Lesson Solving quadratic inequalities

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GENERALITIES.

The standard form of a quadratic inequality in one variable 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 all the values of x that make the inequality true.

All these values of x constitute the solution set of the inequality. Solution sets of quadratic inequalities are expressed in terms of intervals.

Examples of quadratic inequalities:

3x^2 - 7x + 4 > 0  ;  8x^2 - 9x - 17 < 0
(3x - 5)(7x -11) > 0  ;  (x^2 - 5x + 4x)/(x^2 - 3x - 4) > 0

STEPS IN SOLVING QUADRATIC INEQUALITIES. There are 4 steps.

STEP 1. Transform the given inequality into standard form: f(x) = ax^2 + bx + c < 0 (or > 0).

Example: Solve (2x - 3)(x + 4) > -5.
In Step 1, transform it into standard form: f(x) = 2x^2 + 5x - 7 < 0

Example: Solve x(3x - 7) > 5.
In step 1, transform it into standard form f(x) = 3x^2 - 7x - 5 > 0.

STEP 2. Solve the quadratic equation f(x) = 0 to get the 2 real roots x1 and x2. You can use any of 
the 4 methods (factoring, completing the square, quadratic formula, and graphing) or the new Diagonal Sum Method (Amazon e-book 2010). 
Before proceeding solving, make sure that the quadratic equation has 2 real roots. How? Find out if the discriminant D = b^2 - 4ac is positive (> 0). There is no need to calculate the exact value of D. Just use mental math to make sure that D > 0. If D < 0, there are no real roots, refer to Remark 2 below for answers.
Also, find out if the quadratic equation can be factored? How? First, use the Diagonal Sum Method to solve the equation. It usually requires fewer that 3 trials. If this method fails to get the answer, then we can conclude that the equation can not be factored, and the quadratic formula must be used for solving.

STEP 3. Solve the quadratic inequality f(x) < 0 (or > 0), based on the 2 real roots obtained from Step 2. You can use one of these 3 methods to solve quadratic inequalities:

1. Using the number line and test points method
2. Using the algebraic method.
3. Using the graphing method.

STEP 4. Express the answers, or solution set, in terms of intervals. You must master on how to write interval symbols. For examples:

(a, b): open interval between a and b; the 2 end points are not included in the solution set.
[a, b]: closed interval; the end points a and b are included in the solution set.
(-infinity, b]: half closed interval: the end point b is included in the solution set.

METHOD FOR SOLVING QUADRATIC INEQUALITIES.

1. THE NUMBER LINE AND TEST POINT METHOD.

The 2 real roots x1 and x2, obtained from Step 2, are plotted on the number line. They divide the number line into one segment and 2 rays. Always use the origin O as test point. Substitute x = 0 into the quadratic inequality in standard form. If it is true, then the origin is located on the true segment (or the true ray). If one ray is a part of the solutions set, then the other ray is also a part of the solution set, due to the symmetrical property of the parabola graph.

Example 1. Solve: 5x^2 - 34x < 7.
Solution. First step, transform the inequality into standard form: f(x) = 5x^2 - 34z - 7 < 0
Second step, solve f(x) = 0. Use the Diagonal Sum Method to solve it. Roots have opposite signs.
There is unique probable root-pair: (-1/5, 7/1). Its diagonal sum is 35 - 1 = 34 = -b. The 2 real roots are -1/5 and 7.
In third step, plot the 2 real roots -1/5 and 7 on the number line. Substitute x = 0 into the inequality f(x) < 0. It gives: -7 < 0. It is true, then the origin O is located on the true segment: -1/5 - 7. 
In Step 4: the solution set is the open interval (-1/5 , 7).

Example 2. Solve -7x^2 + 30x < 8.
Solution. First step, transform the inequality to standard form f(x) = -7x^2 + 30x - 8 < 0.
In second step, solve f(x) = 0 by using the Diagonal Sum Method. Both roots are positive. There are 3 probable root-pairs: (1/7, 8/1)(2/1, 4/7)(2/7, 4/1). The diagonal sum of the third pair is 28 + 2 = 30 = b. According to the Diagonal Sum Rule, the 2 real roots are 2/7 and 4.
In Step 3, solve f(x) < 0 by plotting the 2 real roots 2/7 and 4 on the number line. Substitute x = 0 into the inequality. It shows: -8 < 0. It is true, then the origin O is located on a true ray. By symmetry, the other ray also belongs to the solution set.
In step 4, express the solution set in the form of intervals: (-infinity, 2/7) and (4, +infinity).

NOTE 1. During tests/exams, time is limited. Students must find the answers as fast as possible. Using the number line and taking the origin O as test point is enough. It is useless time to do more test points or to discuss the sign status of the 2 binomials through factoring. Taking 3 test points for 3 areas is effective for first time lesson learning, however it becomes useless when students have mastered the process.

NOTE 2. By this method, you may use a double-number line or even a triple number line, to solve a system of 2 (or 3) quadratic inequalities in one variable. Each inequality' solution set is separately plotted on its own number line. Through superimposing, the combined solution set of the system can be easily seen.

2. THE ALGEBRAIC METHOD.

This is an European method. It bases on a theorem about the sign status of the trinomial f(x) when x varies along the x-axis. Students study once the theorem' development, then apply it to solve various quadratic inequalities. 
This method may be faster than the number line method since it doesn't require drawing the number line each time. In addition, it helps to set up a Sign Table in solving, by algebraic approach, a system of quadratic inequalities in one variable.

THEOREM ON THE SIGN STATUS OF A TRINOMIAL f(x)= ax^2 + bx + c.

"Between the 2 real roots x1 and x2, f(x) has the opposite sign of the constant a".

This means, between x1 and x2, f(x) > 0 if a is negative (-); and f(x) < 0 if a is positive.

Example 3: The trinomial f(x) = - 6x^2 + 9x - 3 is positive (> 0), opposite to the sign of a = -6, between the 2 real roots 1/2 and 1.

Example 4: The trinomial f(x) = 3x^2 - 4x - 7 is negative (< 0), opposite to the sign of a = 3, between the 2 real roots -1 and 7.

NOTE. You can easily understand this theorem by relating it to the parabola graph of the quadratic function f(x). 
If a is positive (+), the parabola is upward. Between the 2 x-intercepts (real roots), a part of the parabola is below the x-axis, meaning f(x) is negative (< 0) in this interval, opposite to the sign of a.
If a is negative (-), the parabola is downward. Between the 2 x-intercepts (real roots), a part of the parabola is above the x-axis, meaning f(x) is positive (> 0) in this interval, opposite in sign to the constant a.

3. THE GRAPHING METHOD.

You can solve a quadratic inequality f(x) < 0 (or > 0) by graphing the quadratic function f(x). When the parabola graph of f(x) is above the x-axis, f(x) is positive. When the parabola is above the x-axis, f(x) is negative.

You don't have to accurately graph the parabola. Bases on the 2 real roots obtained from solving the equation f(x) = 0, you may just roughly sketch the parabola. Pay attention to if the parabola is upward (a positive) or downward (a negative).
By this method, you may solve a system of 2 (or 3) quadratic inequalities by graphing the two (or 3) parabolas on the same coordinate grid, if possible.

IMPORTANT REMARKS.

REMARK 1. When the inequality has an additional equal sign (greater, lesser, or equal to), the end points are automatically included into the solution set.

REMARK 2. If the Discriminant D = b^2 - 4ac is negative (D < 0), the equation f(x)= 0 doesn't have real roots, f(x) is always positive, or always negative, depending on the sign of constant a.
If a is negative, f(x) is always negative (< 0), and vice versa.

Example 5. Solve: f(x) = 15x^2 - 8x + 7 > 0.
Solution. The Discriminant D = 64 - 460 < 0. There are no real roots. Since a is positive, f(x) is always positive (> 0), regardless of the values of x. The inequality is always true.

Example 6. Solve: f(x) = -4x^2 + 9x - 7 > 0.
Solution. D = 81 - 112 < 0. There are no real roots. Since a is negative (a = -4), f(x) is always negative regardless of the values of x. The inequality is always not true.

EXAMPLES OF SOLVING QUADRATIC INEQUALITIES.

Example 7. Solve: f(x) = 5x^2 + 12x - 9 > 0.
Solution. Use the Diagonal Sum Method to solve f(x) = 0. Roots have opposite signs. There are 2 probable root-pairs: (-1/5, 9/1),(-3/1, 3/5). The diagonal sum of the second pair is: -15 + 3 = -12 = -b. The 2 real roots are -3 and 3/5. Use the origin O as test point to solve f(x) > 0. Substitute x = o into the inequality. It shows: -9 > 0. It is not true, then the origin O is not located on the solution set. The solution set should be the 2 rays as expressed by the 2 open intervals (-infinity, -3) and (3/5, +infinity).

Example 8. Solve: f(x) = 8x^2 - 42x - 11 < (or =) 0. 
Solution. Use the Diagonal Sum Method to solve f(x) = 0. Roots have opposite signs. There are 3 probable root-pairs:(-1/8, 11/1),(-1/2, 11/4),(-1/4, 11/2). The diagonal sum of the third pair is: 44 - 2 = 42 = -b. The 2 real roots are -1/4 and 11/2. Use the origin (x = 0) as text point. It shows: -11 < 0. It is true, then the origin O is located on the true segment -1/4 - 11/2. The solution set is the closed interval [-1/4, 11/2]. Both end points (critical points) are included into the solution set.

Example 9. Solve: f(x) = -10x^2 + 31x - 13 < 0.
Solution. Use the Diagonal Sum Method to solve f(x) = 0. Both real roots are positive. There are 3 probable root-pairs: (1/10, 13/1),(1/2, 13/5),(1/5, 13/2). The diagonal sum of the second pair is : 26 + 5 = 31 = b. Since a is negative (-), the 2 real roots are 1/2 and 13/5. Use the Theorem to solve f(x) < 0. Between the 2 real roots 1/2 and 13/5, f(x) is positive, as opposite in sign to a = -10. The solution set should be the 2 rays where f(x) is negative (< 0). The answer is the 2 open intervals (-infinity, 1/2) and (13/5, +infinity). There is no need to draw the number line.

Example 10. Solve f(x) = 11x^2 - 76x - 7 < 0.
Solution. Solve f(x) = 0. Roots have opposite signs. There are unique probable root-pair: (-1/11, 7/1). Its diagonal sum is: 77 - 1 = 76 = -b. The 2 real roots are -1/11 and 7. Use the Theorem to solve f(x) < 0. Between the 2 real roots, f(x) is negative (< 0) as opposite in sign to constant a. The solution set is the open interval (-1/11, 7). There is no need to draw the number line!