JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer Let \[p,q\in R.\]If \[2-\sqrt{3}\] is a root of the quadratic equation, \[{{x}^{2}}+px+q=0,\] then : [JEE Main 9-4-2019 Morning]

    A) \[{{q}^{2}}+4p+14=0\]   

    B) \[{{p}^{2}}4q12=0\]

    C) \[{{q}^{2}}4p16=0\]          

    D) \[{{p}^{2}}4q+12=0\]

    Correct Answer: B

    Solution :

    In given question \[p,q\in R.\] If we take other root as any real number \[\alpha ,\]then quadratic equation will be \[{{x}^{2}}-(\alpha +2-\sqrt{3})x+\alpha .\left( 2-\sqrt{3} \right)=0\] Now, we can have none or any of the options can be correct depending upon\['\alpha '\] Instead of \[p,q\in R\]it should be \[p,q\in Q\]then other root will be \[2+\sqrt{3}\] \[\Rightarrow p=-\left( 2+\sqrt{3}-2-\sqrt{3} \right)=-4\] and\[q=\left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right)=1\] \[\Rightarrow {{p}^{2}}-4q-12={{(-4)}^{2}}-4-12\]                         \[=16-16=0\] Option [b] is correct


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