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Re: different integers [#permalink]
03 Nov 2011, 22:07
This post received KUDOS
Okay, something is off here for me
What I did is this
X ^ 2 = xy
I took the square root of x ^ 2 so I ended up with
x = square root of (x.y) which means x = square root of (x) * square root of (y)
Therefore, square root of (y) will equal x/square root of (x)
If x is = 0 then it will be undefined. You can't divide by 0.
The same applies to the value of square root of (x) which will equal x/square root of (y). Therefore, y can't equal 0.
Anyone help me here?
First of all, square root of x^2 is |x|, not x. Do not take the square root until and unless you really need to. x^2 = xy You do not divide both sides by x here. You lose out on a solution. What you can very safely do is x^2 - xy = 0 x(x-y) = 0 Now, either x = 0 or x = y or both Since x and y are different, x must be 0.
When x = 0, no matter what the value of y, x^2 is equal to xy since both sides are equal to 0. Since x = 0, you cannot re-write this as y = x^2/x _________________
Re: problem solving [#permalink]
08 Feb 2011, 05:33
This post received KUDOS
if x and y ARE different integers and x^2=xy which of the following must be true?
x=0 y=0 x=-y
a. 1 only b. 2 only c. 3 only d. 1 and 3 e. 1,2 and 3
x^2=xy --> x(x-y)=0 --> either x=0 or x=y but as given that x and y are different numbers than the second option is out and we have: x=0. So only I is always true (in fact because of the same reason that x and y are different numbers II and III are never true).
The question gives you x^2=xy
you can simplify that as x^2-xy=0
x(x-y)=0 the solution will be either x=0 or x=y, since x and y are different integers x cannot equal y, that leaves you x=0 to be the only answer.
Re: If x and y are different integers and x^2 = xy, which of the [#permalink]
19 Jan 2014, 00:07
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