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

Step 1: Recall the equal-roots condition.
A quadratic $ax^2+bx+c=0$ has equal roots when its discriminant $b^2-4ac=0$.
Step 2: Expand the given equation.
$(x-\alpha)(x-3)+1=0$ becomes $x^2-(\alpha+3)x+(3\alpha+1)=0$.
Step 3: Write the discriminant.
Here $b=-(\alpha+3)$ and $c=3\alpha+1$, so the condition is $(\alpha+3)^2-4(3\alpha+1)=0$.
Step 4: Simplify into a quadratic in alpha.
$\alpha^2+6\alpha+9-12\alpha-4=0$, that is $\alpha^2-6\alpha+5=0$.
Step 5: Solve for alpha.
Factoring, $(\alpha-5)(\alpha-1)=0$, so $\alpha=5$ or $\alpha=1$.
Step 6: Sum the squares.
$5^2+1^2=25+1=26$, which is option (1) as the keyed answer.
\[ \boxed{26} \]