If the roots of the equation frac{alpha}{x - | vedclass

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

Question

(A) Given equation: $\frac{\alpha}{x - \alpha} + \frac{\beta}{x - \beta} = 1$Multiplying by $(x - \alpha)(x - \beta)$,we get:$\alpha(x - \beta) + \bet

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given equation: $\frac{\alpha}{x - \alpha} + \frac{\beta}{x - \beta} = 1$Multiplying by $(x - \alpha)(x - \beta)$,we get:$\alpha(x - \beta) + \beta(x - \alpha) = (x - \alpha)(x - \beta)$$\alpha x - \alpha \beta + \beta x - \alpha \beta = x^2 - \alpha x - \beta x + \alpha \beta$$(\alpha + \beta)x - 2\alpha \beta = x^2 - (\alpha + \beta)x + \alpha \beta$Rearranging terms to form a quadratic equation:$x^2 - 2(\alpha + \beta)x + 3\alpha \beta = 0$Let the roots be $k$ and $-k$. For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots is $-b/a$.Sum of roots: $k + (-k) = 0$From the equation,the sum of roots is $2(\alpha + \beta)$.Therefore,$2(\alpha + \beta) = 0$,which implies $\alpha + \beta = 0$.

If the roots of the equation $\frac{\alpha}{x - \alpha} + \frac{\beta}{x - \beta} = 1$ are equal in magnitude but opposite in sign,then $\alpha + \beta = $

Similar Questions

The value of $p$ for which one root of the equation ${x^2} - 30x + p = 0$ is the square of the other,are

If $c > 0$ and the equation $3ax^2 + 4bx + c = 0$ has no real root,then :-

The solution of the equation $x + \frac{1}{x} = 2$ is:

If $\alpha_1 < \alpha_2 < \alpha_3 < \alpha_4 < \alpha_5 < \alpha_6$,then the equation $(x - \alpha_1)(x - \alpha_3)(x - \alpha_5) + 3(x - \alpha_2)(x - \alpha_4)(x - \alpha_6) = 0$ has :-

Determine $p$ such that the equation $x^{2}+5px+16=0$ has no real roots.

Vedclass Test Series

Exam Paper Generator

Online Exam Module