What are the methods and steps to solve quadratic inequalities?

Answer 1

The standard form of a quadratic inequality is : ax^2 + bx + c > 0 (or < 0) . If the inequality is given in other form, transform it into the standard form. Solving for x means finding the values of x that make the inequality true. All these values of x make the solution set of the inequality. Solution sets of quadratic inequalities are expressed in terms of intervals.

Examples of quadratic inequalities:

3x^2 + 7x - 10 > 0 ; (2x - 7)(3x + 5) < 0 ; (x^2 - 3x - 4)/(x^2 -5x -6) < 0

Examples of solution sets in terms of intervals:

(-3 , 4) ; [-2, 7] ; (-infinity , 3] ; [4 , +infinity)

There are 3 methods to solve quadratic inequalities:

1. By using the number-line and test point.

2. By the Algebraic method

3. By graphing

There are 4 steps in solving quadratic inequalities.

Step 1. Transform the inequality into a trinomial f(x) on the left side and leave 0 on the right side: f(x) > 0 or f(x) < 0

Step 2. Solve the quadratic equation f(x) = 0.

There are 4 existing common methods (quadratic formula, factoring, completing the squares, graphing) and the new Diagonal Sum method (factoring shortcut) to choose. I advise you to use first the Diagonal Sum Method that can directly give the 2 real roots without having to factor the equation. (Amazon e-book 2010). If this method fails to get answers, then the quadratic formula must be used.

Step 3. Solve the quadratic inequality, based on the values of the 2 real roots, obtained from Step 2.. There are two methods for solving that will be described later.

Step 4. Express the answer (solution set) in the form of intervals. You must know how to write the correct interval symbols:

(a, b): open interval between a and b, the end points a and b are not included.

[a, b]: closed interval; the end points a and b are included in the solution set.

[a , +infinity); half closed interval; the end point a is included.

Methods for solving quadratic inequalities

There are 2 common methods to select, besides the graphing method that will be described later.

1. The number line and test-point method.

The 2 real roots obtained from Step 2 are plotted on the number line. They divide it into one segment and 2 rays.

Always select the origin as test point. Substitute x = 0 into the inequality.

If It is true, then the origin is located on the true segment (or the true ray). If one ray is the solution set, then the other ray also belongs to it by symmetric property.

If it is not true, then the origin O is not located on the true segment (or the true ray).

2. The algebraic method.

This is an popular European method. It bases on a theorem showing the sign status of f(x) when x varies along the x-axis and passes by the the 2 real roots. Students study once the Theorem and they apply it for solving all quadratic inequalities.

Theorem: Between the 2 real roots, f(x) has the opposite sign of constant a.

It means: Between the 2 real roots, f(x) > 0 if a is negative.

Between the 2 real roots, f(x) < 0 if a is positive.

NOTE 1. You can easily understand this theorem by relating it to the parabola graph of the quadratic function f(x).

If the constant a is positive, the parabola is upward.Between the 2 real roots (x-intercepts), a part of the parabola is below the x-axis. It means f(x) is negative in this interval, opposite in sign to the constant a.

If a is negative, the parabola is downward. Between the 2 x-intercepts (real roots), apart of the parabola is abovethe x-axis, meaning f(x) is positive in this interval, opposite in sign to a.

NOTE 2: If the equation doesn't have real roots (when Discriminant D < 0), f(x) is always positive (or always negative) depending on the sign of a.

If a is negative (< 0), f(x) is always negative (<0) regardless of the values of x.

If a is positive (>9), f(x) is always positive regardless of the values of x

NOTE 3: When the inequality has an additional equal sign (greater or equal to) the end-points are included in the solution set.

NOTE 4. During tests/exams time is limited. Students must find the answers as fast as possible.

When using the test point method, always take the origin as test point, except when zero is a real root. It is useless to take 3 test points for 3 areas of the number line. Never discuss the sign status of the 2 binomials through factoring since it easily leads to errors.

For multiple-choice answers, the algebraic method is the fastest and convenient because its doesn't require to draw the number line each time.

Examples of solving quadratic inequalities.

Example 1: Solve: 15 > 6x^2 + 43x.

Solution: In step 1, transform the inequality into: - 6x^2 - 43x + 15 > 0.

Step 2. Use the Diagonal Sum Method to solve f(x) = 0.

Rule of sign indicates that the roots have opposite signs. a is negative.

Write down the probable roots-pairs: (-3/2, 5/3) (-1/2, 15/3) (-1/3, 15/2).

The diagonal sum of third set is 43. The 2 real roots are (1/3) and (-15/2).

Step 3. Select the algebraic method to solve the inequality f(x) > 0.

According to the theorem, f(x) is positive between the 2 roots (-15/2) and (1/3) as opposite in sign to the constant a = -6.

Step 4. The solution set is the open interval (-15/2, 1/3).

The end-points are not included.

Example 2: Solve x(6x +1) < 15.

Solution. Step 1. Transform the inequality into: f(x) = 6x^2 + x - 15 < 0.

Step 2. Use the Diagonal Sum Method to solve f(x) = 0.

Rule of sign says : 2 roots have opposite signs.

Write down the probable root-pairs: (-3/2, 5/3) and (-3/3, 5/2)

Diagonal sum of the first set: 10 - 9 = 1 = b.

The answer is the opposite set; the 2 real roots are (3/2) and (-5/3).

Step 3. Select the algebraic method to solve f(x) < 0.

Between the 2 roots f(x) is negative as having the opposite sign of a = 6.

Step 4. The solution set is the open interval (-5/3, 3/2).

Example 3. Solve 15x^2 - 8x + 7 > 0.

Solution. The Discriminant D = 64 - 420 < 0 . There are no real roots. Refer to NOTE 2.

Since a is positive, f(x) is always positive regardless of values of x. The inequality f(x) is always true.

Example 4. Solve 9x^2 + 1 < 12x.

Solution. Step 1. Transform into 9x^2 - 12x + 1 < 0.

Step 2. The Diagonal Sum fails to solve it. The quadratic formula must be used.

The 2 real roots are x1 = 0.09 and x2 = 1.24.

Step 3. Solve f(x) < 0 by the Theorem. Between the 2 real roots, f(x) is negative (< 0) as opposite in sign to a = 9.

Step 4. The solution set is the open interval (0.09 , 1.24).

The graphing method.

The concept is simple. When the graph of f(x) is above the x-axis, the trinomial f(x) is positive. When the graph is below the x-axis , f(x) is negative. You don't need to accurately graph the parabola. Based on the 2 real roots obtained from Step 2, you may roughly draw a sketch of the parabola. Pay close attention to if it is upward or downward.

By this graphing method, you can solve a system of 2 (or 3) quadratic inequalities by graphing 2 (or 3) parabolas on the same coordinate grid.

With the help of graphing calculators, this innovative graphing approach can help solving various complex inequalities and systems of complex inequalities in one variable. Complex inequalities in one variable may include cubic, bi-quadratic, rational, radical, exponential and trigonometric inequalities. See last chapter of book titled "New methods for solving quadratic equations and inequalities" (Amazon e-book 2010).