Question

a(a^2+b^2)x^2+b^2x-a=0 Find x using quadratic equation

Answer 1

\[a=a(a^2+b^2) ,b=b^2,c=-a\]
\[b^2-4ac =(b^2)^2-4*(a(a^2+b^2))*(-a) = b^4+4a(a^3+ab^2) = b^4+4a^4+4a^2b^2\]
\[=(b^2)^2+2(b^2)(2a^2)+(2a)^2 = (b^2+2a^2)^2\]



\[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-b^2 \pm \sqrt{(b^2+2a^2)^2}}{2a(a^2+b^2)}\]
\[= \frac{-b^2 \pm b^2+2a^2}{2a(a^2+b^2)} =\frac{-b^2+b^2+2a^2}{2a(a^2+b^2)},\frac{-b^2-b^2-2a^2}{2a(a^2+b^2)}\]
\[= \frac{2a^2}{2a(a^2+b^2)},\frac{-2(b^2+a^2)}{2a(a^2+b^2)} = \frac{a}{a^2+b^2},\frac{-1}{a}\]