Algebra #14

by sjaelee, Aug 17, 2011, 2:47 AM

Quote:
The distinct real numbers x and y satisfy

$x^2=33y+907$ and $y^2=33x+907$.

Find $xy$.

Source: WOOT

We subtract the equations from each other:

$x^2-y^2=33(y-x)$ and $y^2-x^2=33(x-y)$. Difference of squares:

(and multiplying by negative one to the 33 and the term next to it)

$(x+y)(x-y)=-33(x-y)$ and $(y-x)(y+x)=-33(y-x)$. Aha! We can

divide by $(x-y),(y-x)$ since $x,y$ are distinct:

$x+y=-33$, so $y=-(33+x)$. We substitute this into original

expression:

$x^2=33(-33-x)+907$, so $x^2=-33x-182$, so

$x^2+33x+182=0$. Using quadratic formula,

$\frac{-33\pm\sqrt{1089-728}}{2}=(-7,-26)$.

We note that $x+y=-33$ and the two solutions for x, $(-7,-26)$, add

to $-33$, so x and y are interchangable with $(-26,-7)$. Thus,

$-26*-7=182$.
This post has been edited 1 time. Last edited by djmathman, Aug 17, 2011, 2:51 AM

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This was from the Intermediate Algebra book, right?

That's where I solved the problem.

by djmathman, Aug 17, 2011, 2:53 AM

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yeah, thats where I have been getting thses alg problems.... :blush: :oops: :roll:

by sjaelee, Aug 17, 2011, 2:56 AM

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