Question

Solve by factorization
$a^2{x}^2+(a^2+b^2)x+b^2=0,a\ne 0$

Answer 1

Let us simplify the expression,

$a^2{x}^2+(a^2+b^2)x+b^2=0$

$\Rightarrow a^2{x}^2+a^{2}x+b^{2}x+b^2=0$

$\Rightarrow a^{2}x(x+1)+b^2(x+1)=0$

$\Rightarrow (a^{2}x+b^2)(x+1)=0$

$\Rightarrow (a^{2}x+b^2)=0$ or $(x+1)=0$

$\Rightarrow a^{2}x=-b^2$ or $x=-1$

$\Rightarrow x=-\frac{b^2}{a^2}$ or $x=-1$

$\therefore$ Value of $x=-\frac{b^2}{a^2},-1$