Question

If $\frac{x^2-bx}{ax-c}=\frac{m-1}{m+1}$ has roots which are numerically equal but of opposite sings, the value of m must be:

• A. $\frac{a-b}{a+b}$
• B. $\frac{a+b}{a-b}$
• C. c
• D. $\frac{1}{c}$

Answer 1

Answer: (A)

$\frac{a-b}{a+b}$

$⟹\frac{x^2-bx}{ax-c}=\frac{m-1}{m+1}⟹(m+1)x^2-b(m+1)x=(m-1)ax-c(m-1)⟹(m+1)x^2-\left[b\right(m+1)+(m-1\left)a\right]x+c(m-1)=0$

Roots are numerically equal but of opposite sign.

$\therefore$ Sum of roots = 0

$⟹(b+a)m+(b-a)=0$

$\therefore m=\frac{a-b}{a+b}$