Question

If the roots of the quadratic equation $p(q-r)x^2+q(r-p)x+r(p-q)=0,$ are equal, then find the value of $\frac{2}{q}$ where $p,q,r\in R$

• A. $\frac{1}{p}-\frac{1}{r}$
• B. $pr$
• C. $\frac{1}{p}+\frac{1}{r}$
• D. $p+r$

Answer 1

The correct option is C

$\frac{1}{p}+\frac{1}{r}$

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

$\because$ Roots are equal
$\therefore D=0$
$\Rightarrow D=\left(q\right(r-p){)}^2-4(p(q-r)\left)\right(r(p-q)\left)\right)=0$

$\Rightarrow q^2(r^2+p^2-2pr)-4\left(\right(pq-pr\left)\right(pr-qr\left)\right)=0$

$\Rightarrow q^2(r^2+p^2-2pr)-4(p^{2}qr-p{q}^{2}r-p^2{r}^2+pq{r}^2)=0$

$\Rightarrow q^2{r}^2+p^2{q}^2-2p{q}^{2}r-4p^{2}qr+4p{q}^{2}r+4p^2{r}^2+4pq{r}^2=0$

$\Rightarrow q^2{r}^2+p^2{q}^2+4p^2{r}^2-4p^{2}qr+2p{q}^{2}r+4pq{r}^2=0$

$\Rightarrow (pq+qr-2pr{)}^2=0\left[\right(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca]$

$\Rightarrow pq+qr=2pr$
On dividing by $pqr$ both sides, we get

$\frac{1}{r}+\frac{1}{p}=\frac{2}{q}$

Or, $\frac{2}{q}=\frac{1}{p}+\frac{1}{r}$