If all the roots of the equation x^2 - 2ax + a^2 + a - 3 = 0 are less

If all the roots of the equation \(x2−2ax+a2+a−3=0\) are less than 3, then which of the following must be true?

A. a < 2
B. 2 < a < 3
C. 3 < a < 4
D. 4 < a < 5
E. a > 5

sum of the roots = 2a < 6; a < 3. ans seems to be between A & B.

on solving the equation, x = \(a ± \sqrt{3-a}\)

taking first root:
\(a-\sqrt{3-a}<3\)
\(\sqrt{3-a}(-1 - \sqrt{3-a}) < 0\)
\(\sqrt{3-a}(1+\sqrt{3-a}) > 0\)
solving inequality,
\(\sqrt{3-a} < -1\) not possible or
\(\sqrt{3-a} > 0; a < 3\)


taking second root:
\(a + \sqrt{3-a} < 3\)
\(\sqrt{3-a} < 3 -a\); as both sides are positive I can square both sides
\(3-a < (3-a)^2\)
\((3-a) - (3-a)^2 < 0\)
\((3-a)(a-2) < 0\); I know (3-a) is positive; (a-2) must be < 0 or a < 2

Ans: A