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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Condition for common roots, Quadratic expressions and Position of roots

  • question_answer
    If \[\alpha \] and \[\beta \], \[\alpha \] and \[\gamma \], \[\alpha \] and \[\delta \] are the roots of the equations \[a{{x}^{2}}+2bx+c=0\], \[2b{{x}^{2}}+cx+a=0\] and \[c{{x}^{2}}+ax+2b=0\] respectively, where \[a,b\] and \[c\] are positive real numbers, then \[\alpha +{{\alpha }^{2}}\]= [Kerala (Engg.) 2005]

    A) - 1

    B) 0

    C) abc

    D) \[a+2b+c\] 

    E)  abc

    Correct Answer: B

    Solution :

    Here \[\alpha +\beta =-\frac{2b}{a}\] \[\gamma +\alpha =-\frac{c}{2b}\], \[\alpha +\delta =-\frac{a}{c}\] and \[\alpha \beta =\frac{c}{a},\,\alpha \gamma =\frac{a}{2b},\,\,\alpha \delta =\frac{2b}{c}\] Þ \[\alpha +\delta =-\frac{1}{\alpha \beta },\,{{\alpha }^{2}}\beta +\alpha \beta \delta =-1\] .....(i) Þ \[\alpha +\beta =-\frac{1}{\alpha \gamma },\,{{\alpha }^{2}}\gamma +\alpha \beta \gamma =-1\] .....(ii) Þ \[\alpha +\gamma =-\frac{1}{\alpha \delta },\,{{\alpha }^{2}}\delta +\alpha \beta \gamma =-1\] .....(iii) Solve equations (i), (ii) and (iii), we get  \[\alpha =-1\]     \[\alpha +{{\alpha }^{2}}=(-1)+{{(-1)}^{2}}\] = \[-1+1=0\].


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