Type II — root smaller

When the root is on the smaller side, $\sqrt{A(x)}<B(x)$, the solution is a single system of three simultaneous conditions.

Property — $\sqrt{A(x)}<B(x)$

The inequality $\sqrt{A(x)}<B(x)$ is equivalent to the system (all three conditions hold simultaneously): $\begin{cases} A(x) \ge 0 \\ B(x) > 0 \\ A(x) < B^2(x) \end{cases}$

Remark — Why all three conditions

$A(x)\ge 0$ is the domain of existence of the root; $B(x)>0$ is necessary because a quantity that is $\ge 0$ (the root) cannot be smaller than a non-positive number; $A(x) < B^2(x)$ is the squaring, permissible because both members are $\ge 0$ under the two preceding conditions.

Example

$\sqrt{x^2-1}<x$.

System: $\begin{cases}x^2-1\ge 0 \\ x > 0 \\ x^2-1 < x^2\end{cases}$.

• First: $x\le -1 \vee x\ge 1$.
• Second: $x>0$.
• Third: $-1<0$, always true.

Intersection: $x\ge 1$. Solution: $\boxed{x\ge 1}$.

Caution — The direction $\le$ and $\ge$

The formulae above hold for strict inequalities ($>$ and $<$). For the non-strict versions ($\ge$ and $\le$) it is enough to replace the corresponding symbols in the schemes, bearing in mind that the point at which both members are zero (if there is one) becomes a solution (previously it was not).

Links

Topics: Radicals
Concepts: Conditions of existence · Irrational inequality
Methods: Irrational inequality case analysis
Skills: Reasoning by cases · Solving inequalities