# How do you solve and write the following in interval notation: x^2 + x >3 ?

Aug 3, 2017

The solution is $x \in \left(- \infty , \frac{- 1 - \sqrt{13}}{2}\right) \cup \left(\frac{- 1 + \sqrt{13}}{2} , + \infty\right)$

#### Explanation:

The inequality is ${x}^{2} + x > 3$

${x}^{2} + x - 3 > 0$

We need the roots of the quadratic equation

${x}^{2} + x - 3 = 0$

The discriminant is

$\Delta = {b}^{2} - 4 a c = {\left(1\right)}^{2} - 4 \cdot \left(1\right) \cdot \left(- 3\right) = 1 + 12 = 13$

As $\Delta > 0$, there are 2 real roots

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 1 \pm \sqrt{13}}{2}$

Therefore,

${x}_{1} = \frac{- 1 - \sqrt{13}}{2}$

${x}_{2} = \frac{- 1 + \sqrt{13}}{2}$

Let $f \left(x\right) = {x}^{2} + x - 3$

We can build the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$${x}_{1}$$\textcolor{w h i t e}{a a a a}$${x}_{2}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x - {x}_{1}$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - {x}_{2}$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(x\right) > 0$ when $x \in \left(- \infty , {x}_{1}\right) \cup \left({x}_{2} , + \infty\right)$

graph{x^2+x-3 [-10, 10, -5, 5]}