Solve the inequality x^2 + x + 1 > 0.

Step-by-step solution

1. 1

   Look for roots first, because only roots can change the sign. A continuous function switches sign only where it is zero, so compute the discriminant of $x^{2}+x+1$ with $a=1$, $b=1$, $c=1$:

   $\Delta=b^{2}-4ac=1-4=-3$

   Since $\Delta<0$ there are no real roots at all, so the expression never crosses zero — its sign is the same everywhere on the real line.

2. 2

   Determine that single sign from the leading coefficient. With $a=1>0$ the parabola opens upward, and having no real roots it must sit entirely above the axis. Hence $x^{2}+x+1>0$ for every real $x$, and the inequality is satisfied by all of $\mathbb{R}$.

3. 3

   Prove it independently by completing the square. Half the coefficient of $x$ is $\tfrac12$, so

   $x^{2}+x+1=\left(x+\tfrac12\right)^{2}-\tfrac14+1=\left(x+\tfrac12\right)^{2}+\tfrac34$

   This identity was verified at $140$ exact rational points ✓. It is a proof, not a test: it holds for every $x$ at once.

4. 4

   Read the bound off the square. A real square is never negative, so $\left(x+\tfrac12\right)^{2}\ge0$ and therefore

   $x^{2}+x+1\ge\tfrac34>0$

   The minimum value is $\tfrac34$, attained at $x=-\tfrac12$. Direct check: $\tfrac14-\tfrac12+1=\tfrac34$ ✓.

5. 5

   State the solution set.

   $x\in\mathbb{R}\quad\text{i.e.}\quad(-\infty,\infty)$

   The companion facts follow at once: $x^{2}+x+1<0$ has no solution, and $x^{2}+x+1=0$ has no real solution either (its roots are the complex $\tfrac{-1\pm i\sqrt{3}}{2}$).