Question

How do you solve `x^5+9x>=10x^3`?

Answer 1

The answer is `x in [-3,-1] uu [0, 1] uu [3,+ oo [`

Explanation:

Let `f(x)=x^5-10x^3+9x`

Let's rearrange the equation

`x^5-10x^3+9x>=0`

`x(x^4-10x^2+9)>=0`

`x(x^2-1)(x^2-9)>=0`

`x(x+1)(x-1)(x+3)(x-3)>=0`

We can do a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-3` `color(white)(aaaa)` `-1` `color(white)(aaaa)` `0` `color(white)(aaaa)` `1` `color(white)(aaaa)` `3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+3` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaa)` `+` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x+1` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaa)` `+` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaa)` `+` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x-1` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaa)` `-` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x-3` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaa)` `-` `color(white)(aaa)` `-` `color(white)(aaa)` `+`

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

So,

`f(x)>=0` when `x in [-3,-1] uu [0, 1] uu [3,+ oo [`