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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
    If \[{{x}^{2}}+{{y}^{2}}=25,\ xy=12\], then \[x=\] [BIT Ranchi 1992]

    A) {3, 4}

    B) {3, -3}

    C) {3, 4, -3, -4}

    D) {-3, -3}

    Correct Answer: C

    Solution :

    \[{{x}^{2}}+{{y}^{2}}=25\] and\[xy=12\] Þ  \[{{x}^{2}}+{{\left( \frac{12}{x} \right)}^{2}}=25\,\,\Rightarrow {{x}^{4}}+144-25{{x}^{2}}=0\] Þ  \[({{x}^{2}}-16)({{x}^{2}}-9)=0\]Þ \[{{x}^{2}}=16\]and\[{{x}^{2}}=9\] Þ  \[x=\pm \,4\]and\[x=\pm 3\].


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