Question

If the roots of the equation $\dfrac{1}{x+p}+\dfrac{1}{x+q}=\dfrac{1}{r}$ are equal in magnitude but opposite in sign, then the product of the roots will be.
A. $\dfrac{p^{2}+q^{2}}{2}$
B. $\dfrac{-\left ( p^{2}+q^{2} \right )}{2}$
C. $\dfrac{p^{2}-q^{2}}{2}$
D. $\dfrac{-\left ( p^{2}+q^{2} \right )}{2}$

Answer 1

Hint:To solve this problem, we need to take the value of another root to be negative of the first one. Further proceeding, we need to find the value of a sum of the roots which will give us the value of r. Substituting its value in the product of roots expression we will get the ultimate result.

Formula Used:
Sum of roots, $\alpha + \beta = \dfrac{{ - B}}{A}$
Product of roots, $\alpha \beta = \dfrac{C}{A}$

Complete step by step Solution:
Let us first consider the roots of the equation to be $\alpha$ and$ -\alpha$.
Here, the given equation is $\dfrac{1}{x+p}+\dfrac{1}{x+q}=\dfrac{1}{r}$
Which can be written in the general form of a quadratic equation.
So, we can write it as
$\dfrac{x+q+x+p}{x^{2}+px+qx+pq}=\dfrac{1}{r} $
Which implies,
$\dfrac{\left ( 2x+p+q \right )r}{x^{2}+px+qx+pq}=1$
Now,
$2xr+pr+qr=x^{2}+px+qx+pq$
On further simplifying, we get
$x^{2}+x\left ( p+q-2r \right )+pq-pr-qr=0$
Thus, we can say that the above equation is the form of the general quadratic equation which is $ax^{2}+bx+c=0$. Now comparing this equation with the above one we can determine the value of a,b and c.
So, the product of the roots will be
$-\alpha ^{2}=pq-qr-pr=pq-r\left ( p+q \right )$
Also, the sum of the roots will be
$0=p+q-2r$
$r=\dfrac{p+q}{2}$
Substituting the value of r in the equation of product of roots, we get
$-\alpha ^{2}=pq-\dfrac{p+q}{2}\left ( p+q \right )$
$=\dfrac{-1}{2}\left \{ \left ( p+q \right )^{2} -2pq\right \}$
On further simplifying, we get
$=\dfrac{-\left ( p^{2}+q^{2} \right )}{2}$

Hence, the correct option is (D).


Note: It is to be noted that the given equation must be simplified in the form of a general quadratic equation. This needful step can give us the value of a,b, and c which will help determine the product of roots.