Question

The roots of the equation $(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0a\in \mathrm{ℝ}$ are always

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

Answer 1

The correct option is C

real and distinct

Explanation for the correct option:

Given,

$(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0$

Put $x-a=t$

$t(t–1)+(t–1)(t–2)+t(t–2)=0$

$t^2–t+t^2–3t+2+t^2–2t=0$

$3t^2–6t+2=0$

$$\begin{gathered} t=\frac{\left[6\pm \sqrt{36-24}\right]}{6} \\ t=\frac{\left[6\pm 2\sqrt{3}\right]}{6} \\ =\frac{\left[3\pm \sqrt{3}\right]}{3} \\ x=a+\left\{\frac{\left[3\pm \sqrt{3}\right]}{3}\right\} \end{gathered}$$

Hence the correct option is option(c) i.e. real and distinct.