If x^1/2=y, what is the value of x^1/2?

x^1/2=y
Nothing is told whether y is positive integer or not.
(1) This only gives us understanding that y is integer, and nothing about x. Not Suff
(2) Since it is not told that x is integer it can be any number from this range. Not Suff
(1)+(2)
Since y is integer, then x is integer as well, then x=121 from the given range, however we're still lacking the infromation whether y is positive or not.
My answer is E.

Hi Alexey1989x
could u propose any suitable negative value for y???

Thanks

-11
rohit8865

Alexey1989x

Square roots of negative real numbers do not exist in the real numbers. They do exist in the complex numbers, using the imaginary unit i.

And as per my knowledge GMAT does'nt comply with imaginary numbers..

Hope it helps

rohit8865
You're correct.
The question is "what is is value of x^1/2". From both conditions it is not clear whether it is positive or negative.
(1)+(2) gives us the following
x,y are integers
x=121, x^1/2=|11|
35-11=24 - integer, complies with (1)
35-(-11)=46 - integer, complies with (1)
Thus, we cannot determine what is the value of x^1/2.
P.S. I'm not insisting on my approach, however iaw conditions provided I cannot answer the question.
Please correct me if I'm wrong.

I think Alexey has sense.

\(35 - (-11) = 46\)

\((-11)^2 = 121\)

In this case we are not taking square root of a negative integer, we are taking square root of positive \((-y)^2\)

\(\sqrt{x} = +/- y\)

Disagree, question is worded in a way to get us to think y is positive since we can't get negative value from a positive square root (ignoring complex numbers for gmat). If it said X square is equal to Y and y is between 101 - 143, then you're correct it could be either 11 or -11, but that's not the way question is structured. Anytime you're rooting (exception is odd power), integer has to be positive.

I do think that my logic is reasonable since it is not said that y is positive or not.
Bunuel would you please give a feedback why it is not "E"?
Thanks

This question has nothing to do with complex numbers. First, what is complex number? Every number can be expressed in the form \(a + bi\), where \(i\) is our imaginary part:

\(i^2 = -1\)

\(i = \sqrt{-1}\)

Had we got here \(x^2 = -y\) ---> \(x = \sqrt{-y}\) -----> \(x = i*\sqrt{y}\)

then we could talk about complex numbers. But here we have \(\sqrt{x} = y\) not \(\sqrt{-x} = y\)

Next (2) 101 < x < 143 says that \(x=y^2\) is between 101 and 143 not \(y\).

Now, some elementary school's algebra. If you square any number, result will be positive, even if the initial number is negative. And when you are taking square root of any perfect square you also should take into consideration that resultant can be negative. How can you solve simple quadratic equation \(x^2=4\)? Do we need to discard root \(x=-2\)? When you are taking square root you always need to keep in mind negative answer and this has nothing to do with complex roots.

Lol, i am not sure why this is so important to you. Bottom line GMAT rules state square root of a number can't be negative. Open any gmat book you will see the rule. You're conflating x^2 = 4 (two answers) to square root of 4 = (2 or -2), that's not the case. The definition of Square root as defined by Euclid requires that root be positive. That's all, he didn't put negative numbers into the definition because they came to the mathematical scene much later, and many mathematicians didn't like the idea of negative numbers. So per definition Square root of a number can't be negative, that's math convention.

For all real numbers \(\sqrt{x^2} = |x|\). I think you are familiar with properties of absolute values.

And there is only one convention in elementary math - expression UNDER THE RADICAL can't be negative, because we are not dealing with imaginary numbers in GMAT.

Some quote from math book:

"Every positive number a has two square roots: √a, which is positive, and −√a, which is negative. Together, these two roots are denoted ± √a"

To clear couple of things:

1. By default on the GMAT all numbers are real. You should not consider complex numbers at all.

2. Because of (1) the even roots from negative numbers are not defined for the GMAT.

3. For the problem above: for \(\sqrt{x}=y\) to be true for the GMAT x must be a non-negative number, which also means that y is a non-negative number.

P.S. 4. I agree that it would have been better if it were given that x > 0.

Good gracious man! I hope you are not arguing just for the sake of arguing. Anyone have any doubts about this pls refer to MGMAT Algebra book, chapter #4. Or GMATClub Math Book page #14. Clearly states, root can't be negative on "GMAT".

No more arguments,however gmat logic seems vague in this sort of poblem. OG also doesn't specify whether any square root is positive by default. MGMat is pretty comprehensive prep source nevertheless such thing should be described by OG.

Yes, that's true.

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{16}=4\), NOT +4 or -4. In contrast, the equation \(x^2=16\) has TWO solutions, +4 and -4. Even roots have only a positive value on the GMAT.

Odd roots have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).