There are two primary ways to solve quadratic inequalites:
(1) Graph the function and see on what intervals the graph lies above or below the x-axis.
(2) Find the zeros of the function. These divide the domain into at most 3 regions -- check an x-value from each region to determine the sign of y. (If the function is always above or below the x-axis, then there are no zeros. If there is exactly 1 zero then the rest of the function is above (or below) the x-axis.)
Solving quadratic inequalities. There are 3 methods and 4 steps.
Step 1. Bring the inequality to standard form f(x) > 0 (or < 0). Example: The equation x(2x - 5) > -3 will be transformed into f(x) = 2x^2 - 5x + 3 > 0. Step 2. Solve the equation f(x) = 0. Students can use any of the 4 methods (factoring, quadratic formula, completing square, graphing) or the new Diagonal Sum Method (factoring shorcut). Step 3. Use the 2 real roots from step 2 to solve the inequality. There are 3 methods for solving, described below. Step 4. Express the solution set in the form of intervals. Example: open interval (-2, 5), closed interval [-3, 7] when -3 and 7 are included, or half closed interval [4, +infitity)
Three methods for solving quadratic inequalities.
A. The number line and test point method. Plot the 2 real roots (from step 2) on the number line. They divide the line into one segment and 2 rays. Select the origin as test point. Replace x = 0 into the inequality. If it is true then the origin is located on the true segment (or true ray. By symetry, the other ray also belongs to the solution set). Example. Solve 5x^2 - 7x < 12. Step 1. Trans form to standard form: 5x^2 - 7x - 12 < 0. Step 2. Solve f(x) = 0. The 2 real roots are -1 and 12/5. Step 3. Use the number line and test point method. Replace x = 0 into the inequality. We have -12 < 0. It is true, then the orgin is located on the true segment -1 and 12/5. Step 4. The solution set is the open interval (-1, 12/5).
B. The algebraic method. This world wide method is based on a theorem. Students study its development once, then use it to solve various quadratic inequalities. Theorem:"Between the 2 real roots, the trinomial f(x) has opposite sign to the sign of a". This means f(x) is positive between the 2 real roots when a is negative. Example. Solve -9x^2 - 13x + 22 > 0. The 2 real roots are -1 and 22/9. Use the theorem to solve f(x) > 0. Since a is negative, f(x) is positive (f(x) > 0) between the 2 real roots 1 and 22/9. The solution set is the open interval (1, 22/9). Note: To understand this theorem, just look at the parabola graph of f(x). When a is +, the parabola is upward. Between the 2 x-intercepts, a part of the parabola is below the x-axis, meaning f(x) is negative between the 2 real roots, opposite in sign to a that has the +sign.
C The graphing method. To know how, please read book titled:"New methods for solving quadratic equations and inequalities" (Amazon e-book 2010) or the articles about solving quadratic inequalities on wikihow.com and bukisa.com