Question

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

• A. $\frac{\left[r-p\right]}{\left[q-r\right]},1$
• B. $\frac{\left[p-q\right]}{\left[q-r\right]},1$
• C. $\frac{\left[p-q\right]}{\left[q-r\right]},2$
• D. $\frac{\left[q-r\right]}{\left[p-q\right]},2$
• E. $\frac{\left[r-p\right]}{\left[p-q\right]},1$

Answer 1

The correct option is B

$\frac{\left[p-q\right]}{\left[q-r\right]},1$

Explanation for the correct option:

Find the roots of the given equation

If $\alpha ,\beta$ are the roots of the quadratic equation $a{x}^2+bx+c=0$, then

$\alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{d}$

Now comparing the quadratic equation $\left(q-r\right)x^2+\left(r-p\right)x+\left(p-q\right)=0$ with the standard form we get

$\alpha +\beta =\frac{p-r}{q-r}$

$\alpha \beta =\frac{p-q}{q-r}$ …….(i)

Putting $x=1$ in the quadratic equation $f\left(x\right)=\left(q-r\right)x^2+\left(r-p\right)x+\left(p-q\right)$ we get,

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

Clearly one root is $1$

Say $\beta =1$

Putting the value of $\beta$ in equation (i) we get

$\alpha =\frac{p-q}{q-r}$

Therefore the roots are $\frac{\left[p-q\right]}{\left[q-r\right]},1$

Hence, the correct option is $\left(\mathrm{B}\right)$