High School Algebra II Unlocked (2016)

Lesson 3.5. Radical Inequalities in One Variable

The solution process for radical inequalities is similar to the solution process for rational inequalities.

To solve a radical inequality, solve an equation version of the inequality and use the potential solutions as endpoints of intervals to test in the original inequality. Also, neither the radicand nor the value of the radical can be negative for square roots, so use the variable value that makes the radicand 0 as another endpoint demarcating another interval to test.

These rules are
for finding all real
number solutions.
With imaginary
numbers, the radicand
could be negative.

The radicand is the value inside the radical symbol.

Solve the following inequality for x.

≥ x

The radical is a square root, so we know that the radicand must be greater than or equal to 0, so 2x + 8 ≥ 0, which simplifies to x ≥ −4. Keeping this limitation in mind, let’s solve the inequality as an equation.

Let’s plot the points −2 and 4 on a number line. Also, let’s plot the point −4, because we know that x ≥ −4.

Now we’ll test points in each interval.

When x = −5: is equal to , so this is not a solution in the set of real numbers.

When x = −3: ≥ −3 is true, because ≥ −3 is true.

When x = 0: ≥ 0 is true, because ≥ 0 is true.

When x = 5: ≥ 5 is false, because is not greater than or equal to 5.

The solution set is the interval between −4 and −2 and the interval between −2 and 4. Let’s also test each of the endpoints of the intervals.

When x = −4: ≥ −4 is true, because 0 ≥ −4 is true.

When x = −2: ≥ −2 is true, because 2 ≥ −2 is true.

When x = 4: ≥ 2 is true, because 4 ≥ 2 is true.

The radical is already isolated on the left side of the equation, so we can immediately square both sides of the equation.

Hold on! Before you mark (A), test each value in the original equation.

The equation 1 = −1 is not true, so −3 is an extraneous solution. The equation 9 = 9 is true, so 7 is the only solution for y. The answer is (E).

The points −4, −2, and 4 are all included in the solution set. The solution set is shown on the number line below.

The solution set can also be written as −4 ≤ x ≤ 4.

We no longer
need to mark a
closed-circle point
at −2, because it
is not an endpoint
of a solution set. It
is simply included
in the solution
set, along with
all other values
between −4 and 4.

Another way to solve Example 15 is by creating and graphing a system of equations. In this case, instead of just looking for points of intersection, we will look at the graph to see where the value of is greater than or equal to the value of x. The system of equations is shown below.

y =

y = x

We are using a method
similar to the one
used in Example 10, of
graphing a system of
equations to relate the
values of the two sides
of the given equation.
However, instead of
finding points with the
same function value,
here we are comparing
to see where
has an equal or
greater value than x.

Using graphing technology, we can graph the two functions on the same coordinate grid, as shown below.

The two graphs intersect at the point (4, 4). The graph of y = only exists for x ≥ −4. The graph of y = is higher than the graph of y = x from x = −4 to x = 4, where they intersect. So, ≥ x for −4 ≤ x ≤ 4.

If we were solving
the inequality
> x, then the
solution set would be
−4 ≤ x < 4. The
x-value of 4 makes
equal to
x, so it is included in
the solution when the
inequality symbol is
≥ but not when the
inequality symbol is >.