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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2004

  • question_answer
    The roots of the equations\[{{x}^{2}}+ax+b=0\]and \[{{x}^{2}}+bx+a=0\]are common and\[a\ne b,\]then\[a+b\]is equal to

    A)  0               

    B)  1

    C)  \[-1\]              

    D)  None of these

    Correct Answer: C

    Solution :

     Given, equations are\[{{x}^{2}}+ax+b=0\]and\[{{x}^{2}}+bx+a=0\] \[\Rightarrow \] \[(a-b)x+(b-a)=0\] \[\Rightarrow \] \[(a-b)(x-1)=0\] \[\Rightarrow \] \[a=b,x=1\] but            \[a\ne b\] \[\therefore \]\[x=1\] Put\[x=1\]in first equation, \[1+a+b=0\] \[a+b=-1\]


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