# How do I use a sign chart to solve 2x^2+5x-3>0?

##### 1 Answer
Jun 30, 2015

$f \left(x\right)$ is positive when $x < - 3$ and $x > 0.5$

#### Explanation:

We have $f \left(x\right) = 2 {x}^{2} + 5 x - 3$, and we want to find the values of $x$ for which $f \left(x\right) > 0$.

We start by finding the critical numbers.

Set $f \left(x\right) = 2 {x}^{2} + 5 x - 3 = 0$ and solve for $x$.

$\left(2 x - 1\right) \left(x + 3\right) = 0$

$2 x - 1 = 0$ or $x + 3 = 0$

$2 x = 1$

x = ½ = 0.5 or $x = - 3$

The critical numbers are $- 3$ and $0.5$.

Now we check for positive and negative regions.

We have three regions to consider: (a) $x < - 3$; (b) $- 3 < x < 0.5$; and (c) $x > 0.5$.

Case (a): Let $x = - 4$.

Then f(4) = 2×(-4)^2 + 5(-4) -3 =32 – 20 -3 = 9

$f \left(x\right)$ is positive when $x < - 3$.

Case (b): Let $x = 0$.

Then f(0) = 2×0^2 + 5×0 -3 = 0 + 0 -3 = -3

$f \left(x\right)$ is negative when $- 3 < x < 0.5$

Case (c): Let $x = 1$.

Then f(1) = 2×1^2 + 5×1 -3 = 2 +5 – 3 = 4

$f \left(x\right)$ is positive when $x > 0.5$. Answer: $f \left(x\right)$ is positive when $x < - 3$ and $x > 0.5$.