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, show that $p+q=2r$ and that the product of the roots is equal to $\frac{-1}{2}(p^2+q^2)$.

Answer 1

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

Simplify the given expression.

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

$2xr+pr+qr=x^2+px+qx+pq$

$x^2-(2r-p-q)x+pq-pr-qr=0$

The sum of roots.$=\frac{2r-p-q}{1}$$=2r-p-q$

Let the roots of the equation are $x_1$ and $x_2$.

Since, the roots are equal in magnitude and opposite in sign. Therefore,

$x_1=a$ and $x_2=-a$

$a-a=2r-p-q$

$0=2r-p-q$

$2r=p+q$

The product of roots.$=\frac{pq-pr-qr}{1}$$=pq-(p+q)r$ (1)

substitute$2r=p+q$ in equation(1),

$pq-2r^2$

Thus, the product of the roots is $pq-2r^2$.