Question

The roots of $\left(x-a\right)\left(x-a-1\right)+\left(x-a-1\right)\left(x-a-2\right)+\left(x-a\right)\left(x-a-2\right)=0$
$a\in R$ are always

• A. equal
• B. imaginary
• C. real and distinct
• D. rational and equal

Answer 1

Answer: (C)

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