Question

Let $p,q$ and $r$ be real numbers $\left( {p \ne q,r \ne 0} \right)$, such that the roots of the equation $\dfrac{1}{{x + p}} + \dfrac{1}{{x + q}} = \dfrac{1}{r}$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to:
(A) ${p^2} + {q^2} + {r^2}$
(B) ${p^2} + {q^2}$
(C) $2\left( {{p^2} + {q^2}} \right)$
(D) \[\dfrac{{{p^2} + {q^2}}}{2}\]

Answer 1

Hint: First of all we convert the given equation in the standard quadratic equation, i.e., $a{x^2} + bx + c = 0$ and find $a,b,c$ to find out the sum and multiplication of the roots.

Complete step-by-step answer:
Given equation is: $\dfrac{1}{{x + p}} + \dfrac{1}{{x + q}} = \dfrac{1}{r}$
Now solving the given equation by taking LCM,
$\dfrac{{\left( {x + q} \right) + \left( {x + p} \right)}}{{\left( {x + p} \right)\left( {x + q} \right)}} = \dfrac{1}{r}$
$ \Rightarrow r\left( {x + q} \right) + r\left( {x + p} \right) = \left( {x + p} \right)\left( {x + q} \right)$
$ \Rightarrow rx + rq + rx + rp = {x^2} + qx + px + pq$
$ \Rightarrow 2rx + r\left( {p + q} \right) = {x^2} + \left( {p + q} \right)x + pq$
$ \Rightarrow {x^2} + \left( {p + q} \right)x - 2rx + pq - r\left( {p + q} \right) = 0$
$ \Rightarrow {x^2} + \left( {p + q - 2r} \right)x + pq - r\left( {p + q} \right) = 0$….. (1)
On comparing the equation (1) with the standard quadratic equation, i.e., $a{x^2} + bx + c = 0$; we get-
$a = 1,b = \left( {p + q - 2r} \right),c = pq - r\left( {p + q} \right)$
Let $\alpha $ and $\beta $ be the roots of a given equation, which are equal in magnitude but opposite in sign, i.e., $\beta = - \alpha $.
Now, sum of roots, \[\alpha + \beta = \dfrac{{ - b}}{a}\]
$ \Rightarrow \alpha + \left( { - \alpha } \right) = \dfrac{{ - \left( {p + q - 2r} \right)}}{1}$
$ \Rightarrow 0 = - p - q + 2r$
$ \Rightarrow p + q = 2r$
$ \Rightarrow r = \dfrac{{p + q}}{2}$ …. (1)
Multiplication of roots, $\alpha \beta = \dfrac{c}{a}$
$ \Rightarrow \alpha \beta = \dfrac{{pq - r\left( {p + q} \right)}}{1}$
$ \Rightarrow \alpha \beta = pq - \dfrac{{\left( {p + q} \right)}}{2}\left( {p + q} \right)$ [from(1)]
$ \Rightarrow \alpha \beta = pq - \dfrac{{{{\left( {p + q} \right)}^2}}}{2}$
$ \Rightarrow \alpha \beta = \dfrac{{2pq - \left( {{p^2} + {q^2} + 2pq} \right)}}{2}$
$ \Rightarrow \alpha \beta = \dfrac{{ - {p^2} - {q^2}}}{2}$
We have to find the sum of squares of these roots i.e., ${\alpha ^2} + {\beta ^2}$.
We know that, ${\left( {\alpha + \beta } \right)^2} = {\alpha ^2} + {\beta ^2} + 2\alpha \beta $
$ \Rightarrow {\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $
Since $\beta = - \alpha $ $ \Rightarrow $ ${\left( {\alpha + \beta } \right)^2} = 0$
$\therefore {\alpha ^2} + {\beta ^2} = - 2\alpha \beta $
$ \Rightarrow {\alpha ^2} + {\beta ^2} = - 2\left( {\dfrac{{ - {p^2} - {q^2}}}{2}} \right)$
$ \Rightarrow {\alpha ^2} + {\beta ^2} = {p^2} + {q^2}$

Hence, option (B) is the correct answer.

Note: Here roots of the equation are equal in magnitude but opposite in sign, i.e., $\beta = - \alpha $.
Therefore, $\alpha + \beta = \alpha + \left( { - \alpha } \right) = 0$ and $\alpha \beta = \alpha \left( { - \alpha } \right) = - {\alpha ^2}$.