Question

If the roots of the equation $\frac{1}{(x+p)}+\frac{1}{(x+q)}=\frac{1}{r}Wherex\ne -p,x\ne -q,r\ne 0$are equal in magnitude but opposite in sign, then $p+q=$

• A. $r$
• B. $2r$
• C. $\frac{1}{r}$
• D. $\frac{2}{r}$

Answer 1

The correct option is B

$2r$

Step 1: simplify $\frac{1}{(x+p)}+\frac{1}{(x+q)}=\frac{1}{r}$ into general quadratic form $a{x}^2+bx+c$.

$$\begin{gathered} \frac{1}{(x+p)}+\frac{1}{(x+q)}=\frac{1}{r} \\ \frac{(x+q+x+p)}{(x+p)(x+q)}=\frac{1}{r} \\ Takingcrossmultiplication. \\ (2x+p+q)\times r=(x+p)(x+q) \\ 2xr+(p+q)r=x^2+qx+px+pq \\ Rearrangingtheterms \\ {x}^2+qx+px+pq-(p+q)r=0 \\ {x}^2+x(q-2r+p)+pq-(p+q)r=0...\left(1\right) \end{gathered}$$

Step 2: Put the coefficient of $x$ equal to zero.

If the roots of the above equation are equal but opposite in sign, the coefficient of $x$ will be zero.

Compare equation (1) with$a{x}^2+bx+c$, we get the coefficient of $x$ = $(q-2r+p)$.

Hence the coefficient of $x$ is $(q-2r+p)$

$$\begin{gathered} q-2r+p=0 \\ Rearrangingtheterms \\ p+q=2r \end{gathered}$$

Therefore, option (B), $p+q=2r$ is the correct option.