# Irrational Inequalities

When we solve irrational inequalities, we use follwoing fact:

**Fact**. If both sides of inequality on some interval X take only non-negative values then squaring both sides (or taking any even power) will give equivalent inequality (on interval X). Taking odd power of both sides will always give equivalent inequality.

Therefore:

- `sqrt(f(x))<g(x)` is equivalent to the system `{(f(x)<(g(x))^2),(f(x)>=0),(g(x)>0):}`.
- `sqrt(f(x))>g(x)` is equivalent to the set of systems `{(f(x)>=0),(g(x)<0):}`, `{(f(x)>(g(x))^2),(f(x)>=0),(g(x)>=0):}`.

**Example 1**. Solve `sqrt(x^2-x-12)<x`.

This system is equivalent to `{(x^2-12x-12<x^2),(x^2-x-12>=0),(x>0):}`, i.e. `{(x> -1),((x+3)(x-4)>=0),(x>0):}`.

Solving this system we obtain that `x>=4`.

**Example 2**. Solve `sqrt(x^2-4x+2)>x+3`.

This inequality is equivalent to the set of systems `{(x^2-4x+2>=0),(x+3<0):}`, `{(x^2-4x+2>(x+3)^2),(x^2-4x+2>=0),(x+3>=0):}`.

We can neglect second inequality of second system, because it follows from first inequality.

Solving first system, we obtain that `x<-3`, from second system we have that `-3<=x<-0.7`.

Taking union of this intervals will yield solution: `x<-0.7`.