If x is positive, what is the value of x^(1/2)?

(1) (x^1/3)=2
(2) x^2=64

In GMAT, we only consider the +ve root value of a square root. Example: x= \(\sqrt{4}\) -> x=2, whereas \(x^2\) = 4 -> x =+2 0r -2.

Now coming to the question, as we are asked about \(\sqrt{x}\).

St1: \(x^\frac{1}{3}\)=2 -> x=\(2^3\) so, \(\sqrt{x}\) = \(\sqrt{2^3}\). We can definitely get the value of x. Sufficient.

St2: \(x^2\)=64 -> x=+8 0r -8. Note: We cannot have square root of a -ve number so, \(\sqrt{x}\) = \(\sqrt{8}\). We can definitely get the value of x. Sufficient.

Answer:(D).

If x is positive, what is the value of sqrt x

(1) (x^1/3)=2
(2) x^2=64




You can't have the square root of a negative number. Irrational numbers are way beyond the scope of GMAT. If a question ask for \(\sqrt{x}\) you can assume x is positive or zero.

If x is positive, what is the value of \(\sqrt{x}\)?


(1) \(\sqrt[3]{x}=2\) --> take to the third power: x = 8 --> \(\sqrt{x}=\sqrt{8}\). Sufficient.

(2) x^2=64 --> x = 8 or x = -8. Since we are told that x is positive, then x = 8 and \(\sqrt{x}=\sqrt{8}\). Sufficient.

Answer: D.



First of all, we are told that x is positive, so x cannot be \(-2\sqrt{2}\). Next, the square root cannot give a negative result, that is \(\sqrt{4}=2\) NOT +2 and -2. (In contrast the equation x^2 = 4 has TWO solutions x = 2 and x = -2).



Yes, even roots from negative numbers are not defined for the GMAT (\(\sqrt[even]{negative}\) is undefined). So, you don't need complex numbers for the GMAT.

GMAT deals with only real numbers: integers (-3, -2, -1, 0, 1, 2, 3, ...), fractions/decimals (3/2, 4/3, 0.7, 17.5, ...) and irrational numbers (\(\sqrt{3}\), \(\sqrt{2}\), \(\pi\), ...).

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I agree. And I am considering x to be positive (8). However the question asks for sqrt of x. It doesn't suggest that sqrt of x is positive. hence it can be +- 2(2)^1/2. Am I missing something?

When the GMAT gives you a square root symbol, it’s referring to one specific value: the positive square root. 

So from statement 1: x=2^3=8. Sufficient.
Statement 2: x^2=64
IxI=8; x can be +8 or,-8.
As per question stem x is positive. So only x=8 satisfied.
Hence sufficient

Ans: D

Sent from my ASUS_Z010D using GMAT Club Forum mobile app

So, you are saying that \(\sqrt{8}\) could be \(2\sqrt{2}\) or \(-2\sqrt{2}\). That's not true.



Thus, \(\sqrt{8}=2\sqrt{2}\) only.

Hope it helps.
_________________

1) x^1/3=2
which means x=2^3= 8
therefore x=8 and root x = root 8
(sufficient)

2) x^2=64
which means x=root 64, ie, 8
therefore, x=8 and root x= root 8
(sufficient)

KarishmaB

Madam, I did not get the answer. How could the answer be D? As per statement 2, x^2 = 64 which means x=8 and x^(1/2) could be +/- 2*(2)^(1/2). I would use another simple example to explain this, if x = 4 then square root of x could be +/-2.

Much to my surprise this question is of 5% difficulty level. Please help me with this concept. Thanks a lot in advance Madam!

I think you are missing the crucial info from the stem

If x is positive, what is the value of \(\sqrt{x}\)?

So, from x^2=64, when you get that x = 8 or x = -8, you should discard x = -8 because we are explicitly given that x is positive and you'll get only one value: x = 8, thus \(\sqrt{x}=\sqrt{8}=2\sqrt{2}\).

Does this make sense?
_________________

Thanks Bunuel for your reply. I did not miss the condition that x is positive. But i would like to highlight that we are not given that x^(1/2) will be positive. It could be both positive and negative. Since square of neg is also positice. For example, If we are given x=4 then square root of x is +2 and -2.

Please do let me know flaw in my reasoning. Thanks in advance!

Posted from my mobile device

Thanks a lot Bunuel! This is really very helpful.

Hey Bunuel, I still can't wrap my head around this question.

Although i DO agree that only x can be determined to be +8 based on the stem, how can we presume that the sqrt of +8 cannot be negative?

For example, given that x = +4, then sqrt of +4 can be +/-2.

multiplying -2*-2 will return the positive x value +4.


In the case of the question - why would this not apply..?

it could effectively be written as:

\([(-\sqrt{8})^2]^(1/3)\)

EDIT: i believe the confusing part about this questions is that the question is not asking you to TAKE the square root of x, but rather assume that the expression provided is already \(\sqrt{x}\), implying that the number will always be positive.