The method for the larger root — Type II

Consider now the opposite case: $\sqrt{A(x)}\ge B(x).$

The left-hand side, when it exists, is $\ge 0$; the right-hand side may be positive or negative. Let us distinguish.

• Case 1: $B(x)<0$. A non-negative quantity is automatically $\ge$ a negative one, so the inequality is always satisfied within the existence conditions. The only remaining condition is $A(x)\ge 0$.
• Case 2: $B(x)\ge 0$. Both sides are now non-negative: we can square while preserving the direction, obtaining $A(x)\ge (B(x))^2$. The existence condition $A(x)\ge 0$ is in fact implied by the latter (if $A\ge B^2\ge 0$), but we leave it explicit for clarity.

The final solution is the union of the solutions of the two cases.

In brief — Method for $\sqrt{A(x)}\ge B(x)$

$\underbrace{\begin{cases}A(x)\ge 0 \\ B(x)<0\end{cases}}_{\text{Caso 1}} \quad\cup\quad \underbrace{\begin{cases}A(x)\ge 0 \\ B(x)\ge 0 \\ A(x)\ge \bigl(B(x)\bigr)^2\end{cases}}_{\text{Caso 2}}$

The diagram below summarises the structure of the method: from the starting inequality the two cases branch out, and their solutions reunite in a union.

The structure of the method for $\sqrt{A(x)}\ge B(x)$: two cases that reunite in a union.

Warning — The most common mistake

It is easy to forget Case 1 and solve only Case 2: in this way one loses all the solutions in which the right-hand side is negative. One way not to go wrong is to always start from the question: “when the right-hand side is negative, is the inequality trivially true or trivially false?”.

Links

Topics: Irrational inequalities
Concepts: Existence conditions · Irrational inequality
Methods: Irrational inequality by cases · Squaring
Skills: Reasoning by cases · Solving inequalities