# How do you solve 3x^2-5x>8?

Jan 28, 2018

See below.

#### Explanation:

$3 {x}^{2} - 5 x > 8$

Subtract $8$ from both sides:

$3 {x}^{2} - 5 x - 8 > 0$

We now solve the equation:

$3 {x}^{2} - 5 x - 8 = 0$

This will give us the boundary values of $x$.

Factoring:

$\left(3 x - 8\right) \left(x + 1\right) = 0 \implies x = - 1 \mathmr{and} x = \frac{8}{3}$

METHOD 1

From:

$3 {x}^{2} - 5 x - 8 > 0$

Notice that the coefficient of ${x}^{2} > 0$, this means the parabola is in this form:

$\bigcup$

For values greater than $0$ we will be above the $x$ axis.

Since the roots to the equation are on the $x$ axis we can see that for positive $y$ values we will be to the left of $x = - 1$ and to the right of $x = \frac{8}{3}$.

Since this is a $>$ and not a $\ge$ inequality $x = - 1$ and $x = \frac{8}{3}$ are not included points. So the solution in interval notation we be a union of intervals:

$\left(- \infty , - 1\right) \cup \left(\frac{8}{3} , \infty\right)$

METHOD 2

Using the factors of $3 {x}^{2} - 5 x - 8 > 0$ we found earlier, we can see that:

$\left(3 x - 8\right) \left(x + 1\right)$

Will be positive i.e. $> 0$ if both brackets are positive or both brackets are negative. We can use a table to check this using the following inequalities:

$x < - 1$ , $\textcolor{w h i t e}{88} - 1 < x < \frac{8}{3}$ , $\textcolor{w h i t e}{88} \frac{8}{3} < x$ You can see from the table that if we assign $x$ a value satisfying the inequality in each column. it is only for the inequalities $x > - 1$ and $\frac{8}{3} < x$ that the product of the brackets is positive.

So our solution in interval notation is:

(-oo,-1)uu(8/3 , oo)color(white(88) as before.

Graph
graph{y<3x^2-5x-8 [-32.48, 32.46, -16.24, 16.25]}