Solving Nonlinear Inequalities Digital Lesson Solving Nonlinear Inequalities

The symbols , , and  may also be used. A quadratic inequality in one variable is an inequality which can be written in the form ax2 + bx + c > 0 (a  0) for a, b, c real numbers. The symbols , , and  may also be used. Example: x2 – 3x + 7  0 is a quadratic inequality since it can be written 1x2 + (– 3)x + 7  0. Example: 3x2 < x + 5 is a quadratic inequality since it can be written 3x2 + (–1)x + (– 5) < 0. Example: x2 + 3x  x2 + 4 is not a quadratic inequality since it is equivalent to 3x  4  0. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Quadratic Inequality

Example: Which of the values of x are solutions of x2 + 3x  4  0 ? 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 x2 + 3x  4  0 ? x x2 + 3x – 4 x2 + 3x – 4  0 Solution?  1 (1)2 + 3(1) – 4  6  0 true yes (0)2 + 3(0) – 4  4  0 true yes 0.5 (0.5)2 + 3(0.5) – 4  2.25  0 true yes 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. Solution

[ ] The solution set of an inequality is the set of all solutions. Study the graph of the solution set of x2 + 3x  4  0. [ ] -2 -1 1 2 -6 -5 - 4 -3 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. x2 + 3x  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. Solution Set

Solving a Quadratic Inequality 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. Solving a Quadratic Inequality

Example: Solve and Graph the Solution Set Example: Solve and graph the solution set of x2  6x + 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 to zero. x = 1 x = 5 Draw vertical lines indicating the numbers where each factor equals zero. Product is positive. Product is negative. Product is positive. x – 1 x – 5 – – – + + + + + + + + For each region, identify if each factor is positive or negative. – – – – – – – – + + + ( ) Factors 3 4 5 6 7 -1 1 2 Draw the solution set. Rounded parentheses indicate a strict inequality. {x | 1 < x < 5} Solution set in set-builder notation. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Solve and Graph the Solution Set

Example: Solve and Graph the Solution Set Example: Solve and graph the solution set of x2  x  6. x2  x  6  0 Rewrite the inequality so that zero appears on the right. The product of the factors is positive. (x + 2)(x  3)  0 Factor. Numbers where each factor equals zero. x =  2, 3 x + 2 x – 3 – – – – + + + + + + + + + Draw vertical lines where each factor equals zero. – – – – – – – – – – – + + ] [ Indicate positive and negative regions for each factor. 1 2 3 4 - 4 -3 -2 -1 Square brackets are used since the inequality is . Draw solution set. {x | x   2 or x  3} Solution set in set-builder notation. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Solve and Graph the Solution Set

Example: Solve a Cubic Inequality Cubic inequalities can be solved similarly. Example: Solve and graph the solution set of x3 + x2  9x  9 > 0. x2(x + 1)  9(x + 1) > 0 Factor by grouping. (x2  9)(x + 1) > 0 (x + 3)(x  3)(x + 1) > 0 x =  3, +3, 1 Numbers where each factor equals zero. x + 3 x + 1 x – 3 – – – – – – – + + Draw three vertical lines. + + – – + + + + + Indicate positive and negative regions for each of the three factors. – – + + + + + + + ( ) ( 1 2 3 4 - 4 -3 -2 -1 {x | 3 < x < 1 or x > 3} Solution set Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Solve a Cubic Inequality

Example: Solve an Inequality Involving a Rational Function Inequalities involving rational functions can be solved similarly. Example: Solve and graph the solution set of . (x + 1) = 0 (x  2) = 0 Find the numbers for which each factor equals zero. x =  1 x = 2 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 – – – – – + + + + + + + There are two regions where the quotient of the two factors is positive. – – – – – – – – – + + + ] ( 1 2 3 4 - 4 -3 -2 -1 {x | x   1 or x > 2} Solution set Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Solve an Inequality Involving a Rational Function

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

Example: One leg of a right triangle is 2 inches longer than the other 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  4 x + 2 = other leg x x + 2 Area of triangle Solve: x + 4 x – 2 – + + + + + + + + + + + – – – – – – – – – + + + ] [ 1 2 3 4 - 4 -3 -2 -1 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. Example: Word Problem