Question

Using quadratic formula solve the following quadratic equation:

$p^2{x}^2+(p^2-q^2)x-q^2=0,p\ne 0$

• A. The roots are : $\frac{q^2}{p^2},-1$
• B. The roots are : $\frac{q^2}{p^2},-6$
• C. The roots are : $\frac{2q^2}{p^2},-1$
• D. The roots are : $\frac{q^2}{9p^2},-1$

Answer 1

The correct option is A

The roots are : $\frac{q^2}{p^2},-1$

Given equation is: $p^2{x}^2+(p^2-q^2)x-q^2=0,p\ne 0$

Comparing this equation with $a{x}^2+bx+c=0$, we have

$a=p^2,b=p^2-q^{2}andc=-q^2$

$\therefore$Discriminant,$D=b^2-4ac=(p^2-q^2{)}^2-4\times p^2\times -q^2$

$=(p^2-q^2{)}^2+4p^2{q}^2$

$=(p^2+q^2{)}^2$

$\Rightarrow D>0$

So, the given equation has real roots and given by,

$\alpha =\frac{-b+\sqrt{D}}{2a}=\frac{-(p^2-q^2)+(p^2+q^2)}{2p^2}=\frac{q^2}{p^2}$

and, $\beta =\frac{-b-\sqrt{D}}{2a}=\frac{-(p^2-q^2)-(p^2+q^2)}{2p^2}=-1$

$\therefore$The roots are $\frac{q^2}{p^2},-1.$