Question

If the roots of the equation $\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$ are equal in magnitude but opposite in sign, then the product of the roots is :

• A. $-2(p^2+q^2)$
• B. $-(p^2+q^2)$
• C. $-(p^2+q^2)/2$
• D. $-pq$

Answer 1

The correct option is C $-(p^2+q^2)/2$

$given$
$\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$
$\Rightarrow \frac{x+q+x+p}{(x+p)(x+q)}=\frac{1}{r}$
$\Rightarrow \frac{2x+p+q}{x^2+x(p+q)+pq}=\frac{1}{r}$
$\Rightarrow 2xr+pr+qr=x^2+px+qx+pq$
$\Rightarrow x^2+x(p+q-2r)+pq-pr-qr=0$
$lettherootsbe\alpha ,-\alpha$
$sumofroots=\alpha -\alpha -\frac{-b}{a}=-\frac{p+q-2r}{1}=0$
$p+q=2r$
$\Rightarrow r=\frac{p+q}{2}......\left(i\right)$
$productofroots=\frac{c}{a}=\frac{pq-pr-qr}{1}$
$=pq-r(p+q)$
$=pq-\frac{p+q}{2}(p+q)$
$=pq-\frac{{(p+q)}^2}{2}$
$=\frac{2pq-p^2-q^2-2pq}{2}$
$=-\frac{(p^2+q^2)}{2}$