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 $(p - q) + (q - r) + (r - p) = 0$. If the su

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 $(p - q) + (q - r) + (r - p) = 0$. If the sum of the coefficients of a quadratic equation $ax^2 + bx + c = 0$ is $0$,then $x = 1$ is always one of the roots. Let the roots be $\alpha$ and $\beta$. We know that the product of the roots $\alpha \beta = \frac{c}{a}$. Here,$\alpha = 1$,so $1 	imes \beta = \frac{r - p}{p - q}$. 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 ${x^2} + {y^2} = 25$ and ${xy} = 12$,then the possible values of $x$ are:

If the equation $x^2 - m(2x - 8) - 15 = 0$ has equal roots,then $m = ......$

If $f(x) = x^2 + bx + c$ and $f(1+k) = f(1-k)$ for all $k \in R$,for two real numbers $b$ and $c$,then:

If $a, b, c$ are in Arithmetic Progression $(AP)$,then the roots of the equation $ax^2 - 2bx + c = 0$ are

If $72^x \cdot 48^y = 6^{xy}$,where $x$ and $y$ are non-zero rational numbers,then $x+y$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module