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tricky prob

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by quantskillsgmat » Tue Dec 27, 2011 11:41 pm
canot use plug inn as this question is related to must type questions.

If square of x +5(square of y) +square of z=2y(2x+z) then which of the following must be true.
a)x=2y b)x=2z c)2x=z
a)onlya b)only b c) a and b both d)none
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by pemdas » Wed Dec 28, 2011 3:27 am
(x^2)+5(y^2)+(z^2)=2y(2x+z) we need to simplify

(x^2)+5(y^2)+(z^2)-4xy-2yz=0
[spoiler](x-y)^2=x^2+y^2-2xy, (z-y)^2=y^2+z^2-2yz[/spoiler]
(x-y)^2 + (z-y)^2 - 2y(x-y) + y^2 =0
(x-y)*(x-y-2y) + (z-y)^2 +y^2 =0
(x-y)*(x-3y) + (z-y)^2 +y^2 =0

for this whole expression to be turned into 0 we need x=y=z=0

d
quantskillsgmat wrote:If square of x +5(square of y) +square of z=2y(2x+z) then which of the following must be true.
a)x=2y b)x=2z c)2x=z

a)only a
b)only b
c)a and b both
d)none
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by user123321 » Wed Dec 28, 2011 5:22 am
quantskillsgmat wrote:canot use plug inn as this question is related to must type questions.

If square of x +5(square of y) +square of z=2y(2x+z) then which of the following must be true.
a)x=2y b)x=2z c)2x=z
a)onlya b)only b c) a and b both d)none
I think a

x^2 + 5y^2 + z^2 = 4yx + 2yz
=> (x^2 + 4y^2 -4yx) + (y^2 + z^2 - 2yz) = 0
=> (x-2y)^2 + (y-z)^2 = 0
=> x=2y & y=z

so it is A

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by pemdas » Wed Dec 28, 2011 5:36 am
x=2y and y=z, then 2y=2z and x=2z
you marked A, isn't it C both valid?
user123321 wrote:
quantskillsgmat wrote:canot use plug inn as this question is related to must type questions.

If square of x +5(square of y) +square of z=2y(2x+z) then which of the following must be true.
a)x=2y b)x=2z c)2x=z
a)onlya b)only b c) a and b both d)none
I think a

x^2 + 5y^2 + z^2 = 4yx + 2yz
=> (x^2 + 4y^2 -4yx) + (y^2 + z^2 - 2yz) = 0
=> (x-2y)^2 + (y-z)^2 = 0
=> x=2y & y=z

so it is A

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by user123321 » Wed Dec 28, 2011 6:32 am
ahh, nice twist.

you're right, it should be option C.

thanks for correcting.
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by ronnie1985 » Wed Dec 28, 2011 7:10 am
x^2+5y^2+z^2 = 4xy+2yz
=> (x^2-4xy+4y^2) + (z^2-2yz+y^2) = 0
=> (x-2y)^2 + (z-y)^2 = 0
x = 2y and z = y
(a) and (b) are correct = > (C) is the answer.
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