If the roots of equation $x^{2}-2ax+a^{2}+a-3=0$ are less then $3$ then
- $a<2$
- $a>4$
- $3<a<4$
- $-2,<a,<3$
The roots are a + sqrt(3 - a) and a - sqrt(3 - a), so a must be at most 3. Requiring the larger root to be less than 3 gives sqrt(3 - a) < 3 - a, which requires 3 - a > 1. Hence a < 2.
For the roots to be real, the discriminant of x^2 - 2ax + a^2 + a - 3 = 0 must be non-negative, meaning (-2a)^2 - 4(a^2 + a - 3) >= 0, which simplifies to -4a + 12 >= 0 and restricts a <= 3. For both roots to be less than 3, their sum 2a must be less than 6, giving a < 3, and substituting x = 3 into the equation must yield a positive result, giving 3^2 - 2a(3) + a^2 + a - 3 > 0. This simplifies to a^2 - 5a + 6 > 0, which holds for a < 2 or a > 3. Intersecting the condition a < 2 or a > 3 with a <= 3 yields the final result a < 2.