# 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:

1. sqrt(f(x))<g(x) is equivalent to the system {(f(x)<(g(x))^2),(f(x)>=0),(g(x)>0):}.
2. 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.