Question

How do you solve and graph `x^3<=4x^2+3x`?

Answer 1

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

Explanation:

We solve this inequality with a sign chart.

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

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

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

We need the roots of the quadratic equation

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

The discrimenant is

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

As, `Delta>0`, there are 2 real roots

`x_2=(-b+sqrtDelta)/2=1/2(4+sqrt28)=2+sqrt7=4.646`

`x_1=(-b-sqrtDelta)/2=1/2(4+sqrt28)=2-sqrt7=-0.646`

Let `f(x)=x(x-x_1)(x-x_2)`

We can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `x_1` `color(white)(aaaa)` `0` `color(white)(aaaaaa)` `x_2` `color(white)(aaaa)` `+oo`

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

`color(white)(aaaa)` `x` `color(white)(aaaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

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

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(x)<=0` when `x in (-oo, 2-sqrt7] uu [0, 2+sqrt7]`
graph{x^3-4x^2-3x [-12.34, 12.97, -9.01, 3.65]}