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rahulp11
If x > 0 and y > 0, which of the following is equal to \(\frac{1}{\sqrt{x}+\sqrt{x+y}}\)?
(A)\(\frac{1}{y}\)
(B)\(\sqrt{2x + y}\)
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}\)
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\)
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}\)

Source: Total GMAT Math - Jeff Sackmann

I. Algebraic

Whenever you see a term that is something like \(\sqrt{x}+\sqrt{x+y}\) in denominator, remember the formula \(a^2-b^2=(a-b)(a+b)\)


\((\frac{1}{\sqrt{x}+ \sqrt{x+y)=(\frac{\sqrt{x+y}-\sqrt{x{(\sqrt{x+y}+\sqrt{x})(\sqrt{x+y}- \sqrt{x})})=(\frac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y})^2-(\sqrt{x})^2})=(\frac{\sqrt{x+y}-\sqrt{x}}{x+y-x})= (\frac{\sqrt{x+y}-\sqrt{x}}{y})\)



II. Substitute some value for x and y.
Take x as 9 and y as 16..

\(\frac{1}{\sqrt{x}+\sqrt{x+y}}=\) \(\frac{1}{\sqrt{9}+\sqrt{16+9}}=\) \(\frac{1}{3+5}=\frac{1}{8}\)

(A)\(\frac{1}{y}=\frac{1}{16}\)….No
(B)\(\sqrt{2x + y}\)…..Clearly not a fraction
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}=\frac{3}{5}\)…No
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\)…..Negative answer
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}=\frac{5-3}{16}=\frac{1}{8}\)…Yes


E

In the first solution you used the (a^2-b^2) formula for conjugate. However, I not understanding one thing as I took the Conjugate of √x-√x+y.
Shouldn’t we take a= √ x and b= √ x=y?
Can you please explain this part?

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Method isn't full proof. If x=4 y=5 then you'd mistakenly get answer A?

1/2+3 = 1/5

chetan2u
rahulp11
If x > 0 and y > 0, which of the following is equal to \(\frac{1}{\sqrt{x}+\sqrt{x+y}}\)?
(A)\(\frac{1}{y}\)
(B)\(\sqrt{2x + y}\)
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}\)
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\)
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}\)

Source: Total GMAT Math - Jeff Sackmann

I. Algebraic

Whenever you see a term that is something like \(\sqrt{x}+\sqrt{x+y}\) in denominator, remember the formula \(a^2-b^2=(a-b)(a+b)\)


\((\frac{1}{\sqrt{x}+ \sqrt{x+y}})=(\frac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y}+\sqrt{x})(\sqrt{x+y}- \sqrt{x})})=(\frac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y})^2-(\sqrt{x})^2})=(\frac{\sqrt{x+y}-\sqrt{x}}{x+y-x})= (\frac{\sqrt{x+y}-\sqrt{x}}{y})\)



II. Substitute some value for x and y.
Take x as 9 and y as 16..

\(\frac{1}{\sqrt{x}+\sqrt{x+y}}=\) \(\frac{1}{\sqrt{9}+\sqrt{16+9}}=\) \(\frac{1}{3+5}=\frac{1}{8}\)

(A)\(\frac{1}{y}=\frac{1}{16}\)....No
(B)\(\sqrt{2x + y}\).....Clearly not a fraction
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}=\frac{3}{5}\)...No
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\).....Negative answer
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}=\frac{5-3}{16}=\frac{1}{8}\)...Yes


E
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I would recommend the algebraic approach. However, if you go with the second approach, for the numbers you pick you will have to test each option - for x=4 y=5 you will find option A and E both hold true, in that case you will have to check both with different numbers and select the one that holds true for any set of numbers.
Mgerman42
Method isn't full proof. If x=4 y=5 then you'd mistakenly get answer A?

1/2+3 = 1/5

chetan2u
rahulp11
If x > 0 and y > 0, which of the following is equal to \(\frac{1}{\sqrt{x}+\sqrt{x+y}}\)?
(A)\(\frac{1}{y}\)
(B)\(\sqrt{2x + y}\)
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}\)
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\)
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}\)

Source: Total GMAT Math - Jeff Sackmann

I. Algebraic

Whenever you see a term that is something like \(\sqrt{x}+\sqrt{x+y}\) in denominator, remember the formula \(a^2-b^2=(a-b)(a+b)\)


\((\frac{1}{\sqrt{x}+ \sqrt{x+y}})=(\frac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y}+\sqrt{x})(\sqrt{x+y}- \sqrt{x})})=(\frac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y})^2-(\sqrt{x})^2})=(\frac{\sqrt{x+y}-\sqrt{x}}{x+y-x})= (\frac{\sqrt{x+y}-\sqrt{x}}{y})\)



II. Substitute some value for x and y.
Take x as 9 and y as 16..

\(\frac{1}{\sqrt{x}+\sqrt{x+y}}=\) \(\frac{1}{\sqrt{9}+\sqrt{16+9}}=\) \(\frac{1}{3+5}=\frac{1}{8}\)

(A)\(\frac{1}{y}=\frac{1}{16}\)....No
(B)\(\sqrt{2x + y}\).....Clearly not a fraction
(C)\(\frac{\sqrt{x }}{ \sqrt{x+y}}=\frac{3}{5}\)...No
(D)\(\frac{\sqrt{x} -\sqrt{ x+y}}{ y}\).....Negative answer
(E)\(\frac{\sqrt{x+y}-\sqrt{x}}{y}=\frac{5-3}{16}=\frac{1}{8}\)...Yes


E
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