JEE Main & Advanced JEE Main Paper (Held On 12 April 2014)

  • question_answer
    The sum of the roots of the equation, \[{{x}^{2}}+|2x-3|-4=0,\] is:   [JEE Main Online Paper ( Held On 12 Apirl  2014 )

    A) \[2\]

    B) \[-2\]

    C) \[\sqrt{2}\]

    D) \[-\sqrt{2}\]

    Correct Answer: C

    Solution :

                    \[{{x}^{2}}+|2x-3|-4=0\] \[|2x-3|=\left\{ \begin{matrix}    (2x-3)\,\,\,\text{if} & x>\frac{3}{2}  \\    -(2x-3)\,\,\text{if} & x<\frac{3}{2}  \\ \end{matrix} \right.\] for\[x>\frac{3}{2},{{x}^{2}}+2x-3-4=0\] \[{{x}^{2}}+2x-7=0\] \[x=\frac{-2\pm \sqrt{4+28}}{2}=\frac{-2\pm 4\sqrt{2}}{2}=-1\pm 2\sqrt{2}\] Here \[x=2\sqrt{2}-1\]   \[\left\{ 2\sqrt{2}-1<\frac{3}{2} \right\}\] for\[x<\frac{3}{2}\] \[{{x}^{2}}-2x+3-4=0\] \[\Rightarrow \]\[{{x}^{2}}-2x-1=0\] \[\Rightarrow \]\[x=\frac{2\pm \sqrt{4+4}}{2}=\frac{2\pm 2\sqrt{2}}{2}=1\pm \sqrt{2}\] Here \[x=1-\sqrt{2}\left\{ (1-\sqrt{2})<\frac{3}{2} \right\}\] Sum of roots : \[(2\sqrt{2}-1)+(1-\sqrt{2})=\sqrt{2}\]                

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