Question

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

• A. $\frac{p-q}{r-q},1$
• B. $\frac{q-r}{p-q},1$
• C. $\frac{r-p}{p-q},1$
• D. none

Answer 1

The correct option is C $\frac{r-p}{p-q},1$

$(p-q)x^2+(q-r)+(r-p)=0$

$(p-q)x^2+(q-p)x+(p-r)x+(r-p)=0$

$(p-q)x(x-1)+(p-r)(x-1)=0$

$\left(\right(p-q)x+(p-r\left)\right)(x-1)=0$

$x=\frac{-(p-r)}{p-q},x=1$

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