If x is an integer, what is the value of x?

What you are saying is exactly correct : Square root of a number will always be a non-negative entity. The question doesn't wrestle that point.
\(\sqrt{x^2} = |x| \geq{0}.\). However, as you can see, both x = 1 and x = -1 are valid as because |1| = |-1| = 1.

So basically, \((-1)^2 = (1)^2 = 1\), but \(\sqrt{1}\) = 1.

From F. S 1, we know that |23x| is a prime number for both x =1 and x=-1. Thus,no unique value of x is present. Insufficient

From F.S 2, \(2*\sqrt{x^2}\) can be a prime no only if \(x^2 = 1\). Again, \(x^2 = 1\) \(\to\) \(x = \pm1\). Just as above, we get 2 values of x, hence Insufficient.

Even after combining both the fact statements, no unique value of x can be found.

E.

Bunuel,

As per the explanation for the option 2) , x can be either +1 or -1. But since 2 sqrt(x^2) is a prime number, can we not assume that x can't be -1 because a prime number is "an integer that has no integral factors but itself and 1" which will not be the case if x is -1.

If you plug \(x=-1\) into \(2\sqrt{x^2}\) you'll still get a prime number:

\(2\sqrt{x^2}=2\sqrt{(-1)^2}=2\sqrt{1}=2=prime\).

i dont agree with the solution. if mod of 23x is a prime number then as per the definition of prime number 23x is always positive. So how will x have two values?

and as per the statement 2, sqrt of (x2) is always positive. and when it was mentioned it should be prime then how we can take the value of -1.

i dont agree with the solution. There are many problems i did where we took positive value when square root was given.

The solutions above are correct.

For (1):
If x = 1, then |23x| = |23*1| = |23| = 23 = prime.
If x = -1, then |23x| = |23*(-1)| = |-23| = 23 = prime.

For (2):
If x = 1, then \(2\sqrt{x^2}=2\sqrt{1}=2=prime\).
If x = -1, then \(2\sqrt{x^2}=2\sqrt{(-1)^2}=2\sqrt{1}=2=prime\).

Hope it's clear now.

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