Question

Hi! :D

If the sum of the roots of the quadratic equation :
$\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$
is zero, show that the product of the roots is :
$( - \frac { p^{2} + {q}^{2} }{2} )$
​

Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

Quadratic equation :-$\dfrac{1}{x + p} + \dfrac{1}{x + q} = \dfrac{1}{r}$

Solving it further,

$\leadsto { \dfrac{x + q + x + p}{(x + p)(x + q)} = \dfrac{1}{r}}$

$\leadsto { \dfrac{2x + p + q}{(x + p)(x + q)} = \dfrac{1}{r}}$

$\leadsto {(2x + p + q)r = (x + p)(x + q)}$

$\leadsto {2xr + pr + qr = {x}^{2} + x(p + q) + pq}$

$\leadsto {{x}^{2} + x(p + q - 2r) + (pq - r(p + q)) = 0}$

So;$a = 1$,

$b = (p + q - 2r)$and

$c = [pq - r(p + q)]$

Sum of zeros :- $\alpha + \beta = \dfrac{ - b}{a} = \dfrac{ - (p + q - 2r)}{1}$

Product of zeros:- $\alpha \times \beta = \dfrac{c}{a} = \dfrac{pq - r(p + q)}{1}$

It is given that,

Sum of the zeros is equal to 0.

So,

$\leadsto { \dfrac{ - (p + q - 2r)}{1} = 0}$

$\leadsto { - (p + q - 2r) = 0}$

$\leadsto { - p - q + 2r = 0}$

$\leadsto { - p - q = - 2r}$

$\leadsto { - (p + q) = - 2r}$

$\leadsto {p + q = 2r}$

$\leadsto {r = \dfrac{p + q}{2} }$

Substituting the value of "r" in the equation formed by the product of zeros:-

$= \dfrac{pq - (( \dfrac{p + q}{2})( p + q)) }{1}$

$= pq - ( \dfrac{ {(p + q)}^{2} }{2} )$

$= pq - (\dfrac{ {p}^{2} + {q}^{2} + 2pq }{2} )$

$= pq - \dfrac{ {p}^{2} - {q}^{2} - 2pq}{2}$

$= \dfrac{2pq - {p}^{2} - {q}^{2} - 2pq}{2}$

$= \dfrac{ - {p}^{2} - {q}^{2} }{2}$

$=( - \dfrac { p^{2} + {q}^{2} }{2} )$

Hence proved !!!!

Answer 2

Given Equation,

$\large{ \sf{ \frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r} }}$

Any quadratic equation should be of the form ax² + bx + c = 0.

Implies,

$\large{ \hookrightarrow \: \sf{ \frac{(x + p) + (x + q)}{(x + p)(x + q)} = \frac{1}{r} }} \\ \\ \large{ \hookrightarrow \: \sf{r(2x + p + q) = x {}^{2} + (p + q)x + pq }} \\ \\ \large{ \hookrightarrow \: \sf{2xr + pr + qr = x {}^{2} + (p + q)x + pq}} \\ \\ \large{ \hookrightarrow \: \boxed{\sf{ {x}^{2} + (p + q - 2r)x + pq - qr - pr = 0 }}}$

Here,

• a = 1
• b = p + q - 2r
• c = pq - qr - pr

Sum of the Roots is Zero

$\large{ \implies \: \sf{ - (p + q - 2r) = 0}} \\ \\ \large{ \implies \: \sf{2r = p + q}} \\ \\ \large{ \implies \: \boxed{ \boxed{ \sf{r = \frac{p + q}{2} }}}..............(1)}$

Product of Roots will be (pq - qr - pr)

$\large{ \longrightarrow \: \sf{pq - q( \frac{p + q}{2}) - p( \frac{p + q}{2}) \ \ \ \ \ [From (1)]}} \\ \\ \large{ \longrightarrow \: \sf{pq - \frac{pq}{2} - \frac{ {q}^{2} }{2} - \frac{{p}^{2} }{2} - \frac{pq}{2} }} \\ \\ \large{ \longrightarrow \: \sf{pq - \frac{2pq}{2} - \frac{1}{2}(p {}^{2} + q {}^{2}) }} \\ \\ \huge{ \longrightarrow \: \sf{ - \frac{(p {}^{2} + {q}^{2}) }{2} }}$

Henceforth, Proved

Note

• Sum of Roots : - x coefficient/x² coefficient

• Product of Roots : constant term/x² coefficient