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J & K CET Engineering J and K - CET Engineering Solved Paper-2007

  • question_answer
    The    solution    of    the    equation \[2{{x}^{3}}-{{x}^{2}}-22x-24=0\]when two of the roots are in the ratio \[3:4,\] is

    A)  \[3,\,4,\frac{1}{2}\]

    B)  \[-\frac{3}{2},-2,4\]

    C)  \[-\frac{1}{2},\frac{3}{2},2\]

    D)  \[\frac{3}{2},2,\frac{5}{2}\]

    Correct Answer: B

    Solution :

    Given equation is \[2{{x}^{3}}-{{x}^{2}}-22x-24=0\] On putting  \[x=0,1,-1,2,-2\]only \[x=-2\] satisfies this equation. So, \[x=-2\] is a root of this equation and from the given options only [b] has this root.


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