Question

How do you solve the inequality `5x+4<=3x^2` and write your answer in interval notation?

Answer 1

The solution is `x in (-oo, -0.59] uu[2.26, +oo)`

Explanation:

Let's rewrite the inequality

`5x+4<=3x^2`

`3x^2-5x-4>=0`

Let `f(x)=3x^2-5x-4`

The roots of the quadratic equation `3x^2-5x-4=0`, are

`x = ( 5+-sqrt ((-5)^2-(4)* (3) * (-4)) ) /(6)=(5+-sqrt(73))/(6)`

`x_1=(5-sqrt73)/6=-0.59`

`x_2=(5+sqrt73)/6=2.26`

Let's build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaa)` `-oo` `color(white)(aaaaaaa)` `x_1` `color(white)(aaaaaa)` `x_2` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x-x_1` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x-x_2` `color(white)(aaaaa)` `-` `color(white)(aaaa)` ###color(white)(aaaa)#`-` `color(white)(aa)` `0` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `+` `color(white)(aaaa)` `0` `color(white)(aaa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `+`

Therefore,

`f(x)>=0` when `x in (-oo, x_1] uu[x_2, +oo)`

graph{3x^2-5x-4 [-11.39, 11.11, -6.615, 4.635]}