Question

Prove that D = b^2 -4ac

Answer 1

ax^2+bx+c=0

Multiply both sides by a :

\implies a^2x^2+abx+ac=0

Transpose ac to the other side :

\implies a^2x^2+abx=-ac

Add b²/4 on both sides :

\implies a^2x^2+abx+\frac{b^2}{4}=\frac{b^2}{4}-ac

Perfect square on the right hand side :

\implies a^2x^2+2\times a\times \frac{b}{2}\times x+\frac{b^2}{4}=\frac{b^2}{4}-ac

\implies (ax+\frac{b}{2})^2=\frac{b^2-4ac}{4}

Take square root both sides :

\implies ( ax+\frac{b}{2})=\frac{\pm \sqrt{b^2-4ac}}{2}

Transpose b/2 to the other side :

\implies ax=\frac{-b}{2}+\frac{\pm\sqrt{b^2-4ac}}{2}

\implies ax=\frac{-b\pm\sqrt{b^2-4ac}}{2}

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

Hence it is proved that :

Either\:x=\frac{-b+\sqrt{b^2-4ac}}{2a}\\\\Or\:x=\frac{-b-\sqrt{b^2-4ac}}{2a}