JEE Main & Advanced Mathematics Quadratic Equations Question Bank Critical Thinking

  • question_answer 3) 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

    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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