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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2002

  • question_answer
    If\[{{x}^{2}}+ax+10=0\]and\[{{x}^{2}}+bx-10=0\]have a common root, then\[{{a}^{2}}-{{b}^{2}}\]is equal to:

    A)  10                                         

    B)                         20

    C)                         30                                         

    D)                         40

    E)                         50

    Correct Answer: D

    Solution :

    Let\[\alpha \]be the common root for both the equations\[{{x}^{2}}+ax+10=0\]and\[{{x}^{2}}+bx-10=0\] \[\therefore \]  \[{{\alpha }^{2}}+a\alpha +10=0\]                     ...(i) and        \[{{\alpha }^{2}}+b\alpha -10=0\]                     ...(ii) From Eqs. (i) and (ii), we get \[(a-b)\alpha +10+10=0\] \[\Rightarrow \]               \[(a-b)\alpha =-20\] \[\Rightarrow \]               \[\alpha =\frac{-20}{a-b}\] \[\because \]\[\alpha \]is also the root of\[{{x}^{2}}+bx-10=0\] \[\therefore \]  \[\frac{400}{{{(a-b)}^{2}}}+b\left( \frac{-20}{a-b} \right)-10=0\]                 \[40-2b(a-b)={{(a-b)}^{2}}\] \[\Rightarrow \]               \[{{a}^{2}}+{{b}^{2}}-2{{b}^{2}}=40\] \[\Rightarrow \]               \[{{a}^{2}}-{{b}^{2}}=40\]


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