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Let `f(x) = x^2+bx+c` and `g(x) = x^2+qx+r` , where `b, c, q, r in RR` and `c != r` Let...

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roshan-rox | (Level 1) Valedictorian

Posted September 25, 2013 at 7:07 AM via web

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Let `f(x) = x^2+bx+c` and `g(x) = x^2+qx+r` ,

where `b, c, q, r in RR` and `c != r`

Let `alpha, beta` be the roots of `g(x) = 0`

Show that `f(alpha) * f(beta) = (c-r)^2- (b-q) * (cq - br).`

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted September 25, 2013 at 7:54 AM (Answer #1)

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`g(x) = x^2+qx+r = 0` with roots `alpha` and `beta`

So we can say;

`alpha+beta = -q`

`alphaxxbeta = r`


`= (alpha^2+balpha+c)(beta^2+b(beta)+c)`



`= r^2+br(-q)+c(q^2-2r)+b^2r+c^2-bcq`

`= c^2-2cr+r^2-bqr+cq^2+b^2r-bcq`

`= (c-r)^2-bqr+b^2r+cq^2-bcq`

`= (c-r)^2-br(q-b)+cq(q-b)`

`= (c-r)^2-(q-b)(cq-br)`

`= (c-r)^2-(b-q)(cq-br)`

So the required answer is obtained.


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degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted September 25, 2013 at 11:31 PM (Answer #2)

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There's kind of a cool way to solve it with less factoring, which is nice for people like me who aren't the quickest at finding factorizations.

Note that `f(x)-g(x)=(b-q)x+(c-r).` Then since `g(alpha)=g(beta)=0,` we have

`f(alpha)=f(alpha)-g(alpha)=(b-q)alpha+(c-r)`  and


So you can already see where the `b-q`and `c-r` terms come from in the answer (actually those terms were what led me to try subtracting the functions in the first place). Now multiply these (you have to do a little bit of factoring at times too) and use the fact that `alpha*beta=r` and `alpha+beta=-q,` and the answer just pops out. Since the question was already solved though, I'll leave the straightforward details of this to you.

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