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

Hint:

For any equation of the form $(x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = 0$, the roots are always real.

Answer 1

The Correct Option is C

Step 1: Substitution
Let $x - a = t$. The equation becomes $t(t-1) + (t-1)(t-2) + t(t-2) = 0$.
Step 2: Simplify to Quadratic
$t^2 - t + t^2 - 3t + 2 + t^2 - 2t = 0 \Rightarrow 3t^2 - 6t + 2 = 0$.
Step 3: Solve for t
Using the quadratic formula: $t = \frac{6 \pm \sqrt{36 - 24}}{2(3)} = \frac{6 \pm 2\sqrt{3}}{6} = \frac{3 \pm \sqrt{3}}{3}$.
Step 4: Final Roots
$x = a + \frac{3 \pm \sqrt{3}}{3}$. Since the discriminant ($D = 12$) is greater than zero, the roots are real and distinct.
Final Answer: (c)