# How do I use a sign chart to solve x^2+5x-9<0?

Jun 8, 2018

${x}^{2} + 5 x - 9 < 0$ when $x \in \left(- 6.4051 , 1.4051\right)$

#### Explanation:

First, we want to find what the zeros of the function are.

$\text{Let } f \left(x\right) = {x}^{2} + 5 x - 9$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- 5 \pm \sqrt{{5}^{2} - \left(4\right) \left(- 9\right)}}{\left(2\right) \left(1\right)}$

$x = \frac{- 5 \pm \sqrt{25 + 36}}{2}$

$x = \frac{- 5 \pm \sqrt{61}}{2}$

$x = 1.4051 \text{, } x = - 6.4051$

Now, make a number line that includes these values, like the one below.

To make the sign chart, plug in values that fall within the specified interval - I tend to pick easy numbers for which I can do calculations in my head.

The original equation is an inequality. We need to find where $f \left(x\right) < 0$. This occurs where $f \left(x\right)$ is negative, which is in the interval from $\left(- 6.4051 , 1.4051\right)$. This interval makes sense because when we graph it, the parabola drops below the x-axis in this interval, as shown below.

graph{x^2+5x-9 [-28.86, 28.87, -14.43, 14.43]}