Question

In the quadratic equation, $pq{x}^2-(p^2+q^2)x+pq=0,$ then find values of $x$.

• A. $pq,qp$
• B. $\frac{p}{q},\frac{q}{p}$
• C. $p,q$
• D. $p^2+q^2,\frac{p^2}{q^2}$

Answer 1

The correct option is B $\frac{p}{q},\frac{q}{p}$

In the given equation $a=pq,b=-(p^2+q^2),c=pq$
Finding the roots using formula $x=\frac{-b\pm \sqrt{(b^2-4ac)}}{2a}$
$\Rightarrow x=\frac{(p^2+q^2)\pm \sqrt{(-(p^2+q^2){)}^2-4(p^2{q}^2)}}{2pq}$
$x=\frac{(p^2+q^2)\pm \sqrt{(p^2+q^2{)}^2-4(p^2{q}^2)}}{2pq}$
$=\frac{-(p^2+q^2)\pm \sqrt{(p^2-q^2{)}^2}}{2pq}$ [Since, $(a+b{)}^2-4ab=(a-b{)}^2$]
$=\frac{(p^2+q^2)\pm (p^2-q^2)}{2pq}$
$x=\frac{-(p^2+q^2)+(p^2-q^2)}{2pq}$ or $x=\frac{-(p^2+q^2)-(p^2-q^2)}{2pq}$
$x=\frac{p^2-q^2+(p^2-q^2)}{2pq}$ or $x=\frac{p^2+q^2-p^2+q^2)}{2pq}$
$x=\frac{2p^2}{2pq}$ or $x=\frac{2q^2}{2pq}$
Substituting and simplifying we get, $x=\frac{p}{q}$ and $\frac{q}{p}$