A line y = x + 5 intersects a parabola y = x^2 + (a-2)x + 3a at two distinct points A and B . If the point (1, 6) lies strictly on the line segment AB , then the set of all possible values of a is :

Step-by-step solution

To find the intersection points A and B , we equate the equations of the line and the parabola: x^2 + (a-2)x + 3a = x + 5 x^2 + (a-3)x + 3a - 5 = 0 Let the roots of this quadratic equation be x₁ and x₂ , which represent the x -coordinates of points A and B . Since the point (1, 6) lies strictly on the line segment connecting A and B , its x -coordinate must lie strictly between the x -coordinates of A and B . Therefore, we must have x₁ For a quadratic expression f(x) = x^2 + (a-3)x + 3a - 5 with a positive leading