If the roots of the equation $\frac{x^2 - bx}{ax - c} = \frac{m - 1}{m + 1}$ are equal in magnitude but opposite in sign,then $m = \dots$

  • A
    $\frac{a + b}{a - b}$
  • B
    $\frac{a - b}{a + b}$
  • C
    $\frac{b - a}{b + a}$
  • D
    None of these

Explore More

Similar Questions

If $\alpha, \beta, \gamma$ are the roots of $x^3-x+1=0$,then $\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}=$

If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2x + 3 = 0$,then find the equation whose roots are $\frac{\alpha - 1}{\alpha + 1}$ and $\frac{\beta - 1}{\beta + 1}$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-12x^2+kx-18=0$ and one of them is thrice the sum of the other two roots,then $\alpha^2+\beta^2+\gamma^2-k=$

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-ax^2+bx-c=0$,then $\Sigma \alpha^2(\beta+\gamma) = $

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + 2x - 5 = 0$ and the equation $x^3 + bx^2 + cx + d = 0$ has roots $2\alpha + 1, 2\beta + 1, 2\gamma + 1$,then the value of $|b + c + d|$ is (where $b, c, d$ are constants):

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo