Digital Lesson Solving Nonlinear Inequalities A quadratic inequality

Digital Lesson Solving Nonlinear Inequalities

A quadratic inequality in one variable is an inequality which can be written in the form ax 2 + bx + c > 0 (a 0) for a, b, c real numbers. The symbols , , and may also be used. Example: x 2 – 3 x + 7 0 is a quadratic inequality since it can be written 1 x 2 + (– 3)x + 7 0. Example: 3 x 2 < x + 5 is a quadratic inequality since it can be written 3 x 2 + (– 1)x + (– 5) < 0. Example: x 2 + 3 x x 2 + 4 is not a quadratic inequality since it is equivalent to 3 x 4 0. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2

A solution of a quadratic inequality in one variable is a number which, when substituted for the variable, results in a true inequality. Example: Which of the values of x are solutions of x 2 + 3 x 4 0 ? x 1 0 0. 5 x 2 + 3 x – 4 0 Solution? ( 1)2 + 3( 1) – 4 6 0 true yes (0)2 + 3(0) – 4 4 0 true yes 2. 25 0 true yes (0. 5)2 + 3(0. 5) – 4 1 (1)2 + 3(1) – 4 0 0 true yes 2 (2)2 + 3(2) – 4 6 0 false no 3 (3)2 + 3(3) – 4 14 0 false no Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 3

The solution set of an inequality is the set of all solutions. Study the graph of the solution set of x 2 + 3 x 4 0. [ -6 -5 - 4 -3 -2 -1 0 ] 1 2 The solution set is {x | 4 x 1}. The values of x for which equality holds are part of the solution set. These values can be found by solving the quadratic equation associated with the inequality. x 2 + 3 x 4 = 0 Solve the associated equation. (x + 4)(x 1) = 0 Factor the trinomial. x = 4 or x = 1 Solutions of the equation Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 4

To solve a quadratic inequality: 1. If necessary, rewrite the quadratic inequality so that zero appears on the right, then factor. 2. On the real number line, draw a vertical line at the numbers that make each factor equal to zero. 3. For each factor, place plus signs above the number line in the regions where the factor is positive, and minus signs where the factor is negative. 4. Observe the sign of the product of the factors for each region, to determine which regions will belong to the solution set. 5. Express the solution set using set-builder notation and a graph on a real number line. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 5

Example: Solve and graph the solution set of x 2 6 x + 5 < 0. The product of the factors is negative. (x 1)(x 5) < 0 Factor. x 1 = 0 x 5 = 0 Solve for each factor equal x = 1 x = 5 to zero. x – 1 x – 5 Factors Draw vertical lines indicating Product is the numbers where each factor positive. negative. positive. equals zero. – – – + + + – – – – -1 0 ( 1 2 3 4 ) 5 {x | 1 < x < 5} Copyright © by Houghton Mifflin Company, Inc. All rights reserved. + + + 6 7 For each region, identify if each factor is positive or negative. Draw the solution set. Rounded parentheses indicate a strict inequality. Solution set in set-builder notation. 6

Example: Solve and graph the solution set of x 2 x 6 0 Rewrite the inequality so that zero appears on the right. (x + 2)(x 3) 0 Factor. x = 2, 3 Numbers where each The product of the factors is positive. factor equals zero. x + 2 – – + + + + + x – 3 – – – + + ] - 4 -3 -2 -1 0 1 2 [ 3 4 Square brackets are used since the inequality is . {x | x 2 or x 3} Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Draw vertical lines where each factor equals zero. Indicate positive and negative regions for each factor. Draw solution set. Solution set in set-builder notation. 7

Cubic inequalities can be solved similarly. Example: Solve and graph the solution set of x 3 + x 2 9 x 9 > 0. x 2(x + 1) 9(x + 1) > 0 (x 2 9)(x + 1) > 0 Factor by grouping. (x + 3)(x + 1) > 0 x = 3, +3, 1 x – 3 – – x + 1 – – ( – – – + + + – – + + + + ) - 4 -3 -2 -1 0 1 2 ( Draw three vertical lines. Indicate positive and negative regions for each of the three factors. 3 4 {x | 3 < x < 1 or x > 3} Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Numbers where each factor equals zero. Solution set 8

Inequalities involving rational functions can be solved similarly. Example: Solve and graph the solution set of . (x + 1) = 0 (x 2) = 0 x = 1 x = 2 Find the numbers for which each factor equals zero. Note that 2 will not be part of the solution set since the expression is not defined when the denominator is zero. x + 1 – – – + + + + x – 2 – – – – – + + + ] - 4 -3 -2 -1 0 1 ( 2 3 4 {x | x 1 or x > 2} Copyright © by Houghton Mifflin Company, Inc. All rights reserved. There are two regions where the quotient of the two factors is positive. Solution set 9

Example: Solve and graph the solution set of . The quotient is negative. x + 2 = 0 x = 2 Factor. (x 1)(x + 3) = 0 x = 1, 3 Expression is undefined at these points. x + 2 – – – + + + + + x 1 – – – – + + + x + 3 – – + + + + + ) ( - 4 -3 -2 -1 0 ) 1 2 3 4 {x | x < 3 or 2 < x < 1} Solution set Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 10

Example: One leg of a right triangle is 2 inches longer than the other. How long should the shorter leg be to ensure that the area of the triangle is greater than or equal to 4? x = shorter leg x + 2 = other leg x Area of triangle 4 x + 2 Solve: x + 4 – + + + x – 2 – – – – – + + + ] - 4 -3 -2 -1 0 1 [ 2 3 4 Since length has to be positive, the answer is x 2. The shorter leg should be at least 2 inches long. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 11