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Find all values of p, such that the roots of $$f(x) = x^3 + 8x^2 + px + 12$$ satisfy $x_1 + x_2 = x_3$

So I was trying to use the Viet formulas, and I get: $$x_1 + x_2 + x_3 = -8$$ $$x_1x_2 + x_1x_3 + x_2x_3 = p$$ $$x_1x_2 x_3 = -12$$ $$x_1 + x_2 = x_3$$ But I can't find an easy way to get a formula for p. Any tips?

dxiv
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Simon Jachson
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3 Answers3

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Since $x_1 + x_2 + x_3 = -8$ and $x_1 + x_2 = x_3$, then $x_3 = -4$. Then $$f(-4) = (-4)^3 + 8 (-4)^2 + p*(-4) + 12 = 76 - 4p = 0 $$ Then $p = 19$.

Paul
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Apart from the other answer, if you want to use Vieta's formulas, you can see that $x_1x_2=\frac{-12}{-4}=3$. Using this in $x_1x_2 + x_1x_3 + x_2x_3 = p$, we get $$x_1x_2 + (x_1 + x_2)x_3 = p\\\implies x_1x_2 + x_3^2 = p\\\implies 3+16=p$$

So, the answer is $p=19$.

aarbee
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We know $x_1+x_2+x_3=-8$ and $x_1+x_2=x_3$. Thus we get $x_3=-4$.

$$ f(x)=x^3+8x^2+px+12=0 $$

Put $x$ as $-4$ and find the value of $p$