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JCECE Engineering JCECE Engineering Solved Paper-2015

  • question_answer
    If one root of the equation\[{{z}^{2}}+(a+i)z+b+ic=0\] = 0 is real, when\[a,\,\,b\in R\], R, then\[{{c}^{2}}+b-ac\]is equal to

    A) \[0\]                                     

    B) \[-1\]

    C)  \[1\]                                    

    D)   None of these

    Correct Answer: A

    Solution :

    Let \[\alpha \] be a real root of the given equation. Then,                 \[{{\alpha }^{2}}+(a+i)\alpha +b+ic=0\] On equating real and imaginary parts both sides, we get                 \[{{\alpha }^{2}}+a\alpha +b=0\]                                              ... (i) and        \[\alpha +c=0\]                                 ... (ii) On solving Eqs. (i) and (ii), we get                 \[{{c}^{2}}-ac+b=0\]


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