2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    If \[\alpha \] and \[\beta \] are roots of the equation\[4{{x}^{2}}+2x-1=0,\]then the value of\[{{\alpha }^{2}}+{{\beta }^{2}}\]is:

    A)  2                            

    B)         \[\frac{3}{4}\]

    C)  \[3\]                    

    D)         \[\frac{1}{4}\]

    E)  \[\frac{1}{2}\]

    Correct Answer: B

    Solution :

    \[\because \]\[\alpha \]and\[\beta \]are the roots of\[4{{x}^{2}}+2x-1=0,\] then\[\alpha +\beta =-\frac{1}{2},\alpha \beta =-\frac{1}{4}\] \[\therefore \]  \[{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \]                 \[=\frac{1}{4}+\frac{2}{4}=\frac{3}{4}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner