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

It helped indeed, thank you very much!

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.

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

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.

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.

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!

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)

√[1−√(2−√x)] = √[1−√(2−√5)] = √[1−√(2−2.2)] = √[1−√(-ve)] i.e. NON-REAL Number
Hence, Correct Option

Answer: Option E

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

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)

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

As far as I know, √0 = 0

hey GMATinsight but two negative signs next to each other give positive value √[1−[color=#0000ff]√(-ve)
so it should be positive i.e. 1 - (- any value) = positive

whats flaw in my reasoning ?

IanStewart maybe you can help

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.

√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.

[/square_root]

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