An algebra problem by Naren Bhandari

Given that

$p(q-r)x^2 + q(r-p)x + r(p-q) = 0$ Comparing above equation with quadratic equation $ax^2 + bx + c = 0$ we get,

$a = p(q-r)$

$b = q(r-p)$

$c = r(p-q)$

Since the roots are equal so it determinant $D = 0$ i.e; $D = b^2 - 4ac = 0$

Now substituting the value of $a$ , $b$ and $c$ we get

$q(r-p))^2 - 4(pr(q-r)(p-q)) = 0$

$q^2(r^2 - 2pr + p^2) - 4(pr(pq- q^2 - pr + qr)) = 0$

$q^2r^2 - 2prq^2 + p^2q^2 - 4(p^2rq - prq^2 - p^2r^2 + pqr^2 )= 0$

$q^2r^2 + (-2prq^2 + 4prq^2) + p^2q^2 - 4(p^2rq + pqr^2) + 4p^2r^2= 0$

$(qr + pq)^2 - 4(qr + pq)pr + (2pr)^2 = 0$

$(qr +pq -2pr)^2 = 0$

$qr + pq = 2pr$

Dividing both $pqr$ we get,

$\frac{qr}{pqr} + \frac{pq}{pqr} = \frac{2pr}{pqr}$

$\frac{1}{p} + \frac{1}{q} = \frac{2}{q}$

$\begin{aligned} p(q-r)x^2 + q(r-p)x + r(p-q) & = 0 & \small \color{#3D99F6} \text{Divide both sides by } pr \\ \left(\frac qr -1 \right)x^2 + q \left(\frac 1p -\frac 1r \right)x + \left(1-\frac qp \right) & = 0 \end{aligned}$

For the equation to have two equal roots, the discriminant $b^2 - 4ac = 0$ $\implies b^2 = 4ac$ :

$\begin{aligned} q^2 \left(\frac 1p -\frac 1r \right)^2 & = 4 \left(\frac qr -1 \right)\left(1-\frac qp \right) \\ q^2 \left(\frac 1{p^2} - \frac 2{pr} + \frac 1{r^2} \right) & = 4 \left(\frac qr - \frac {q^2}{pr} - 1 + \frac qp \right) \\ q^2 \left(\frac 1{p^2} + \frac 2{pr} + \frac 1{r^2} \right) & = 4 q \left( \frac 1r + \frac 1p \right) - 4 \\ q^2\left(\frac 1p + \frac 1r \right)^2 & = 4 q \left( \frac 1r + \frac 1p \right) - 4 \end{aligned}$

$\begin{aligned} \implies q^2\left(\frac 1p + \frac 1r \right)^2 - 4 q \left( \frac 1p + \frac 1r \right) + 4 & = 0 & \small \color{#3D99F6} \text{A quadratic equation of }q \left( \frac 1p + \frac 1r \right) \\ \left(q \left( \frac 1p + \frac 1r \right) - 2\right)^2 & = 0 \\ \implies q \left( \frac 1p + \frac 1r \right) & = 2 \\ \frac 1p + \frac 1r & = \boxed{\dfrac 2q} \end{aligned}$

The question is too simple.check that $x=1$ is the root so product= $1*1$ .so $r(p-q)=p(q-r)$ .now divide by $pqr$ so $1/p+1/r=2/q$