Question

How do you solve `x^3<=4x^2+3x` using a sign chart?

Answer 1

The answer is `x in ]-oo,2-sqrt7]uu [0,2+sqrt7]`

Explanation:

Let's rewrite and factorise the inequality

`x^3-4x^2-3x<=0`

`x(x^2-4x-3))<=0`

We calculate the roots of the equation `x^2-4x-3=0`

`Delta=(-4)^2-4*1*(-3)=16+12=28`

`x_1=(4+sqrt28)/2=(4+2sqrt7)/2=2+sqrt7`

`x_2=(4-sqrt28)/2=(4-2sqrt7)/2=2-sqrt7`

Let `f(x)=x(x-(2+sqrt7))(x-(2-sqrt7))`

Now, we build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaaa)` `-oo` `color(white)(aaaa)` `2-sqrt7` `color(white)(aaaa)` `0` `color(white)(aaaa)` `2+sqrt7` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x-x_2` `color(white)(aaaaaaa)` `-` `color(white)(aaaaaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `x-x_1` `color(white)(aaaaaaa)` `-` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaaaaaa)` `-` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaaaaa)` `+`

Therefore,

`f(x)<=0` when `x in ]-oo,2-sqrt7]uu [0,2+sqrt7]`