Disequazione irrazionale $\sqrt{4-x}\le x+2$

Text

Solve the irrational inequality $\sqrt{4-x}\le x+2.$

Solution

An inequality of the form $\sqrt{A}\le B$ is equivalent to the system $A\ge 0,\ B\ge 0,\ A\le B^2$.

• $4-x\ge 0\implies x\le 4$;
• $x+2\ge 0\implies x\ge -2$;
• $4-x\le (x+2)^2=x^2+4x+4\implies 0\le x^2+5x\implies x(x+5)\ge 0\implies x\le -5\ \vee\ x\ge 0$. Intersecting with $-2\le x\le 4$ (which excludes $x\le -5$): $\boxed{0\le x\le 4.}$