Question

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

• A. $\frac{p-q}{q-r},1$
• B. $\frac{q-r}{p-q},1$
• C. $\frac{r-p}{p-q},1$
• D. $\frac{r-p}{q-r},1$

Answer 1

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

$(q-r)x^2+(r-p)x+(p-q)=0$
Sum of the coefficients, $q-r+r-p+p-q=0$. Hence, 1 is a root.
Synthetic division
$1|(q-r)+(r-p)+(p-q)$
$|\underset{–}{0+q-r+q-p}$
$q-r+q-p|\underset{–}{0}$
Quotient $(q-r)x+(q-p)=0$ gives $x=\frac{p-q}{q-r}$.