Chapter 3.5, Problem 41AYU

Explain why the inequality $x^2\text{ + }x\text{ + 1 > 0}$ has all real numbers as the solution set.

To determine

To find: To prove the inequality $x^2+x+1>0$ has all real numbers as the solution set.

Answer to Problem 41AYU

Solution:

$x^2+x+1>0$

The above inequality is true for all real numbers $x$.

Therefore the given inequality has all real numbers as the solution set.

Explanation of Solution

Given:

The inequality is: $x^2+x+1>0$

Calculation:

$x^2+x+1>0$

The above inequality is true for all real numbers $x$.

Therefore the given inequality has all real numbers as the solution set.