Question

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

Answer 1

See below.

Explanation:

`3x^2-5x>8`

Subtract `8` from both sides:

`3x^2-5x-8>0`

We now solve the equation:

`3x^2-5x-8=0`

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

Factoring:

`(3x-8)(x+1)=0=> x=-1 and x= 8/3`

METHOD 1

From:

`3x^2-5x-8>0`

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

`uuu`

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=8/3`.

Since this is a `>` and not a `>=` inequality `x=-1` and `x=8/3` are not included points. So the solution in interval notation we be a union of intervals:

`(-oo,-1)uu(8/3,oo)`

METHOD 2

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

`(3x-8)(x+1)`

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` , `color(white)(88)-1< x< 8/3` , `color(white)(88) 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 `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]}