Question

Solve the following quadratic equation using quadratic formula: $a^2{x}^2+(a^2-b^2)x-b^2=0$

• A. $\{-1,\frac{b^2}{a^2}\}$
• B. $\{1,\frac{b^2}{a^2}\}$
• C. $\{-1,\frac{-b^2}{a^2}\}$
• D. None of these

Answer 1

The correct option is A $\{-1,\frac{b^2}{a^2}\}$

$\Rightarrow$ The given quadratic equation is $a^2{x}^2+(a^2-b^2)x-b^2=0$, comparing it with $a{x}^2+bx+c=0$

$\Rightarrow$ We get, $a=a^2,b=(a^2-b^2),c=-b^2$

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

$=\frac{-(a^2-b^2)\pm \sqrt{(a^2-b^2{)}^2-4(a^2\left)\right(-b^2)}}{2\left(a^2\right)}$

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

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

$=\frac{-a^2+b^2\pm \sqrt{(a^2+b^2{)}^2}}{2a^2}$

$=\frac{-a^2+b^2\pm (a^2+b^2)}{2a^2}$

$\Rightarrow$ $x=\frac{-a^2+b^2+a^2+b^2}{2a^2}$ and $x=\frac{-a^2+b^2-a^2-b^2}{2a^2}$

$\Rightarrow$ $x=\frac{2b^2}{2a^2}$ and $x=\frac{-2a^2}{2a^2}$

$\therefore$ $x=\frac{b^2}{a^2}$ and $x=-1$