Question

The value of m for which the equation $\dfrac{a}{x+a+m}+\dfrac{b}{x+b+m}=1$ has roots equal in magnitude but opposite in sign is
(a) $\dfrac{a+b}{a-b}$
(b) 0
(c) $\dfrac{a-b}{a+b}$
(d) $2\left( a-b \right)\left( a+b \right)$

Answer 1

Hint: Take the L.C.M and simplify the terms by cross-multiplication to form a quadratic equation. Assume one of the roots as ‘k’, therefore the other root is ‘-k’. Use the relation: sum of roots of a quadratic equation is $\dfrac{-B}{A}$, where B is the coefficient of x and A is the coefficient of ${{x}^{2}}$, to find the value of m.

Complete step-by-step solution -
We have been given the equation $\dfrac{a}{x+a+m}+\dfrac{b}{x+b+m}=1$. Let us simplify this equation.
Taking L.C.M we get,
$\begin{align}
  & \dfrac{ax+ab+am+bx+ab+bm}{\left( x+m+a \right)\left( x+m+b \right)}=1 \\
 & \Rightarrow \dfrac{ax+ab+am+bx+ab+bm}{{{x}^{2}}+{{m}^{2}}+2xm+bx+bm+ax+am+ab}=1 \\
\end{align}$
By cross-multiplication we get,
$ax+ab+am+bx+ab+bm={{x}^{2}}+{{m}^{2}}+2xm+bx+bm+ax+am+ab$
Cancelling common terms we get,
$\begin{align}
  & ab={{x}^{2}}+{{m}^{2}}+2xm \\
 & \Rightarrow {{x}^{2}}+{{m}^{2}}+2xm-ab=0 \\
\end{align}$
It is given that the roots of the above quadratic equation are equal in magnitude but opposite in sign. Therefore, let us assume one root as ‘k’, therefore, the other root must be ‘-k’.
We know that, sum of roots of a quadratic equation is $\dfrac{-B}{A}$, where B is the coefficient of x and A is the coefficient of ${{x}^{2}}$. Therefore,
$\begin{align}
  & k+\left( -k \right)=\dfrac{-2m}{1} \\
 & \Rightarrow 0=-2m \\
 & \Rightarrow m=0 \\
\end{align}$
Hence, option (b) is the correct answer.

Note: It is important to convert the equation into its quadratic form so that we can apply the formula for the sum of the roots of the quadratic equation. One may note that we haven’t applied the formula for the product of roots of the quadratic equation. This is because if we will do so, then the variable ‘k’ will appear in the equation whose value is unknown. Hence, we will not reach any conclusion. You may note that we can find the value of ‘k’ by discriminant method but that will be a lengthy process.