Chapter 3.5, Problem 41AYU

To determine

To Explain:

The inequality equation $x^2+x+1>0$ has solution set of all real numbers.

Answer to Problem 41AYU

The inequality equation $x^2+x+1>0$ has solution set of all real numbers because it has only one interval.

Explanation of Solution

Given Information:

  $x^2+x+1>0$

The given equation is:

  $x^2+x+1>0$

One comparing it with the standard quadratic equation: $a{x}^2+bx+c>0$ , where

  $a=1,b=1,c=1$

Now, the given quadratic equation contains a set of all real numbers as solution set when it satisfies the following condition:

  $b^2-4ac<0$

Put all values:

  $\begin{array}{l}{b}^2-4ac<0\\ {\left(1\right)}^2-4\left(1\right)\left(1\right)<0\\ 1-4<0\\ -3<0\end{array}$

So, the given equation hasnegative discriminate value. When the value of discriminate is negative, that means there are no real roots for the given quadratic equation.

When the equation has no real roots, there is no critical number for equation which directly shows that for the given there is only one interval; $\left(-\infty ,\infty \right)$ .

So, when the quadratic equation contains only one interval it implies that the set of solution is nothing but all real numbers.