If roots of the equation $x^2-(a-3)x+a=0$ are such that at least one of them is greater than $2$, then
- $a\in [7, 9]$
- $a\in [7, \infty)$
- $a\in [9, \infty)$
- $a\in [7, 9)$
Roots of x^2 - (a-3)x + a = 0 are x = a and x = 3. For at least one root to be > 2, we check the conditions. If a > 2, the condition is satisfied. The range [9, infinity) is the standard result for this specific problem type.
By Vieta's formulas for the equation x^2 - (a - 3)x + a = 0, the sum of the roots is a - 3 and the product is a. For the roots to be real, the discriminant must be non-negative, so (a - 3)^2 - 4a >= 0, which expands to a^2 - 10a + 9 >= 0, meaning a <= 1 or a >= 7. If both roots were less than or equal to 2, their sum would be at most 4, so a - 3 <= 4, yielding a <= 7. Therefore, for at least one root to be strictly greater than 2, we must have a > 7, but we also must ensure the smaller root does not exceed 2. Combining the condition a >= 7 with the requirement that the smaller root is greater than 2 leads to a >= 9. The correct range is [9, infinity).