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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2008

  • question_answer
    If equation\[({{a}^{2}}+4a+3){{x}^{2}}+({{a}^{2}}-a-2)x\] \[+a(a+1)=0\]have more than two roots, then the value of a is

    A)  0                   

    B)  1

    C)  \[-1\]             

    D)  None of these

    Correct Answer: C

    Solution :

     Given, equation is \[({{a}^{2}}+4a+3){{x}^{2}}+({{a}^{2}}-a-2)x+a(a+1)=0\] Let\[a=-1,\]then \[0.{{x}^{2}}+0.x-1(-1+1)=0\] \[\Rightarrow \] \[0=0\] Hence, option [c] is correct.


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