Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    The roots of \[(x-a)(x-a-1)+(x-a-1)(x-a-2)\]\[+(x-a)(x-a-2)=0\], \[a\in R\]are always

    A)  equal                                   

    B)  imaginary

    C)  real and distinct

    D)         rational and equal

    Correct Answer: C

    Solution :

    Given,                 \[(x-a)(x-a-1)(x-a-1)(x-a-2)\]                                                 \[+(x-a)(x-a-2)=0\] Let\[x-a=t\], then                 \[t(t-1)+(t-1)(t-2)+t(t-2)=0\] \[\Rightarrow \]               \[{{t}^{2}}-t+{{t}^{2}}-3t+2+{{t}^{2}}-2t=0\] \[\Rightarrow \]               \[3{{t}^{2}}-6t+2=0\] \[\Rightarrow \]               \[t=\frac{6\pm \sqrt{36-24}}{2(3)}=\frac{6\pm 2\sqrt{3}}{2(3)}\] \[\Rightarrow \]               \[x-a=\frac{3\pm \sqrt{3}}{3}\] \[\Rightarrow \]               \[x=a+\frac{3\pm \sqrt{3}}{3}\] Hence,\[x\]is real and distinct.


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