Question

If the roots of the equation $\frac{a}{x+a+k}+\frac{b}{x+b+k}=2$ are equal in magnitude but opposite in sign, then the value of $k$ is:

• A. $\frac{(a+b)}{4}$
• B. $\frac{(a+b)}{3}$
• C. $\frac{(a+b)}{2}$
• D. $\frac{-(a+b)}{4}$

Answer 1

The correct option is D $\frac{-(a+b)}{4}$

On taking the LCM and simplifying the equation we get,

$a(x+b+k)+b(x+a+k)=2(x+b+k)(x+a+k)$

$ax+ab+ak+bx+ab+bk=2(x^2+bx+kx+ax+ab+ak+kx+kb+k^2)$

$2x^2+2bx+2ax+4kx+2k^2+2ak+2bk=ax+ak+bx+bk$

$2x^2+bx+ax+4kx+2k^2+ak+bk=0$

$2x^2+x(a+b+4k)2k^2+ak+bk=0$

Now let the roots of the equation be $p$ and $q$.

So,

$p+q=0$ (because roots are same in magnitude but opposite in sign)

Also, we know that,

$p+q=\frac{-(a+b+4k)}{2}$

so,

$\frac{-(a+b+4k)}{2}=0$

$-(a+b+4k)=0$

$a+b+4k=0$

$4k=-(a+b)$

$k=\frac{-(a+b)}{4}$