Question

The roots of the quadratic polynomial $(q-r)x^2+(r-p)x+(p-q)$ are

• A. $\frac{(r-p)}{(q-r)},\frac{1}{2}$
• B. $\frac{(p-q)}{(q-r)},1$
• C. $\frac{(q-r)}{(p-q)},1$
• D. $\frac{(r-p)}{(p-q)},\frac{1}{2}$

Answer 1

The correct option is B $\frac{(p-q)}{(q-r)},1$

Given quadratic equation $(q-r)x^2+(r-p)x+(p-q)=0$,

Here sum of the coefficients are $0$,

$\therefore$ one of the root is $1$,

Let the other root be $\alpha$,

Consider product of the roots from the equation,

i.e., $\alpha .1=\frac{p-q}{q-r}⟹\alpha =\frac{p-q}{q-r}$,

$\therefore$ The roots of the quadratic equation are $1,\frac{p-q}{q-r}$