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
Correct Answer: C
Solution :
\[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] 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.
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