Let a and b be real numbers. The simultaneous inequalities x^2 - 5x + a < 0 and x^2 + 3x + b < 0 have solution set 1 < x < 2. Find the pair (a, b).

Step-by-step solution

1. 1

   Translate the setup. Each inequality has an upward parabola on the left, so each solution set is an open interval between that quadratic's two roots. The system's solution is the intersection of the two intervals, and we are told

   $\left(\alpha_1, \beta_1\right) \cap \left(\alpha_2, \beta_2\right) = (1, 2)$

2. 2

   Use Vieta to pin the root sums. The coefficient of $x$ fixes each sum regardless of $a$ and $b$:

   $\alpha_1 + \beta_1 = 5, \qquad \alpha_2 + \beta_2 = -3$

   This is the constraint that makes the problem solvable — the intervals cannot be chosen freely.

3. 3

   Match the endpoints of the intersection. The intersection $(1,2)$ has its left endpoint $1$ and right endpoint $2$ contributed by one interval each. Try $\alpha_1 = 1$: then $\beta_1 = 5 - 1 = 4$, so the first interval is $(1, 4)$ and supplies the left endpoint. The second must supply the right endpoint, so $\beta_2 = 2$ and $\alpha_2 = -3 - 2 = -5$, giving $(-5, 2)$.

4. 4

   Read off a and b from the products. Vieta's product rule gives the constant term:

   $a = \alpha_1\beta_1 = 1 \times 4 = 4, \qquad b = \alpha_2\beta_2 = (-5)(2) = -10$

5. 5

   Verify the factorisations and the intersection. $x^2 - 5x + 4 = (x-1)(x-4) < 0$ on $(1,4)$, and $x^2 + 3x - 10 = (x+5)(x-2) < 0$ on $(-5,2)$. Their intersection is $(1,4) \cap (-5,2) = (1,2)$, exactly as required.

6. 6

   Check that no other assignment works. If instead the second interval supplied the left endpoint, then $\alpha_2 = 1$ and $\beta_2 = -4$ — impossible, since $\beta_2 > \alpha_2$ is required. So $(a,b) = (4,-10)$ is the unique solution.