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A tricky one.

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A tricky one. [#permalink] New post 02 Oct 2011, 02:26
Hi everybody.
I really don't get it, can you explain me pls?


If x^2=y+5, y= z-2 and z=2x, is x^3+y^2+z divisible by 7?
1. x>0
2. y=4
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Re: A tricky one. [#permalink] New post 02 Oct 2011, 02:54
We have:
x^2=y+5
y=z-2
z=2x

z=2x
y=2x-2
x^2=2x+3

x^2-2x-3=0
x=3 or x=-1

If x=3, then y=2x-2=4 and z=2x=6
Therefore x^3+y^2+z=27+16+6=49

If x=-1, then y=2x-2=-4 and z=2x=-2
Therefore x^3+y^2+z=-3+16-2=13

If (1) is true, then the only possible x is 3, and the result is divisible by 7.
If (2) is true, then the only possible x is 3, and the result is again divisible by 7.

So, the answer is (D)
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Re: A tricky one. [#permalink] New post 02 Oct 2011, 03:08
Fine, thank you, but I still don't understand how this:
bagrettin wrote:

z=2x
y=2x-2
x^2=2x+3


Turned into this:
bagrettin wrote:
x^2-2x-3=0
x=3 or x=-1



Explain in further detail please.
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Re: A tricky one. [#permalink] New post 02 Oct 2011, 03:18
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x^2=2x+3
x^2-2x-3=0
x^2-2x+1-4=0
(x-1)^2-4=0
(x-1)^2=4
x-1=2 or x-1=-2
So, x=3 or x=-1
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Re: A tricky one. [#permalink] New post 02 Oct 2011, 03:22
brilliant, thanks a lot.
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Joined: 03 Mar 2011
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Re: A tricky one. [#permalink] New post 02 Oct 2011, 03:23
No problem) Thank you!
You are always welcome) Best regards!!!
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Re: A tricky one.   [#permalink] 02 Oct 2011, 03:23
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A tricky one.

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