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If x – y = p, then 2x^2 − 4xy + 2y^2 =

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Re: If x – y = p, then 2x^2 − 4xy + 2y^2 = [#permalink]
Is it possible to divide the equation by 2 first and then applying (a+b)(a-b) formula
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Re: If x – y = p, then 2x^2 − 4xy + 2y^2 = [#permalink]
santro789 wrote:
Is it possible to divide the equation by 2 first and then applying (a+b)(a-b) formula

You can factor out 2 first: $$2x^2 − 4xy + 2y^2 =2(x^2-2xy+y^2)=2(x-y)^2=2p^2$$.

You cannot divide by 2 (so you cannot get rid of the 2 altogether) because we need the value of $$2x^2 − 4xy + 2y^2$$ not the value of $$x^2 − 2xy + y^2$$.

Hope it helps.
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Re: If x – y = p, then 2x^2 − 4xy + 2y^2 = [#permalink]
Bunuel wrote:
If $$x – y = p$$, then $$2x^2 − 4xy + 2y^2 =$$

(A) p

(B) 2p

(C) 4p

(D) p^2

(E) 2p^2

Take 2 common $$2x^2 − 4xy + 2y^2 =$$

2 * (x-y)^2

2 * p^2

E
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If x – y = p, then 2x^2 − 4xy + 2y^2 = [#permalink]
Bunuel wrote:
santro789 wrote:
Is it possible to divide the equation by 2 first and then applying (a+b)(a-b) formula

You can factor out 2 first: $$2x^2 − 4xy + 2y^2 =2(x^2-2xy+y^2)=2(x-y)^2=2p^2$$.

You cannot divide by 2 (so you cannot get rid of the 2 altogether) because we need the value of $$2x^2 − 4xy + 2y^2$$ not the value of $$x^2 − 2xy + y^2$$.

Hope it helps.

I divided by 2 in order to get x^2 -2xy + y^2, this is the formula a^2 -2ab +b^2 ---> (a-b)(a-b). Since a-b=P this results in P^2, but we divided by 2 so we need to multiply to get back and get 2P^2

This is allowed right?
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Re: If x – y = p, then 2x^2 − 4xy + 2y^2 = [#permalink]
Bunuel wrote:
If $$x – y = p$$, then $$2x^2 − 4xy + 2y^2 =$$

(A) p

(B) 2p

(C) 4p

(D) p^2

(E) 2p^2

Source: Nova GMAT
Difficulty Level: 600

Squaring the first equation, we have:

x^2 - 2xy + y^2 = p^2

Multiplying by 2, we have:

2x^2 + 2y^2 - 4xy = 2p^2

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Re: If x y = p, then 2x^2 4xy + 2y^2 = [#permalink]
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Re: If x y = p, then 2x^2 4xy + 2y^2 = [#permalink]
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