JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    If \[\alpha \] and \[\beta \] are roots of the equation \[{{x}^{2}}+\operatorname{p}x+\frac{3\operatorname{p}}{4}=0,\] such that \[\left| \alpha -\beta  \right|\]=\[\sqrt{10},\]then p belongs to the set:     JEE Main  Online Paper (Held On 22 April 2013)

    A)  {2, -5}                   

    B)  {-3, 2}

    C)  {-2, 5}                   

    D)  {3, -5}

    Correct Answer: C

    Solution :

     Given quadratic eqn. is \[{{x}^{2}}+px+\frac{3p}{4}=0\] So, \[\alpha +\beta =-p,\alpha \beta =\frac{3p}{4}\] Now, given \[|\alpha -\beta |\,=\sqrt{10}\] \[\Rightarrow \]\[\alpha -\beta =\pm \sqrt{10}\] \[\Rightarrow \]\[{{(\alpha -\beta )}^{2}}=10\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta =10\] \[\Rightarrow \]\[{{(\alpha +\beta )}^{2}}-4\alpha \beta =10\] \[\Rightarrow \]\[{{p}^{2}}-4\times \frac{3p}{4}=10\Rightarrow {{p}^{2}}-3p-10=0\] \[\Rightarrow \]\[p=-2,5\Rightarrow p\in \{-2,5\}\]

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