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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 equation\[({{a}^{2}}+{{b}^{2}}){{t}^{2}}-2(ac+bd)t+({{c}^{2}}+{{d}^{2}})=0\] are equal, then  [MP PET 1996]

    A) \[ab=dc\]

    B) \[ac=bd\]

    C) \[ad+bc=0\]

    D) \[\frac{a}{b}=\frac{c}{d}\]

    Correct Answer: D

    Solution :

    Accordingly,  \[{{\{2(ac+bd)\}}^{2}}=4({{a}^{2}}+{{b}^{2}})({{c}^{2}}+{{d}^{2}})\] Þ \[4{{a}^{2}}{{c}^{2}}+4{{b}^{2}}{{d}^{2}}+8abcd=4{{a}^{2}}{{c}^{2}}+4{{a}^{2}}{{d}^{2}}\]   \[+4{{b}^{2}}{{c}^{2}}+4{{b}^{2}}{{d}^{2}}\] Þ \[4{{a}^{2}}{{d}^{2}}+4{{b}^{2}}{{c}^{2}}-8abcd=0\,\,\,\Rightarrow 4{{(ad-bc)}^{2}}=0\] Þ \[ad=bc\Rightarrow \frac{a}{b}=\frac{c}{d}\].


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