Question

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

• A. $-\frac{1}{2}(p^2+q^2)$
• B. $\frac{p^2+q^2}{2}$
• C. $\frac{p+q}{2}$
• D. $\frac{1}{2}(p+q{)}^2$

Answer 1

The correct option is A $-\frac{1}{2}(p^2+q^2)$

Given that the equation

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

has roots equal in magnitude but opposite in sign $\Rightarrow$ the sum of roots $=0$

We know that if ${ax}^2+bx+c=0$

then sum of roots $=-b/a$

Now, $\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$

$\Rightarrow (2n+p+q)r=x^2+(p+q)x+pq$

$\Rightarrow x^2+(p+q-2r)x+pq-(p+q)r=0$

Sum of roots $=0$

$\Rightarrow \frac{-(p+q-2r)}{1}=0\Rightarrow p+q=2r\Rightarrow r=\frac{p+q}{2}$

$\therefore$ Products of roots $=pq-(p+q)r$

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

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

$=-\left(\frac{p^2+q^2}{2}\right)$