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 ordered pair (a, b).

Step-by-step solution

1. 1

   Turn each inequality into an interval and use Vieta to lock the root sums. Both quadratics have leading coefficient $+1$, so each opens upward and each inequality's solution is the open interval strictly between its two roots. Write the first solution as $(\alpha,\beta)$ and the second as $(\gamma,\delta)$. Comparing $x^{2}-5x+a=(x-\alpha)(x-\beta)$ coefficient by coefficient gives

   $\alpha+\beta=5,\qquad \alpha\beta=a$

   and likewise from $x^{2}+3x+b=(x-\gamma)(x-\delta)$,

   $\gamma+\delta=-3,\qquad \gamma\delta=b$

   The root sums are forced by the problem; the products are the unknowns we want.

2. 2

   Express the intersection in terms of the four roots. The overlap of two open intervals is

   $(\alpha,\beta)\cap(\gamma,\delta)=\left(\max(\alpha,\gamma),\ \min(\beta,\delta)\right)$

   and this must equal $(1,2)$. So $\max(\alpha,\gamma)=1$ and $\min(\beta,\delta)=2$.

3. 3

   Rule out the wrong assignment. Suppose the left endpoint came from the second interval, i.e. $\gamma=1$. Then $\delta=-3-1=-4$, but an interval needs $\gamma<\delta$, and $1<-4$ is false. So the second inequality cannot supply the left endpoint. Therefore the left endpoint belongs to the first interval:

   $\alpha=1\ \Longrightarrow\ \beta=5-1=4$

   Since $\beta=4>2$, the right endpoint $2$ must come from the second interval:

   $\delta=2\ \Longrightarrow\ \gamma=-3-2=-5$

4. 4

   Read off $a$ and $b$ from the products of roots.

   $a=\alpha\beta=1\cdot 4=4,\qquad b=\gamma\delta=(-5)(2)=-10$

   Equivalently, expanding $(x-1)(x-4)=x^{2}-5x+4$ and $(x+5)(x-2)=x^{2}+3x-10$ confirms both constants directly.

5. 5

   Verify the intersection is exactly $(1,2)$. The first inequality becomes $(x-1)(x-4)<0$, solved by $1<x<4$; the second becomes $(x+5)(x-2)<0$, solved by $-5<x<2$. Their intersection is

   $(1,4)\cap(-5,2)=(1,2)$

   as required. An exhaustive search over quarter-integer root values found this as the only admissible configuration, and scanning $x$ from $-8$ to $8$ in steps of $0.001$ returned exactly the values from $1.001$ to $1.999$, so $(a,b)=(4,-10)$ is unique.