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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    Let a, b be the solutions of\[{{x}^{2}}+px+1=0\]and c, d be the solutions of\[{{x}^{2}}+qx+1=0\]. If \[(a-c)(b-c)\]and\[(a+d)(b+d)\]are the solutions of\[{{x}^{2}}+ax+\beta =0,\]then \[\beta \] equals:

    A)  \[p+q\]               

    B)         \[p-q\]

    C)  \[{{p}^{2}}+{{q}^{2}}\]         

    D)         \[{{p}^{2}}-{{q}^{2}}\]

    E)  \[{{q}^{2}}-{{p}^{2}}\]

    Correct Answer: E

    Solution :

    \[\because \]a and b are the solutions of \[{{x}^{2}}+px+1=0,\] then\[a+b=-p\]and \[ab=1\] Also c and d are solution of \[{{x}^{2}}+qx+1=0,\] then\[c+d=-q\]and\[cd=1\]. Now,\[(a-c)(b-c)\]and\[(a+d)(b+d)\]are the roots of \[{{x}^{2}}+ax+\beta =0\] \[\therefore \]\[(a-c)(b-c)(a+d)(b+d)=\beta \] \[\Rightarrow \]\[1+p(c-d)-{{p}^{2}}-pcd(c-d)+{{(c+d)}^{2}}\]                 \[-2cd+{{(cd)}^{2}}=\beta \] \[\Rightarrow \]               \[{{q}^{2}}-{{p}^{2}}=\beta \]


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