A quadratic inequality has the form ax
+ bx + c > 0 (or <, or , or ). The highest exponent is 2. The next few
examples show how to solve quadratic inequalities.
Solving Quadratic Inequalities
Solve the quadratic inequality x
- x < 12.
Write the inequality with 0 on one side, as x
- x - 12 < 0. This inequality is solved with values of x that
make x - x - 12 negative (< 0). The quantity x
- x - 12 changes from positive to negative or from negative to
positive at the points where it equals 0. For this reason, first
solve the equation x
- x - 12 = 0.
x - x - 12 = 0
(x - 4)(x + 3) = 0
x = 4 or x = -3
Locating -3 and 4 on a number line, as shown in Figure 3,
determines three intervals A, B, and C.
Decide which intervals include numbers that make x
- x - 12 negative by substituting any number from each interval
in the polynomial. For example,
choose -4 from interval A: (-4)
- (-4) - 12 = 8 > 0;
choose 0 from interval B: 0
- 0 - 12 = -12 < 0;
choose 5 from interval C: 5
- 5 - 12 = 8 > 0.
Only numbers in interval B satisfy the given inequality, so
the solution is (-3, 4). A graph of this solution is shown in
Solving Polynomial Inequalities
Solve the inequality x(x-1)(x+3) 0.
This is not a quadratic inequality. If the three factors are
multiplied, the highest-degree term is x. However, it can be solved in the same way as a
quadratic inequality because it is in factored form. First solve
the corresponding equation.
x(x - 1)(x + 3) = 0
x = 0 or x - 1 = 0 or x + 3 = 0
x = 0 or x = 1 or x = -3
These three solutions determine four intervals on the number
line: , (-3, 0), (0, 1) and . Substitute a number from each interval into the
original inequality to determine that the solution includes the
numbers less than or equal to -3 and the numbers that are equal
to or between 0 and 1. See Figure 5.
In interval notation, the solution is
* The symbol indicates the union of two sets, which
includes all elements in either set.