Question

If \(\alpha\in R\) and equation \[ (x-\alpha)(x-3)+1=0 \] has equal roots, then sum of squares of all values of \(\alpha\) is

• A. 13
• B. 25
• C. 26
• D. 20

Hint:

Equal roots always mean discriminant becomes zero.

Answer 1

The Correct Option is A

Concept: For equal roots of quadratic equation: \[ D=b^2-4ac=0 \]

Step 1: Expand equation.
\[ (x-\alpha)(x-3)+1=0 \] \[ x^2-(\alpha+3)x+3\alpha+1=0 \]

Step 2: Apply equal roots condition.
\[ (\alpha+3)^2-4(3\alpha+1)=0 \] \[ \alpha^2+6\alpha+9-12\alpha-4=0 \] \[ \alpha^2-6\alpha+5=0 \] \[ (\alpha-5)(\alpha-1)=0 \] So \[ \alpha=5,1 \]

Step 3: Find sum of squares.
\[ 5^2+1^2 \] \[ =25+1 \] \[ =26 \] Hence \[ \boxed{26} \]