Question

Using quadratic formula solve the following quadratic equation:

$m^2{x}^2+(m^2-n^2)x-n^2=0,m\ne 0$

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

Answer 1

The correct option is A The roots are : $\frac{n^2}{m^2},-1$

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

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

$a=m^2,b=m^2-n^{2}andc=-n^2$

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

$=(m^2-n^2{)}^2+4m^2{n}^2$

$=(m^2+n^2{)}^2$

$\Rightarrow D>0$

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

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

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

$\therefore$The roots are $\frac{n^2}{m^2},-1.$