Quadratic Inequalities: Problems and Examples

Before you continue studying these quadratic inequalities and problems, be sure to understand the 2 main techniques to solve them, that is, the Test Point Method and the Factor Method. The examples below will primarily use the Test Point to show you how the solution is affected by the type of quadratic function (does it intersect the x-axis at 2, 1 or no points?) you're working with, and the type of inequality sign that is used.





Example 1 – Quadratic inequality graph intersects x-axis at 2 points (positive "a")
Solve for:
(a) x2 – 2x – 1 ≥ 2
(b) x2 – 2x – 1 > 2
(c) x2 – 2x – 1 ≤ 2
(d) x2 – 2x – 1 < 2

Example 1 Solution
(a) First, solve for the roots.


So we know that the roots on the x-axis are x = 3, and x = -1. From the second line, we see that we're looking for the parts of the graph that are touching and are above the x-axis. We know that the roots of the quadratic are the point touching the x-axis. However, we still need to find out where the other parts of the graph are above the x-axis.

From the roots, we can see that the x-axis can be divided into 3 domains, i.e. (-∞, -1), (-1, 3), (3, +∞). We'll need to substitute any number per domain to determine if the graph is above or below the x-axis in that domain.

Domain (-∞, -1):
We're going choose -2 for convenience.
x2 – 2x – 3
= (-2)2 – 2(-2) – 3
= 4 + 4 – 3
= 5 (positive)

Domain (-1, 3):
Substituting 0 is usually easy, so we'll use that.
x2 – 2x – 3
= 0 – 0 – 3
= -3 (negative)

Domain (3, +∞):
We'll select 4 here for convenience.
x2 – 2x – 3
= (4)2 – 2(4) – 3
= 16 – 8 – 3
= 5 (positive)

The solution is then x ≤ -1, and x ≥ 5, i.e. the graph is either on the x-axis, or above it.

Here's the graph of x2 – 2x – 3 = 0, in case you're not convinced that's the answer ☺.



(b) You can solve this with the exact steps we did for part (a) just now, but this time we're looking for parts of the graph that are above AND not touching the x-axis.

From the domains test, we know that x < 1 and x > 3 is above the x-axis.

(c) Similarly, the steps to solve this are identical to (a), up to and including the domains test. However, we are now looking for parts of the graph that are below or touching the x-axis (≤ 0). From the testing of the domains, we know that (-1, 3) is below the graph. Hence, the solution is that for -1 ≤ x ≤ 3, the graph is below or touching the x-axis.

(d) As before, the steps to solve this are identical to(a). Since we are now looking for parts of the graph that are below the x-axis (< 0), the solution is similar to (c). However, the solution is that for -1 < x < 3, the graph is below the x-axis.


Example 2 – Quadratic inequality graph intersects x-axis at 2 points (negative "a")
Solve for:
(a) -x2 + 2x + 1 ≥ -2
(b) -x2 + 2x + 1 > -2
(c) -x2 + 2x + 1 ≤ -2
(d) -x2 + 2x + 1 < -2

Example 2 Solution
(a) First, solve for the roots of the quadratic.


So we know that the graph intersects the x-axis at x=3 and x = -1. As with Example 1, we know that the domain (i.e., the x-axis) can be split into three parts. That is, (-∞, -1), (-1, 3), (3, +∞). These need be tested to see if the portion of the quadratic graph in those domains are above or below the x-axis.

Domain (-∞, -1):
We substitute -2 into the quadratic in its standard form for convenience.
-x2 + 2x + 3
= -(-2)2 + 2(-2) + 3
= -4 – 4 + 3
= -5 (negative)

Domain (-1, 3):
Substitute 0 into the quadratic.
-x2 + 2x + 3
= 0 + 0 + 3
= 3 (positive)

Domain (3, +∞):
Use 4 as a test. You can use any other number if it's easier for you to work out in your head.
-x2 + 2x + 3
= -(4)2 + 2(4) + 3
= -16 + 8 + 3
= -5 (negative)

From the domain test, we know that the graph is above or touching the x-axis at -1 ≤ x ≤ 3.

Checking against the plotted graph, we see that the solution is correct.



(b) Solution is almost identical to (a), except we exclude the points that touch the x-axis. So the solution is -1 < x < 3.

(c) Now we're looking for the portion of the graph below or touching the x-axis. From the domain test in (a), we can see that the solution is x ≤ -1, x ≥ 3.

(d) The solution is x < -1, x > 3.


Example 3 – Quadratic inequality graph intersects x-axis at 1 point (positive "a")
Solve for:
(a) x2 – 2x + 1 ≥ 0
(b) x2 – 2x + 1 > 0
(c) x2 – 2x + 1 ≤ 0
(d) x2 – 2x + 1 < 0

Example 3 Solution
(a) First, find the roots of the quadratic equation, ignoring the inequality sign.


We can see from the roots that the graph touches the x-axis only at one point, that is, x = 1. Since "a" is +ve, it's a "smiley" graph. When plotted, it looks like the one below:

There are 2 regions we need to test, i.e x < 1, and x > 1.

x < 1:
We take x = 0 to test for convenience.
(x – 1)(x – 1)
= (-1)(-1)
= 1 (+ve)

Hence, the domain x < 1 is above the x-axis.



x > 1:
We take x = 2 to test for convenience.
(x – 1)(x – 1)
= (1)(1)
= 1 (+ve)

Hence, the domain x > 1 is also above th x-axis.

The solution for the inequality is therefore x ≥ 1 and x ≤ 1.

(b) The solution is x > 1 and x < 1.

(c) From (a), we can clearly see that all of the graph is above the x-axis, except at x = 1, where it touches the x-axis. Hence, the solution is x = 1.

(d) There is no solution, since the graph is either touching or above the x-axis for all values of x.


Example 4 – Quadratic inequality graph intersects x-axis at 1 point (negative "a")
Solve for:
(a) -x2 + 2x – 1 ≥ 0.
(b) -x2 + 2x – 1 > 0.
(c) -x2 + 2x – 1 ≤ 0.
(d) -x2 + 2x – 1 < 0.

Example 4 Solution
(a) First, solve for the roots.


We can see that the roots are real and repeated, i.e. x = 1. Then we test the domains to determine which domain is above or below the x-axis.

x > 1:
We take x = 2 to test for convenience.
(- x + 1)(x – 1)
= (-1)(1)
= -1 (-ve)

x < 1:
We take x = 0 to test for convenience.
(- x + 1)(x – 1)
= (1)(-1)
= -1 (-ve)

Hence, the solution is x = 1, since the graph touches the x-axis at that point. All other points on the graph are below the x-axis.

We can double-check this with the graph of the quadratic below.



(b) Based on the domain analysis in (a), there is no solution since none of the points on the graph are above the x-axis.

(c) The solution is "all values of x", since all points of the quadratic graph are either below the x-axis, or touching the x-axis.

(d) The solution is x < 1, x > 1, that is, all the points of the quadratic graph except the one touching the x-axis.


Example 5 – Quadratic inequality graph does not intersect x-axis (positive "a")
Solve for:
(a) x2 + 1 ≥ 0
(b) x2 + 1 > 0
(c) x2 + 1 ≤ 0
(d) x2 + 1 < 0

Example 5 Solution
(a) When we attempt to find the root for the quadratic inequality, we find that there is no real roots. A good way to do this is to use the root formula.



Since "a" is positive, we know that this quadratic equation is a smiley. Since the graph never never intersects the x-axis (it has no real roots), it is always above the x-axis for all values of x. Hence, the solution for this quadratic inequality is "all values of x".

(b) The solution is also "all values of x".

(c) No solution, since none of the points of the quadratic graph are touching the x-axis or below it.

(d) No solution, since none of the points of the quadratic graph are below the x-axis.


Example 6 – Quadratic inequality does not intersect x-axis (negative "a")
Solve for:
(a) -x2 – 1 ≥ 0
(b) -x2 – 1 > 0
(c) -x2 – 1 ≤ 0
(d) -x2 – 1 < 0

Example 6 Solution
(a) As with example 5, you will find that there are no real roots for this equation. This is because the roots are imaginary (let's not worry about that for now). Plotting the graph, we can see that all the values of the quadratic graph are below the x-axis.

Hence, there is no solution for this quadratic inequality. (b) As with (a), there is no solution for this quadratic inequality. (c) The solution is "all values of x", since all the quadratic graph points are below the x-axis. (d) The solution is "all values of x", since all the quadratic graph points are below the x-axis.

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