If the roots of the equation frac{x^2 - bx}{a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation: $\frac{x^2 - bx}{ax - c} = \frac{m - 1}{m + 1}$.Cross-multiplying,we get: $(m + 1)(x^2 - bx) = (m - 1)(ax - c)$.Expanding the

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the equation: $\frac{x^2 - bx}{ax - c} = \frac{m - 1}{m + 1}$.Cross-multiplying,we get: $(m + 1)(x^2 - bx) = (m - 1)(ax - c)$.Expanding the terms: $(m + 1)x^2 - (m + 1)bx = (m - 1)ax - (m - 1)c$.Rearranging into the standard quadratic form $Ax^2 + Bx + C = 0$:$(m + 1)x^2 - [b(m + 1) + a(m - 1)]x + c(m - 1) = 0$.Expanding the coefficient of $x$: $b(m + 1) + a(m - 1) = bm + b + am - a = m(a + b) - (a - b)$.So,the equation is $(m + 1)x^2 - [m(a + b) - (a - b)]x + c(m - 1) = 0$.Since the roots are equal in magnitude but opposite in sign,their sum must be zero.For a quadratic equation $Ax^2 + Bx + C = 0$,the sum of roots is $-B/A$.Setting the sum of roots to zero: $m(a + b) - (a - b) = 0$.Therefore,$m(a + b) = a - b$,which gives $m = \frac{a - b}{a + b}$.

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

Similar Questions

${x \in R : |x - 2| = x^2} = $

Solve the given two equations and select the correct answer from the given options.
$I.$ $7x = 4y + 85$
$II.$ $y = \sqrt[3]{17576}$

Let $x_1, x_2, x_3 \in R - \{0\}$,$x_1 + x_2 + x_3 \neq 0$ and $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$. Then $\frac{1}{x_1^n + x_2^n + x_3^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{x_3^n}$ holds good for:

Number of integral values of $a$ for which both roots of the quadratic equation $x^2 - (2a + 3)x + a^2 + 3a = 0$ lie in the interval $(0, 4)$ is:

The number of solutions for the equation ${x^2} - 5|x| + 6 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module