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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Solution of quadratic equations and Nature of roots

  • question_answer
    The roots of the given equation \[(p-q){{x}^{2}}+(q-r)x+(r-p)=0\] are                    [RPET 1986; MP PET 1999; Pb. CET 2004]

    A) \[\frac{p-q}{r-p},1\]

    B) \[\frac{q-r}{p-q},1\]

    C) \[\frac{r-p}{p-q},1\]

    D) \[1,\frac{q-r}{p-q}\]

    Correct Answer: C

    Solution :

    Given equation is  \[(p-q){{x}^{2}}+(q-r)x+(r-p)=0\]        \[x=\frac{(r-q)\pm \sqrt{{{(q-r)}^{2}}-4(r-p)(p-q)}}{2(p-q)}\] Þ \[x=\frac{(r-q)\pm (q+r-2p)}{2(p-q)}\Rightarrow x=\frac{r-p}{p-q},1\]


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