M01-19

That's not correct. How did you get that from (√x+√y)(√x-√y) = 9 both factors are either 3 or -3?

Forgive my ignorance but I am a little bit confused about what happens here
(1) x+y=4+2\sqrt{xy}

x-2\sqrt{xy}+y=4;

(\sqrt{x}-\sqrt{y})^2=4; <----HERE where did the 2 go?

And also I am new to this forum so I am unfamiliar with how to format, and this is just what it looked like when I pressed paste. Sorry !

Thanks for any help, I am very grateful

Sabine

Apply \(a^2-2ab+b^2=(a-b)^2\):

\(x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2\).

P.S. Writing Mathematical Formulas on the Forum: rules-for-posting-please-read-this-before-posting-133935.html#p1096628

I think this is a high-quality question and I agree with explanation.

Hi Bunuel,

if we apply

\((a-b)^2\) to the denominator, where does the how does it lead to \(( \sqrt{x} - \sqrt{y} ) ( \sqrt{x} + \sqrt{y} ) :\)

why not simply (x+y) (x-y)

First of all we are applying \(a^2-b^2=(a-b)(a+b)\) to the expression in the denominator, not \((a-b)^2=a^2-2ab+b^2\).

In the denominator we have x - y, which can be written as \(x-y=(\sqrt{x})^2-(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})\).

Hope it's clear.

Hi Bunuel and chetan2u VeritasPrepKarishma

I could'nt understand why the second solution √x−√y= - 2 is not valid when x−y>0. kindly explain if it has to solved algebraically

We are given that x > y > 0, so x > y, so √x > √y, so √x − √y > 0 and cannot equal to -2.

Another way to look at it (if that helps):
√x and √y have to be positive (principal square roots). So the only way √x−√y= - 2, if √x is 2 less than √y.
But if x - y > 0 i.e. x > y, then that is not possible.

Hi Bunuel,

Can you please elaborate on how st 2 is not sufficient ?

x - y = 9
( \sqrt{x} + \sqrt{y} ) ( \sqrt{x} - \sqrt{y} ) = 9 = 9 . 1 OR 3. 3
Since it's given the x> y >0, sum of two numbers cannot be equal to difference of the same two numbers. So 3,3 is out. So 9, 1 is still left.
Both numbers positive, one greater than the other,
so
\sqrt{x} + \sqrt{y} = 9
\sqrt{x} - \sqrt{y} = 1
this can be solved and we get the ans which obviously i wrong as per OA.

What's wrong here and why is 2 not sufficient?

Thanks.

You are assuming that \(\sqrt{x} + \sqrt{y}\) and \(\sqrt{x} - \sqrt{y}\) are integers, which is not given. Why cannot \(\sqrt{x} + \sqrt{y}\) be 81 and \(\sqrt{x} - \sqrt{y}\) be 1/9? x - y = 9 has infinitely many solutions, so you cannot get the single numerical value of \(\sqrt{x} - \sqrt{y}\)

Fell for the trap. Thanks Bunuel as always for the quick response - eye opener!

I think this is a high-quality question and I agree with explanation.

Hi @VeritasPrepKarishma,
Can you please explain why
√x and √y have to be positive (principal square roots).

\(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.


The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT (and generally in math) provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

From the question data, x and y are both positive numbers and x>y
The value of the expression needs to be evaluated.

From statement I alone, x + y = 4 + 2√xy.

Transposing the variable terms onto the LHS, we have x + y - 2√xy. = 4
The expression on the LHS represents the simplified form of\( (√x - √y)^2\). Therefore, the equation can be rewritten as \((√x - √y)^2\) = 4.

Taking the square root on both sides, we have (√x - √y) = ± 2; since the question says that both x and y are positive, (√x - √y) cannot be equal to – 2, therefore, (√x - √y) = 2.

In the expression given in the question stem, factoring out common terms, we have,
\(\frac{√2 (√x + √y )}{ (x-y)}\)

The denominator can be expressed as the product of (√x - √y) (√x + √y ).

Therefore, \(\frac{√2 (√x + √y )}{ (x-y)}\) = \(\frac{√2 (√x + √y )}{ (√x - √y) (√x + √y )}\)

Cancelling off (√x + √y ) and substituting the value of (√x - √y), value of the expression = \(\frac{√2 }{ 2}\) = \(\frac{1}{ √2}\)
Statement I alone is sufficient to answer the question. Answer options B, C and E can be eliminated.

From statement II alone, x – y = 9.

If x = 25 and y = 16, the value of the expression equal to √2
If x = 26 and y = 17, the value of the expression is not equal to √2
Statement II alone is insufficient to find a unique value for the given expression. Answer option D can be eliminated.

The correct answer option is A.

Hope that helps!
Aravind BT

vikastgmat
√x is by definition called the principal square root or the positive square root. The negative square root is given by -√x.

So if we are given \(\sqrt{4}\), it means we are given the positive square root which will be 2 only.

So a question such as "what is the value of
\(\sqrt{4} + 3\)" would be answered with 5.

If the negative square root has to be written, we need to write: "what is the value of
\(-\sqrt{4} + 3\)" and this would be answered with 1.

PS - Tag me at VeritasKarishma