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Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    If\[b<0\], then the roots\[{{x}_{1}}\]and\[{{x}_{2}}\]of the equation \[2{{x}^{2}}+6x+b=0\], satisfy the condition\[\left( \frac{{{x}_{1}}}{{{x}_{2}}} \right)+\left( \frac{{{x}_{2}}}{{{x}_{1}}} \right)<K\], where\[K\]is equal to

    A)  2                                            

    B)  -2

    C)  0                            

    D)         4

    Correct Answer: B

    Solution :

    The discriminant of the quadratic equation\[2{{x}^{2}}+6x+b=0\]is given by\[D=36-8b>0\]. Therefore, the given equation has real roots. we have,                 \[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{x}_{2}}}{{{x}_{1}}}=\frac{x_{1}^{2}+x_{2}^{2}}{{{x}_{1}}\cdot {{x}_{2}}}=\frac{{{({{x}_{1}}+{{x}_{2}})}^{2}}-2{{x}_{1}}{{x}_{2}}}{{{x}_{1}}\cdot {{x}_{2}}}\] \[=\frac{{{(-3)}^{2}}-2(b/2)}{(b/2)}=\frac{18}{b}-2<-2\]     \[[\because \,\,b<0]\]


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