Introduction We have looked at a variety of methods to solve quadratic equations. The solutions to quadratic equations are points. Quadratic inequalities can be written in the form ax 2 + bx + c < 0, ax 2 + bx + c ≤ 0, ax 2 + bx + c > 0, or ax 2 + bx + c ≥ 0. The solutions to quadratic inequalities are written as intervals. An interval is the set of all real numbers between two given numbers. The two numbers on the ends are the endpoints. The endpoints might or might not be included in the interval depending on whether the interval is open, closed, or half-open : Solving Quadratic Inequalities

Key Concepts If the right side of an inequality is 0, we can use logic to determine the sign possibilities of the factors. Another method of solving quadratic inequalities is to set the right side equal to 0 if it isn’t already, replace the inequality sign with an equal sign, and solve the equation. Recall that multiplying or dividing both sides of an inequality by a negative number requires reversing the inequality symbol : Solving Quadratic Inequalities

Key Concepts, continued Use these solutions to create regions on a number line and test points in each region to solve the inequality. The solutions to a quadratic inequality can be one interval or two intervals. If the quadratic equation has only complex solutions, the expression is either always positive or always negative. In these cases, the inequality will have no solution or infinitely many solutions : Solving Quadratic Inequalities

Key Concepts, continued Solutions of quadratic inequalities are often graphed on number lines. Recall that the solutions to an inequality graphed on a number line are represented by a darkened line. The endpoints of the solution interval are represented by either an open dot or a closed dot. Graph the endpoints as an open dot if the original inequality symbol is. Graph endpoints as a closed dot if the original inequality symbol is ≤ or ≥ : Solving Quadratic Inequalities

Common Errors/Misconceptions writing inequalities with signs facing in opposite directions (for example, 3 5) forgetting that answers should be inequalities forgetting to reverse the inequality signs when multiplying or dividing by a negative number using instead of ≤ or ≥, and vice versa improperly representing the sections of the number line : Solving Quadratic Inequalities

Guided Practice Example 1 For what values of x is (x – 2)(x + 10) > 0? : Solving Quadratic Inequalities

Guided Practice: Example 1, continued 1.Determine the sign possibilities for each factor. The expression will be positive when both factors are positive or both factors are negative : Solving Quadratic Inequalities

Guided Practice: Example 1, continued 2.Determine when both factors are positive. x – 2 is positive when x > 2. x + 10 is positive when x > –10. Both factors are positive when x > 2 and x > –10, or when x > : Solving Quadratic Inequalities

Guided Practice: Example 1, continued 3.Determine when both factors are negative. x – 2 is negative when x < 2. x + 10 is negative when x < –10. Both factors are negative when x < 2 and x < –10, or when x < –10. (x – 2)(x + 10) > 0 when x > 2 or x < – : Solving Quadratic Inequalities ✔

Guided Practice: Example 1, continued : Solving Quadratic Inequalities

Guided Practice Example 2 Solve x 2 + 8x + 7 ≤ 0. Graph the solutions on a number line : Solving Quadratic Inequalities

Guided Practice: Example 2, continued 1.Replace the inequality sign with an equal sign. x 2 + 8x + 7 = : Solving Quadratic Inequalities

Guided Practice: Example 2, continued 2.Solve the equation. The equation factors easily. Find two numbers whose product is 7 and whose sum is 8. The numbers are 1 and 7 because (1)(7) = 7 and = 8. Use these numbers to find the factors. The factors are (x + 1) and (x + 7) : Solving Quadratic Inequalities

Guided Practice: Example 2, continued Write the expression as the product of factors. (x + 1)(x + 7) = 0 Use the Zero Product Property to solve. x + 1 = 0 x + 7 = 0 x = –1 x = – : Solving Quadratic Inequalities

Guided Practice: Example 2, continued 3.Use a number line and test points to solve the inequality. The sign of the original inequality was ≤. Therefore, we are looking for solutions that are ≤ 0. Plot the solutions to the equation on a number line. This divides the number line into three regions: numbers less than –7 and –1; numbers in between –7 and –1; and numbers greater than –7 and – : Solving Quadratic Inequalities

Guided Practice: Example 2, continued Test one point from each region of our number line. Pick any point from the region that is easy to work with. Test x = –10 from Region 1: (–10) 2 + 8(–10) + 7 = 27 Set up an inequality that compares the result of your test x-value with 0. Refer back to the sign used in the original inequality— in this case, ≤. 27 ≤ 0 is not true. Therefore, points in Region 1 (x < –7) are not solutions : Solving Quadratic Inequalities

Guided Practice: Example 2, continued Test x = –2 from Region 2: (–2) 2 + 8(–2) + 7 = 4 – = –5 Set up an inequality that compares the result of your test x-value with 0. –5 ≤ 0 is true. Therefore, points in Region 2 (–7 ≤ x ≤ –1) are solutions : Solving Quadratic Inequalities

Guided Practice: Example 2, continued Test x = 0 from Region 3: (0) 2 + 8(0) + 7 = 7 Set up an inequality that compares the result of your test x-value with 0. 7 ≤ 0 is not true. Therefore, points in Region 3 (x > –1) are not solutions : Solving Quadratic Inequalities

Guided Practice: Example 2, continued From the test points, we can see that the expression is less than or equal to 0 when x is between –7 and – : Solving Quadratic Inequalities

Guided Practice: Example 2, continued 4.Solve the inequality and graph the solutions on a number line. x 2 + 8x + 7 ≤ 0 when –7 ≤ x ≤ – : Solving Quadratic Inequalities ✔

Guided Practice: Example 2, continued : Solving Quadratic Inequalities