JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Morning)

  • question_answer
    The sum of the solutions of the equation \[\left| \sqrt{x}-2 \right|+\sqrt{x}\left( \sqrt{x}-4 \right)+2=0,\]\[(x>0)\]is equal to :     [JEE Main 8-4-2019 Morning]

    A) 4                                 

    B) 9

    C) 10

    D) 12

    Correct Answer: C

    Solution :

    \[\left| \sqrt{x}-2 \right|+\sqrt{x}\left( \sqrt{x}-4 \right)+2=0\]           \[\left| \sqrt{x}-2 \right|+{{\left( \sqrt{x} \right)}^{2}}-4\sqrt{x}+2=0\]           \[{{\left| \sqrt{x}-2 \right|}^{2}}+\left| \sqrt{x}-2 \right|-2=0\]           \[\left| \sqrt{x}-2 \right|=-2\](not possible) or \[\left| \sqrt{x}-2 \right|=1\]             \[\sqrt{x}-2=1,-1\]             \[\sqrt{x}=3,1\]             \[x=9,1\]             Sum = 10                               


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