Question

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

Answer 1

Answer: (C)

real and distinct

Answer 2

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