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Bunuel
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If (x + y)^2 = 16 and x^2 - y^2 = 16, what is the value of 2x^2?

(A) 8
(B) 16
(C) 18
(D) 32
(E) 50

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

Each of the expressions given is equal to 16, so we can set them equal to each other. Note that we have the “square of a sum” and the “difference of two squares” special products. We will put them both in distributed form, then simplify:

\((x + y)^2 = x^2 - y^2\)
\(x^2 + 2xy + y^2 = x^2 - y^2\)
\(2xy + y^2 = -y^2\)
\(2xy + 2y^2 = 0\)

Since 2y(x + y) = 0, it must be true that either 2y or (x + y) is equal to 0. However, (x + y) cannot equal 0, since we are told that (x + y)^2 = 16 . So it must be that 2y= 0. Thus y = 0.

If we plug 0 in for y in x^2 – y^2 = 16, we get x^2 = 16. Thus 2x^2 = 2(16) = 32.

The correct answer is D.


(x+y)(x-y) = 16
&
(x+y)(x+y) = 16

divide the 2 equations.

(x-y)/(x+y) = 1

therefore x+y = x-y . Implies y = 0. Hence x^2 - y^2 = 16 implies x^2 = 16(since y =0) , therefore 2x^2 = 32
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Bunuel
If (x + y)^2 = 16 and x^2 - y^2 = 16, what is the value of 2x^2?

(A) 8
(B) 16
(C) 18
(D) 32
(E) 50

\((x + y)^2 = 16 ; --------------> (x+y)(x+y)=16\)

\(x^2 - y^2 = 16; --------------> (x+y)(x-y)=16\)

\((x+y)(x+y) = (x+y)(x-y)\)

\((x+y)=(x-y)\)

\(y=0\)

If \(y=0\) then x can only be \(-+4\)

therefore \(2*x^2= 2*16 = 32\)

ANSWER is D
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Bunuel
If (x + y)^2 = 16 and x^2 - y^2 = 16, what is the value of 2x^2?

(A) 8
(B) 16
(C) 18
(D) 32
(E) 50

Simplifying the second equation, we have:

(x - y)(x + y) = 16

Both equations are equal to 16, so we can set them equal to each other:

(x - y)(x + y) = (x + y)^2

x - y = x + y

0 = 2y

y = 0

Since y = 0, (x + y)^2 = 16 means x^2 = 16, so 2x^2 = 2(16) = 32.

Answer: D
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