The roots of the given equation (p - q){x^2} | vedclass

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

Question

(C) Given equation is $(p - q)x^2 + (q - r)x + (r - p) = 0$.Notice that the sum of the coefficients is $(p - q) + (q - r) + (r - p) = p - q + q - r +

Vedclass · Accepted Answer

$\frac{r - p}{p - q}, 1$

Vedclass · Answer

(C) Given equation is $(p - q)x^2 + (q - r)x + (r - p) = 0$.Notice that the sum of the coefficients is $(p - q) + (q - r) + (r - p) = p - q + q - r + r - p = 0$.If the sum of the coefficients of a quadratic equation $ax^2 + bx + c = 0$ is zero,then $x = 1$ is always one of the roots.Let the roots be $\alpha$ and $\beta$. We know that the product of the roots $\alpha \cdot \beta = \frac{c}{a}$.Here,$\alpha = 1$,so $1 \cdot \beta = \frac{r - p}{p - q}$.Therefore,the roots are $1$ and $\frac{r - p}{p - q}$.

The roots of the given equation $(p - q){x^2} + (q - r)x + (r - p) = 0$ are

Similar Questions

If $2 + i\sqrt{3}$ is a root of the equation $x^2 + px + q = 0$,where $p$ and $q$ are real,then $(p, q) = $

The number of solutions of $\frac{\log 5 + \log (x^2 + 1)}{\log (x - 2)} = 2$ is

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-82x+781=0$
$II.$ $y^{2}=5041$

If $\alpha$ and $\beta$ are the roots of the equation $7x^{2}-3x-2=0$,then the value of $\frac{\alpha}{1-\alpha^{2}}+\frac{\beta}{1-\beta^{2}}$ is equal to

Number of rational roots of the equation $x^{2016} - x^{2015} + x^{1008} + x^{1003} + 1 = 0$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module