NOT defined as a real number?

A. 16
B. 12
C.10
D. 9
E. 4

First see whether  4-\sqrt{x}  could be negative for some value of  x  so you should test max value of  x :  4-\sqrt{x_{max}}=4-\sqrt{16}=0 . As it's not negative then see whether  1-\sqrt{4-\sqrt{x}}  can be negative for some value of  x , so you should test min value of  x  to maximize  4-\sqrt{x} :  1-\sqrt{4-\sqrt{x_{min}}}=1-\sqrt{4-\sqrt{4}}=1-1.41=-0.41<0 .

Answer E.

Hope it's clear.

Safiya · Answer

It helped indeed, thank you very much!

tonebeeze · Answer

Bunuel -

So the approach to problems such as this one is to work your way from the most inner root, out toward the main root? Always keeping track of whether the underlying roots are (1) negative (2) are larger than the main root.

tonebeeze · Answer

Great visual explanation. I now understand. Thanks for taking the time to help.

Stiv · Answer

I don't understand how did you came from  \sqrt{1-\sqrt{2-\sqrt{5}}} to 2-\sqrt{5}<0  ?
I would appreciate if you explain, because I'm obviously missing something.

Bunuel · Answer

If  {x=5}  then the expression becomes: \sqrt{1-\sqrt{2-\sqrt{5}}} . The expression under the second square root is  2-\sqrt{5} . Now, since  2-\sqrt{5}<0  then the square root from this expression is not a real number.

Hope it's clear.

Stiv · Answer

O yes,yes,yes... thanks! Sometimes I just look at numbers and don't see anything no matter how obvious it is. I need a break!

GMATinsight · Answer

CONCEPT : Any Number is considered NON-REAL if the number under the square root is negative (i.e.√(-ve)) .

Such questions require us to check the smallest or highest values among option because at the extremes only the number will result in Real or NON-Real

Here, We need to check the highest value because x is being subtracted from other numbers

@x=5 (Option E)

√  = √  = √  = √ √(-ve) ] i.e.  NON-REAL Number 
Hence, Correct Option

Answer: Option E

JeffTargetTestPrep · Answer

We need to find the value of x that leaves us with a negative value under the root sign, which would not be defined as a number.

We see that if x = 5, then 2 - √5 < 0, so √(2 - √5) will not be a real number.

Answer: E

Hovkial · Answer

The quantity root{2 - root(x)} will be undefined when the quantity 2 - root(x) becomes negative. This will become negative when root(x) is greater than 2. The only value of x that will satisfy this condition is 5.

It is helpful to recall that the value of the positive root of 5 is slightly greater than 2. The positive root of 4 is 2, so the positive root of 5 will be somewhat greater.
 
ANSWER: (E)

Hoozan · Answer

EducationAisle  If I plug in 1 or 0 I land up with √0. Does √0 give us a real value? What does √0 = ?

EducationAisle · Answer

As far as I know, √0 = 0

dave13 · Answer

hey GMATinsight but two negative signs next to each other give positive value √ √(-ve) so it should be positive i.e. 1 - (- any value) = positive whats flaw in my reasoning ?

IanStewart maybe you can help

IanStewart · Answer

You seem to be suggesting that  -\sqrt{-x}   is equal to  \sqrt{-(-x)} , but that is absolutely never true unless x is zero. You can never move negative signs in and out of square roots, because then at some point you have a negative number under your square root, and that is mathematically illegal.

And I'd add: the post you quote is not generally correct when it says "Such questions require us to check the smallest or highest values among option". It happens to be true in this particular question that one of the extreme choices is the right answer, but it is very easy to design a similar question where the right answer is in the middle of the choices, e.g.:

For which of the following values of x is the expression  \sqrt{x^2 - 5}  undefined?
A) -10
B) -5
C) 0
D) 5
E) 10

The right answer here is C.

IanStewart · Answer

√0 is equal to 0, so yes, it produces a real value. When it comes to square roots (or any other even roots), it's only square roots of strictly negative numbers that are undefined/not real.

coucoudanger · Answer

We just need to choose the value of x for which \sqrt{\2-\sqrt{x} <0.
Looking to the answer choices :  \sqrt{\2-\sqrt{5} <0.
Correct answer : E

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For which of the following values of x is (1 - (2 - x^(1/2))^(1/2))

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For which of the following values of x is (1 - (2 - x^(1/2))^(1/2))

Prep Toolkit

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