Square Root Inequalities

Domain First

A closed dot means the endpoint is included ($\ge$ or $\le$). An open dot means it is not included ($>$ or $<$).

Root Less Than a Number

Start with the simplest type: a root compared with a constant $k$. Think about what the root can be. It is never negative, so it can never be smaller than a negative number.

Root Greater Than a Number

Now flip the direction. A root is $\ge 0$, so it automatically beats any negative number.

The curve is $y = \sqrt{3x + 6}$ and the line is $y = 3$. Where is the curve above or on the line?

Systems and Quadratic Radicands

Often you must satisfy several conditions at once, or the radicand is a quadratic. In both cases, solve each condition separately and then intersect the solution sets. (Some problems write "$\sqrt{x} \ge 2$ and $\sqrt{x} < 5$" as a system of inequalities in one unknown.)

When the Other Side Has an $x$

If the inequality is $\sqrt{P(x)} \lessgtr g(x)$ and $g(x)$ contains $x$, you cannot just square. The sign of $g(x)$ changes from one $x$ to another, and squaring only keeps the order when both sides are nonnegative. So split into cases by the sign of $g(x)$.

Some of these inequalities have a left side that is a quadratic radicand, like $\sqrt{x^2 - 4x + 3} \le x - 1$. The method is identical: domain, sign of the right side, square, intersect. Be careful with isolated single points: the domain may be $x \le 1$ or $x \ge 3$ and the sign condition may be $x \ge 1$, which leaves the point $x = 1$ together with $x \ge 3$.

Check It on a Graph

Solving $\sqrt{x + 2} \ge x$ means asking: for which $x$ is the graph of $y = \sqrt{x + 2}$ above or touching the graph of $y = x$? The curve and the line meet where the two sides are equal, and those crossing points are the boundaries of the solution.

The orange segment on the $x$-axis is where the curve is on or above the line. It stops at $x = 2$, exactly where the two graphs cross. Use the interactive graph to test other inequalities this way.

Nested Roots and "For Every $x$" Conditions

Three situations come up again and again.

Nested roots. Work from the outside in, and keep the domain of every layer. For $\sqrt{3 - \sqrt{x + 1}}$, the inner root needs $x + 1 \ge 0$, and the outer radicand needs $3 - \sqrt{x + 1} \ge 0$.

The squeeze. Sometimes one side is forced to be $\le 0$ while the other is $\ge 0$. Then the only possibility is both equal to zero.

Nested inequalities such as $\sqrt{x - \sqrt{1 - x}} \ge 1$ often end in a squeeze like this one. Square once (the right side is $1 > 0$) to remove the outer root, and then use the rule from the previous section on the remaining root.

A condition for every $x$. To make $\sqrt{x^2 + kx + 4}$ defined for all real $x$, the quadratic must never be negative. Its parabola opens upward, so it needs at most one point on the $x$-axis: discriminant $k^2 - 16 \le 0$, so $-4 \le k \le 4$. If the root must be strictly positive for all $x$ (as in "$\sqrt{\ldots} > 0$ for every $x$"), the parabola must not touch the axis at all: discriminant $< 0$, so $-4 < k < 4$.

Roots in Denominators

When the root is in a denominator, the radicand must be strictly positive (see the first section). To compare, remember that for positive numbers, taking reciprocals reverses the inequality: if $a > 0$ and $b > 0$, then $\dfrac{1}{a} > \dfrac{1}{b}$ exactly when $a < b$.

Modeling With Square Roots

Many real formulas have a square root: the swing time of a pendulum, the speed from skid marks, the time it takes something to fall. To use them with inequalities:

1. Name the variable and write the inequality from the words.

2. Add the physical restrictions (lengths and counts cannot be negative).

3. Solve, intersecting with the domain.

4. Answer in the context of the problem, with units.

For the pendulum, note that $2\pi\sqrt{\dfrac{L}{\pi^2}}$ simplifies before you start solving.

Distance and Cost Models

Distances use the Pythagorean Theorem, so a root of a quadratic appears naturally. Cost models often combine a root with a linear term, and then you must isolate the root before squaring.

A radicand like $t^2 - 12t + 52$ can be rewritten by completing the square: $(t - 6)^2 + 16$. That shows it is always positive, so the domain is all real numbers, and you only need the squared inequality. Also remember that a time $t$ cannot be negative.

Summary