Type I — root greater

When the root is on the greater side, $\sqrt{A(x)}>B(x)$, the solution is the union of two cases.

Property — $\sqrt{A(x)}>B(x)$

The inequality $\sqrt{A(x)}>B(x)$ is equivalent to the union of two cases:

• Case (a): $B(x)<0$. The first member (the root) is $\ge 0$ whenever it exists, so it is automatically greater than a negative number. It is always true, provided the root exists: $\begin{cases}A(x)\ge 0 \\ B(x) < 0\end{cases}$
• Case (b): $B(x)\ge 0$. Both members are non-negative, so one can square while keeping the direction: $\begin{cases}B(x)\ge 0 \\ A(x) > B^2(x)\end{cases}$ (The condition $A(x)\ge 0$ is not needed explicitly: it is implied by $A(x)>B^2(x)\ge 0$.)

The final solution is $S_a\cup S_b$.

The two-branch scheme of Type I: the two cases are solved and their solutions are united.

Example

$\sqrt{x-1} > x-3$.

Case (a): $\begin{cases}x-1\ge 0 \\ x-3 < 0\end{cases} \iff 1\le x<3$. Solution zone $[1;\,3)$.

Case (b): $\begin{cases}x-3\ge 0 \\ x-1 > (x-3)^2\end{cases}$. The second: $x-1 > x^2-6x+9 \iff x^2-7x+10 < 0 \iff 2<x<5$. Intersecting with $x\ge 3$: $3\le x<5$.

Union: $\boxed{1\le x < 5}$.

Links

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