Answer
$x \lt \frac{-1 - \sqrt{5}}{2}~~$ or $~~x \gt \frac{-1 + \sqrt{5}}{2}$
The solution set is $~~(-\infty,\frac{-1 - \sqrt{5}}{2})\cup (\frac{-1 + \sqrt{5}}{2}, \infty)$
Work Step by Step
$x^2 + x \gt 1$
$x^2 + x -1 \gt 0$
We can use the quadratic formula to find the values of $x$ such that $x^2+x-1 = 0$:
$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$x = \frac{-1 \pm \sqrt{(1)^2-4(1)(-1)}}{(2)(1)}$
$x = \frac{-1 \pm \sqrt{5}}{2}$
$x = \frac{-1 - \sqrt{5}}{2}, \frac{-1 + \sqrt{5}}{2}$
When $~~x=\frac{-1 - \sqrt{5}}{2}~~$ or $~~x=\frac{-1 + \sqrt{5}}{2},~~$ then $~~x^2+x-1 = 0$
When $x \lt \frac{-1 - \sqrt{5}}{2},~~$ then $~~x^2+x-1 \gt 0$
When $\frac{-1 - \sqrt{5}}{2} \lt x \lt \frac{-1 + \sqrt{5}}{2},~~$ then $~~x^2+x-1 \lt 0$
When $x \gt \frac{-1 + \sqrt{5}}{2},~~$ then $~~x^2+x-1 \gt 0$
Solution:
$x \lt \frac{-1 - \sqrt{5}}{2}~~$ or $~~x \gt \frac{-1 + \sqrt{5}}{2}$
The solution set is $~~(-\infty,\frac{-1 - \sqrt{5}}{2})\cup (\frac{-1 + \sqrt{5}}{2}, \infty)$
