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If $a (p+q)^2+2apq+c=0 $ and $a (p+r)^2+2apr+c=0 $ then find $ q.r $ in terms of $ p, a, c$

My try: I tried to equate both equations and got the relations that either $ a=0, q=r, q+r+4p=0 $

Then I tried to square $q+r+4p=0 $ and got $ q.r=16\frac {p^2}{q^2+r^2}$ but I was unable to get value of $q^2+r^2$ in terms of $a , c $

Any move further towards answer will be useful, also I would like to know how did you think in order to get the solution

Kartik Watwani
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1 Answers1

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If $a\ne0$, $\,q,r$ are roots of the same quadratic equation: $$a(p+x)^2+2apx+c=ax^2+4apx+ap^2+c=0. $$ Now it is well-known the sum and the product of the roots (whether sum anreal or complex) are respectively: $$S=-4p,\quad P=\smash[t]{\frac{ap^2+c}a}=p^2+\frac ca$$ The sum of the squares is: $\,S^2-2P=2\Bigl(7p^2-\dfrac ca\Bigr)$.

Bernard
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