What is a quadratic inequality?

Answer 1

The standard form of a quadratic inequality in one variable is f(x) = ax^2 + bx + c > 0 (or < 0). If the inequality is written in another form, transform it into the standard form, with a trinomial in x on the left side and 0 on the right side.

Solving the inequality for x means finding the values of x that make the inequality true. These values of x constitute the solution set of the inequality. The solution set is expressed in terms of intervals. By convention, we use these following symbols, x1 and x2 being the 2 real roots of the equation f(x) = 0:

- (x1 , x2) for the open interval between x1 and x2. The end points x1 and x2 are not included in the solution set.

- [x1 , x2] for the closed interval between x1 and x2. The end points are included in the solution set.

Note. When the inequality has an additional equal sign (greater, less, or equal to), the 2 end points are automatically included in the solution set. The solution set may be a closed interval [x1 , x2]. Or, the solution set may be half closed intervals: (-infinity, x1] and [x2 , + infinity).

There are 3 common methods to solve a quadratic inequality:

1. Using the number-line

2. Using the algebraic method.

3. By graphing.

Approaches in solving quadratic inequalities.

There are 2 approaches in solving quadratic inequalities depending on the chosen method. The test point approach is convenient for the number-line or the graphing method. The second approach uses a theorem on the sign of a trinomial f(x), if the algebraic method is chosen.

Examples of solving quadratic inequalities.

Example 1. Solve: f(x) = 12x^2 + 5x - 72 < 0.

Solution. First, solve the equation f(x) = 0. I advise you to use the Diagonal Sum Method to solve it. Roots have opposite signs. Write down the all-options-line. After elimination of no-fitted options, the remainder probable roots-set is (-8/3 & 9/4) that give as 2 real roots: -8/3 and 9/4. Next, solve the inequality f(x) < 0.

Plot the two real roots -8/3 and 9/4 on the number line. Use the origin as test point. Substitute x = 0 into the inequality. You get: -72 < 0. It is true, then the origin is located on the true segment. The solution set is the open interval: (-8/3 , 9/4). The 2 end points are not included in the solution set.

Example 2. Solve f(x) = 45x^2 + 74x - 55 > 0.

Solution. Use the Diagonal Sum Method to solve f(x) = 0. It gives 2 real roots:

-11/5 and 5/9. Plot these 2 values on the number-line. Substitute x = 0 into the inequality f(x) > 0. You get: -55 > 0. It is not true!. Then, the origin is on the "wrong" segment. The solution set are the two rays. The solution set are: (-infinity , -11/5) or (5/9 , +infinity). The 2 end points are not included.