If a(p + q)^2 + 2bpq + c = 0 and a(p + r)^2 + | vedclass

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Question

(A) Consider the equation $a(p + x)^2 + 2bpx + c = 0$. Expanding this,we get $a(p^2 + 2px + x^2) + 2bpx + c = 0$. Rearranging the terms as a quadratic

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$p^2 + \frac{c}{a}$

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(A) Consider the equation $a(p + x)^2 + 2bpx + c = 0$. Expanding this,we get $a(p^2 + 2px + x^2) + 2bpx + c = 0$. Rearranging the terms as a quadratic in $x$: $ax^2 + 2x(ap + bp) + ap^2 + c = 0$,which simplifies to $ax^2 + 2xp(a + b) + ap^2 + c = 0$. Since $q$ and $r$ are the roots of this equation,the product of the roots $qr$ is given by the constant term divided by the coefficient of $x^2$. Thus,$qr = \frac{ap^2 + c}{a} = p^2 + \frac{c}{a}$.

If $a(p + q)^2 + 2bpq + c = 0$ and $a(p + r)^2 + 2bpr + c = 0$,then $qr$ =

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