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

Step-by-step solution

1. 1

   Turn the interval endpoint into information about the roots. The parabola $y=x^2+x-b$ opens upward ($a=1>0$), so the set where $y>0$ is everything outside the two roots: $(-\infty,r_1)\cup(r_2,+\infty)$. Being told that $(2,+\infty)$ is one of those pieces forces

   $r_2=2$

   The endpoint of the solution interval is the larger root.

2. 2

   Use the root to pin down $b$. Since $x=2$ satisfies $x^2+x-b=0$:

   $2^2+2-b=0\ \Longrightarrow\ 6-b=0\ \Longrightarrow\ b=6$

3. 3

   Rewrite and factor the now-explicit inequality.

   $x^2+x-6>0$

   Two numbers with product $-6$ and sum $+1$ are $3$ and $-2$, so

   $(x+3)(x-2)>0$

   The roots are $x=-3$ and $x=2$.

4. 4

   Read off both intervals from the sign chart. A product of two factors is positive when the factors share a sign. For $x<-3$ both are negative (product positive); for $-3<x<2$ they differ (product negative); for $x>2$ both are positive. Hence

   $(-\infty,-3)\cup(2,+\infty)$

   The given piece is $(2,+\infty)$, so the other one is $(-\infty,-3)$.

5. 5

   Test one value from each region to be sure. At $x=-4$: $(-4)^2+(-4)-6=16-10=6>0$ $\checkmark$. At $x=0$: $0+0-6=-6<0$, correctly excluded $\checkmark$. At $x=3$: $9+3-6=6>0$ $\checkmark$. The endpoints are open because $x=-3$ and $x=2$ give exactly $0$, and the inequality is strict.