Algebra #3

by sjaelee, Aug 11, 2011, 10:58 AM

Quote:
Assume that a,b,c are the roots of

$2x^3-2x^2-x+4$. Compute

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$.

First, we note that by Vieta's, the sum of the roots $a+b+c=--2/2=1$,

so $b+c=1-a$ and same for b and c. Therefore, we want

$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}$. Expanding, we have

$\frac{(a+b+c)-2(ab+bc+ac)+3(abc)}{(1-a)(1-b)(1-c)}$: the

denominator simplifies to $1-(a+b+c)+(ab+bc+ac)-(abc)$.

Yay! We see the first, second, third symmetric sums. Time to plug in:

$\frac{1-2(-1/2)+3(-2)}{1-(1)+(-1/2)-(-2)}=\frac{-4}{3/2}=\frac{-8}{3}$.

Thus, we have $-8/3$.
This post has been edited 1 time. Last edited by djmathman, Aug 11, 2011, 11:26 AM

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  • legendary problem writer

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  • Roses are red,
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  • hello :)

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  • Do you have a link to your main blog that you started after graduating from high school, I couldn't find it. @dj I met you IRL at Awesome Math summer Program several years ago.

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