If (2, +infinity) is one solution interval of x^2 + x - b > 0, what is the other solution interval?

Step-by-step solution

1. 1

   Use the shape of an upward parabola. Since the coefficient of $x^2$ is $+1$, the graph of $y=x^2+x-b$ opens upward, so $y>0$ holds outside the two roots. The solution set therefore has the form $(-\infty,r_1)\cup(r_2,+\infty)$ with $r_1<r_2$.

2. 2

   Read the given interval as a root. The stated interval $(2,+\infty)$ must be the right-hand branch, so its endpoint $x=2$ is the larger root. Endpoints of the solution set of a strict quadratic inequality are exactly the roots of the equation.

3. 3

   Solve for b. Substituting $x=2$ into $x^2+x-b=0$: $4+2-b=0$, so $b=6$. The inequality is $x^2+x-6>0$.

4. 4

   Factor and find the second root. $x^2+x-6=(x+3)(x-2)$, so the roots are $-3$ and $2$. The product $(x+3)(x-2)$ is positive when both factors share a sign.

5. 5

   State the missing branch. Both factors are negative when $x<-3$, so the other solution interval is $(-\infty,-3)$ and the full solution set is $(-\infty,-3)\cup(2,+\infty)$.

6. 6

   Numerical check at x = -4 and x = 0. At $x=-4$: $16-4-6=6>0$, so $-4$ belongs to the solution set. At $x=0$ (between the roots): $0+0-6=-6<0$, correctly excluded.