Let p,q and r be real numbers (p≠ q,r≠ 0), such that the roots of the equation 1 x+p+ 1 x+q= 1 are equal in magnitude but opposite in sign, then the…

Given quadratic equation is $\frac{1}{x+p }+\frac{1}{x+q}=\frac{1}{r}$x+p1​+x+q1​=r1​.

Let $\alpha$α and $\beta$β be the roots of given equation.

$$\Rightarrow (2x+p+q)r=(x+p)(x+q)$$⇒(2x+p+q)r=(x+p)(x+q)

$$\Rightarrow {x}^{2}+(p+q-2r)x+pq-pr-qr=0$$⇒x2+(p+q−2r)x+pq−pr−qr=0

Now, sum of roots $(\alpha +\beta )=\frac{-b}{a}=-(p+q-2r)$(α+β)=a−b​=−(p+q−2r)

$\Rightarrow -(p+q-2r)=0$⇒−(p+q−2r)=0 ($\because$∵Given that roots are equal in magnitude and opposite in sign)

$$\Rightarrow p+q=2r...(1)$$⇒p+q=2r...(1)

Product of roots $(\alpha \beta )=\frac{c}{a}=pq-pr-qr$(αβ)=ac​=pq−pr−qr

Now, ${\alpha }^{2}+{\beta }^{2}={(\alpha +\beta )}^{2}-2\alpha \beta$α2+β2=(α+β)2−2αβ

$$=0-2[pq-pr-qr]=-2pq+2r(p+q)$$=0−2[pq−pr−qr]=−2pq+2r(p+q)

$=-2pq+{(p+q)}^{2}={p}^{2}+{q}^{2}$=−2pq+(p+q)2=p2+q2 $($($\because$∵ from $(1)$(1)$)$)

$$={p}^{2}+{q}^{2}$$=p2+q2