Worked exercise — Root versus absolute value

Statement

Classify and solve: $\sqrt{x^2+4}\ge |x-1|.$

Solution

It is a shortcut: on the left-hand side a root, on the right an absolute value, both $\ge 0$. Moreover $x^2+4>0$ always, so no CE restricts the domain. We square directly (the modulus disappears: $|x-1|^2=(x-1)^2$): $x^2+4 \ge (x-1)^2 = x^2-2x+1 \implies 4 \ge -2x+1 \implies 2x \ge -3 \implies x\ge -\tfrac32.$

Solution: $\boxed{\,x\ge -\tfrac32\,}$.

Links

Topics: Irrational inequalities
Concepts: Irrational inequality · Squaring · Absolute value
Methods: Non-negative sides shortcut
Skills: Solving inequalities
Exercise type: Solving an inequality