If the roots of the equation $x ^ { 2 } - 2 a x + a ^ { 2 } + a - 3 = 0$ are real and greater than $3 ,$ then
- $a \in \phi$
- $a > 3$
- $a < 2$ or $a > 3$
- $a \leq 3$
For real roots, D >= 0: (2a)^2 - 4(a^2 + a - 3) >= 0 => 4a^2 - 4a^2 - 4a + 12 >= 0 => -4a + 12 >= 0 => a <= 3. For roots > 3, f(3) > 0 and vertex > 3. f(3) = 9 - 6a + a^2 + a - 3 = a^2 - 5a + 6 > 0 => (a-2)(a-3) > 0 => a < 2 or a > 3. Vertex = 2a/2 = a > 3. Intersection of a <= 3, (a < 2 or a > 3), and a > 3 is empty.
For the roots of x^2 - 2ax + a^2 + a - 3 = 0 to be real, the discriminant must be non-negative, so (-2a)^2 - 4(1)(a^2 + a - 3) >= 0. Simplifying this gives 4a^2 - 4a^2 - 4a + 12 >= 0, which means -4a + 12 >= 0 or a <= 3. If a root is strictly greater than 3, then both roots must be greater than 3 because the axis of symmetry is x = a <= 3, meaning no roots can satisfy the condition of being greater than 3. Since a cannot be greater than 3 while also being less than or equal to 3, the set of valid values for a is the empty set.