Given,
⇒ 1 p + 1 q + 1 x = 1 x + p + q ( 1 p + 1 q ) + ( 1 x − 1 x + p + q ) = 0 ( q + p p q ) + ( x + p + q − x x ( x + p + q ) ) = 0 ( q + p p q ) + ( p + q x ( x + p + q ) ) = 0 ( p + q ) ( 1 p q + 1 x ( x + p + q ) ) = 0 ( p + q ) ( x ( x + p + q ) + p q x p q ( x + p + q ) ) = 0 ( p + q ) ( x 2 + x ( p + q ) + p q x p q ( x + p + q ) ) = 0 ∴ ( x 2 + x ( p + q ) + p q x p q ( x + p + q ) ) = 0 x 2 + x ( p + q ) + p q = 0 \Rightarrow \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{x + p + q} \\[1em] \Big(\dfrac{1}{p} + \dfrac{1}{q}\Big) + \Big(\dfrac{1}{x} - \dfrac{1}{x + p + q}\Big) = 0 \\[1em] \Big(\dfrac{q + p}{pq}\Big) + \Big(\dfrac{x + p + q - x}{x(x + p + q)}\Big) = 0 \\[1em] \Big(\dfrac{q + p}{pq}\Big) + \Big(\dfrac{p + q}{x(x + p + q)}\Big) = 0 \\[1em] (p + q)\Big(\dfrac{1}{pq} + \dfrac{1}{x(x + p + q)}\Big) = 0 \\[1em] (p + q)\Big(\dfrac{x(x + p + q) + pq}{xpq(x + p + q)} \Big) = 0 \\[1em] (p + q)\Big(\dfrac{x^2 + x(p + q) + pq}{xpq(x + p + q)}\Big) = 0 \\[1em] \therefore \Big(\dfrac{x^2 + x(p + q) + pq}{xpq(x + p + q)}\Big) = 0 \\[1em] x^2 + x(p + q) + pq = 0 \\[1em] ⇒ p 1 + q 1 + x 1 = x + p + q 1 ( p 1 + q 1 ) + ( x 1 − x + p + q 1 ) = 0 ( pq q + p ) + ( x ( x + p + q ) x + p + q − x ) = 0 ( pq q + p ) + ( x ( x + p + q ) p + q ) = 0 ( p + q ) ( pq 1 + x ( x + p + q ) 1 ) = 0 ( p + q ) ( x pq ( x + p + q ) x ( x + p + q ) + pq ) = 0 ( p + q ) ( x pq ( x + p + q ) x 2 + x ( p + q ) + pq ) = 0 ∴ ( x pq ( x + p + q ) x 2 + x ( p + q ) + pq ) = 0 x 2 + x ( p + q ) + pq = 0
Comparing above equation with ax2 + bx + c = 0 we get,
a = 1, b = (p + q), c = pq
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting value in above equation we get,
⇒ x = − ( p + q ) ± ( p + q ) 2 − 4. ( 1 ) . p q 2 ( 1 ) = − ( p + q ) ± p 2 + q 2 + 2 p q − 4 p q 2 = − ( p + q ) ± ( p − q ) 2 2 = − ( p + q ) ± ( p − q ) 2 = − ( p + q ) + ( p − q ) 2 or − ( p + q ) − ( p − q ) 2 = − p − q + p − q 2 or − p − q − p + q 2 = − 2 q 2 or − 2 p 2 = − q or − p . \Rightarrow x = \dfrac{-(p + q) \pm \sqrt{(p + q)^2 - 4.(1).pq}}{2(1)} \\[1em] = \dfrac{-(p + q) \pm \sqrt{p^2 + q^2 + 2pq - 4pq}}{2} \\[1em] = \dfrac{-(p + q) \pm \sqrt{(p - q)^2}}{2} \\[1em] = \dfrac{-(p + q) \pm (p - q)}{2} \\[1em] = \dfrac{-(p + q) + (p - q)}{2} \text{ or } \dfrac{-(p + q) - (p - q)}{2} \\[1em] = \dfrac{-p - q + p - q}{2} \text{ or } \dfrac{-p - q - p + q}{2} \\[1em] = \dfrac{-2q}{2} \text{ or } \dfrac{-2p}{2} \\[1em] = -q \text{ or } -p. ⇒ x = 2 ( 1 ) − ( p + q ) ± ( p + q ) 2 − 4. ( 1 ) . pq = 2 − ( p + q ) ± p 2 + q 2 + 2 pq − 4 pq = 2 − ( p + q ) ± ( p − q ) 2 = 2 − ( p + q ) ± ( p − q ) = 2 − ( p + q ) + ( p − q ) or 2 − ( p + q ) − ( p − q ) = 2 − p − q + p − q or 2 − p − q − p + q = 2 − 2 q or 2 − 2 p = − q or − p .
Hence, x = -q or -p.
