Question

Using quadratic formula, find the value of x.
$p^2{x}^2+(p^2-q^2)x-q^2=0,p\ne 0$

• A. $\frac{q^2}{p^2},-1$
• B. $-\frac{q^2}{p^2},-1$
• C. $0,-1$
• D. $\frac{q^2}{p^2},1$

Answer 1

The correct option is A $\frac{q^2}{p^2},-1$

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

Comparing this equation with $a{x}^2+bx+c=0$, we have,
$a=p^2,b=p^2-q^{2}andc=-q^2$.
Now, discriminant,
$b^2-4ac=(p^2-q^2{)}^2-4\times p^2\times -q^2$

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

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

Now finding roots,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow x=\frac{-(p^2-q^2)\pm \sqrt{(p^2+q^2{)}^2}}{2p^2}$

$\Rightarrow x=\frac{-(p^2-q^2)+(p^2+q^2)}{2p^2}orx=\frac{-(p^2-q^2)-(p^2+q^2)}{2p^2}$

$\Rightarrow x=\frac{q^2}{p^2}orx=-1$