Question

Solution of $(x^2 + 1) > 0$ is

• A. $x \in (-\infty, -1) \cup (1, \infty)$
• B. $x \in (-1, 1)$
• C. $x \in \mathbb{R}$
• D. No solution

Answer 1

Answer: (C)

$x \in \mathbb{R}$

Answer 2

The inequality is $x^2 + 1 > 0$. The quadratic $x^2 + 1$ has a leading coefficient $a=1 > 0$ and discriminant $\Delta = 0^2 - 4(1)(1) = -4 < 0$. Since $a > 0$ and $\Delta < 0$, the quadratic $x^2 + 1$ is positive for all real values of $x$. Thus, the solution is $x \in \mathbb{R}$.