Question

The roots of
\[ (x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0 \] where \(a\in \mathbb{R}\) are always

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

Hint:

After substitution, check discriminant. If \(\Delta>0\), roots are always real and distinct independent of parameter shift.

Answer 1

The Correct Option is C

Step 1: Substitute \(y=x-a\) to simplify.
Then equation becomes:
\[ y(y-1)+(y-1)(y-2)+y(y-2)=0 \]
Step 2: Expand each term.
\[ y(y-1)=y^2-y \]
\[ (y-1)(y-2)=y^2-3y+2 \]
\[ y(y-2)=y^2-2y \]
Step 3: Add all terms.
\[ (y^2-y)+(y^2-3y+2)+(y^2-2y)=0 \]
\[ 3y^2-6y+2=0 \]
Step 4: Find discriminant.
\[ \Delta = (-6)^2-4(3)(2)=36-24=12>0 \]
So roots are real and distinct.
Step 5: Convert back to \(x\).
\[ x = a+y \]
Since \(y\) has two real distinct roots, \(x\) also has two real distinct roots for all real \(a\).
Final Answer:
\[ \boxed{\text{real and distinct}} \]