• # question_answer If$a{{x}^{2}}+bx+c=0$, then x = [MP PET 1995] A) $\frac{b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ B) $\frac{-b\pm \sqrt{{{b}^{2}}-ac}}{2a}$ C) $\frac{2c}{-b\pm \sqrt{{{b}^{2}}-4ac}}$ D) None of these

$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ Since $\frac{2c}{-b+\sqrt{{{b}^{2}}-4ac}}\,\,.\,\,\frac{-b-\sqrt{{{b}^{2}}-4ac}}{-b-\sqrt{{{b}^{2}}-4ac}}$ =$\frac{2c\,(-b-\sqrt{{{b}^{2}}-4ac})}{4ac}=\frac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$ Similarly $\frac{2c}{-b-\sqrt{{{b}^{2}}-4ac}}\times \frac{-b+\sqrt{{{b}^{2}}-4ac}}{-b+\sqrt{{{b}^{2}}-4ac}}$ =$\frac{2c\,(-b+\sqrt{{{b}^{2}}-4ac})}{4ac}=\,\frac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$ Aliter : On rationalising the given equation $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, we get option C correct.