Question

If the sum of the roots of the equation $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\left[x+a\right]$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\left[x+b\right]$}\right.=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$k$}\right.$ is $0$, then their product is:

• A. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(a^2+b^2\right)$
• B. $-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(a^2+b^2\right)$
• C. ${\left(\raisebox{1ex}{$a+b$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2$
• D. None

Answer 1

The correct option is B

$-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(a^2+b^2\right)$

Explanation for the correct option.

Step 1: Express the given equation as a quadratic equation.

$\begin{array}{rcl}\frac{1}{x+a}+\frac{1}{x+b}& =& \frac{1}{k}\\ \Rightarrow \frac{x+a+x+b}{x^2+\left(a+b\right)x+ab}& =& \frac{1}{k}\\ \Rightarrow \left(2x+a+b\right)k& =& x^2+\left(a+b\right)x+ab\\ & \Rightarrow & x^2+\left(a+b-2k\right)x+ab-\left(a+b\right)k=0\end{array}$

Step 2: Apply the formula of sum of roots.

It is given that the sum of roots is $0$.

The sum of roots of a quadratic equation is $\mathbf{-}\frac{\mathbf{b}}{\mathbf{a}}$.

$\begin{array}{rcl}& \Rightarrow & \frac{-\left(a+b-2k\right)}{1}=0\\ \Rightarrow 2k-a-b& =& 0\\ \Rightarrow k& =& \frac{a+b}{2}\end{array}$

Step 3: Find the product of roots.

The product of roots of a quadratic equation $\mathbf{=}\mathbf{-}\frac{\mathbf{c}}{\mathbf{a}}$

$$\begin{gathered} =\frac{ab-\left(a+b\right)k}{1} \\ =ab-\left(a+b\right)\left(\frac{a+b}{2}\right) \\ =ab-\frac{{\left(a+b\right)}^2}{2} \\ =\frac{2ab-a^2-2ab-b^2}{2} \\ =-\frac{a^2+b^2}{2} \end{gathered}$$

Hence, option B is correct.