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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 the roots of the equation \[({{p}^{2}}+{{q}^{2}}){{x}^{2}}\]\[-2q(p+r)x\]+ \[({{q}^{2}}+{{r}^{2}})=0\] be real and equal, then \[p,q,r\]will be  in

    A) A.P.

    B) G.P.

    C) H.P.

    D) None of these

    Correct Answer: B

    Solution :

    Given equation is\[({{p}^{2}}+{{q}^{2}}){{x}^{2}}-2q(p+r)x+({{q}^{2}}+{{r}^{2}})=0\] Roots are real and equal, then \[4{{q}^{2}}{{(p+r)}^{2}}-4({{p}^{2}}+{{q}^{2}})({{q}^{2}}+{{r}^{2}})=0\] Þ\[{{q}^{2}}({{p}^{2}}+{{r}^{2}}+2pr)-({{p}^{2}}{{q}^{2}}+{{p}^{2}}{{r}^{2}}+{{q}^{4}}+{{q}^{2}}{{r}^{2}})=0\]    Þ \[{{q}^{2}}{{p}^{2}}+{{q}^{2}}{{r}^{2}}+2p{{q}^{2}}r-{{p}^{2}}{{q}^{2}}-{{p}^{2}}{{r}^{2}}-{{q}^{4}}-{{q}^{2}}{{r}^{2}}=0\] Þ \[2p{{q}^{2}}r-{{p}^{2}}{{r}^{2}}-{{q}^{4}}=0\]Þ\[{{({{q}^{2}}-pr)}^{2}}=0\]   Hence\[{{q}^{2}}=pr\]. Thus p, q, r in G.P.


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