Respuesta :

It is given that:

[tex]\begin{gathered} x^2+5x<2 \\ x^2+5x-2<0 \end{gathered}[/tex]

Solve for x to get:

[tex]\begin{gathered} x^2+5x+\frac{25}{4}-2-\frac{25}{4}<0 \\ (x+\frac{5}{2})^2-(\frac{\sqrt{33}}{2})^2<0 \\ (x+\frac{5+\sqrt[]{33}}{2})(x+\frac{5-\sqrt[]{33}}{2})<0 \end{gathered}[/tex]

The two values are less than 0 if either one of them is negative so it follows:

[tex]\begin{gathered} x+\frac{5+\sqrt[]{33}}{2}<0\text{ and }x+\frac{5-\sqrt[]{33}}{2}>0 \\ OR \\ x+\frac{5+\sqrt[]{33}}{2}>0\text{ and }x+\frac{5-\sqrt[]{33}}{2}<0 \end{gathered}[/tex]

The graph is given as below:

The points are x=-5.372 and x=0.372 which is the same as:

[tex]x=-\frac{5+\sqrt[]{33}}{2}=-5.372,x=-\frac{5-\sqrt[]{33}}{2}=-0.372[/tex]

Ver imagen ReanaC626342
ACCESS MORE
EDU ACCESS