Question

How do you solve `2u^2+5u>=12` using a sign chart?

Answer 1

The solution is `u in (-oo,-4] uu [3/2, +oo)`

Explanation:

Lets 's factorise the function

`f(u)=2u^2+5u-12=(2u-3)(u+4)`

The equation we have to solve is

`f(u)=(2u-3)(u+4)>=0`

Let's build the sign chart

`color(white)(aaaa)` `u` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-4` `color(white)(aaaaaaaaaa)` `3/2` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `u+4` `color(white)(aaaaaa)` `-` `color(white)(aa)` `0` `color(white)(aaaa)` `+` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `2u-3` `color(white)(aaaaa)` `-` `color(white)(aa)` ###color(white)(aaaaa)#`-` `color(white)(aaaa)` `0` `color(white)(aaa)` `+`

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

Therefore,

`f(u)>=0` when `u in (-oo,-4] uu [3/2, +oo)`

graph{2x^2+5x-12 [-29.27, 28.45, -13.66, 15.22]}