JEE Main & Advanced JEE Main Paper Phase-I (Held on 08-1-2020 Morning)

  • question_answer
    If the equation, \[{{x}^{2}}+bx+45=0(b\in R)\] has conjugate complex roots and they satisfy \[\left| z+1 \right|=2\sqrt{10},\] Then [JEE MAIN Held On 08-01-2020 Morning]

    A) \[{{b}^{2}}-b=42\]

    B) \[{{b}^{2}}-b=30\]

    C) \[{{b}^{2}}+b=12\]

    D) \[{{b}^{2}}+b=72\]

    Correct Answer: B

    Solution :

    As \[b\in R\] let roots be \[\alpha \pm i\beta \] \[\Rightarrow 2\alpha =-b\] and \[{{\alpha }^{2}}+{{\beta }^{2}}=45\] Also \[\left| \alpha +i\beta +1 \right|=2\sqrt{10}\] \[\Rightarrow {{(\alpha +1)}^{2}}+{{\beta }^{2}}=40\] \[\Rightarrow 45+2\alpha +1=40\] \[\alpha =-3\] and \[b=6\] \[\Rightarrow {{b}^{2}}-b=30\]


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