Q51 / 160 · AP EAMCET 2009 · Mathematics

The roots $\begin{aligned} & (x-a)(x-a-1)+(x-a-1)(x-a-2) \\ & +(x-a)(x-a-2)=0, a \in R \text { are always } \end{aligned}$

  • A
    equal
  • B
    imaginary
  • Correct answer
    C
    real and distinct
  • D
    rational and equal

Solution

Given, $\begin{aligned} & (x-a)(x-a-1)+(x-a-1)(x-a-2) \\ & +(x-a)(x-a-2)=0 \end{aligned}$ Let $x-a=t$, then $t(t-1)+(t-1)(t-2)+t(t-2)=0$ $\begin{array}{lc}\Rightarrow & t^2-t+t^2-3 t+2+t^2-2 t=0 \\ \Rightarrow & 3 t^2-6 t+2=0 \\ \Rightarrow & t=\frac{6 \pm \sqrt{36-24}}{2(3)}=\frac{6 \pm 2 \sqrt{3}}{2(3)} \\ \Rightarrow & x-a=\frac{3 \pm \sqrt{3}}{3} \\ \Rightarrow & x=a+\frac{3 \pm \sqrt{3}}{3}\end{array}$ Hence, x is real and distinct.

Hence, the correct answer is option: 'C'.

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