12. Inequalities and Linear Programming (a) How to solve the quadratic inequalities in one unknown by using graphical method? (i) a > 0, solve ax2+bx+c > 0. Using the graph of y = ax2+bx+c, find the range of values of x by reading the points lying above the x-axis. y x y = ax2+bx+c The x-intercepts of the quadratic graph y = ax2+bx+c are  and β. For y > 0, or x <  x >β the solution of the inequality is above the x-axis, i.e. the blue curve.  O β

12. Inequalities and Linear Programming (a) How to solve the quadratic inequalities in one unknown by using graphical method? (i) a > 0, solve ax2+bx+c < 0. Using the graph of y = ax2+bx+c, find the range of values of x by reading the points lying below the x-axis. y x y = ax2+bx+c The x-intercepts of the quadratic graph y = ax2+bx+c are  and β. For y < 0, the solution of the inequality is below the x-axis, i.e. the orange curve.  O β  < x <β

12. Inequalities and Linear Programming (a) How to solve the quadratic inequalities in one unknown by using graphical method? (ii) a < 0, solve ax2+bx+c > 0. Using the graph of y = ax2+bx+c, find the range of values of x by reading the points lying above the x-axis. ∵ The quadratic graph is vertically y x y = ax2+bx+c a < x <β inverted compared with (i), ∴ The range of values of x is O  β opposite to the case a > 0. For y > 0, the solution of the inequality is above the x-axis, i.e. the blue curve.

12. Inequalities and Linear Programming (a) How to solve the quadratic inequalities in one unknown by using graphical method? (ii) a < 0, solve ax2+bx+c < 0. Using the graph of y = ax2+bx+c, find the range of values of x by reading the points lying below the x-axis. ∵ The quadratic graph is vertically y x y = ax2+bx+c inverted compared with (i), ∴ The range of values of x is O  β opposite to the case a > 0. For y < 0, or x <  x >β the solution of the inequality is below the x-axis, i.e. the orange curve.

12. Inequalities and Linear Programming (b) For algebraic method, how to use the graphs or tables to find the solutions of the quadratic inequalities in one unknown? E.g. (i) Solve the inequality x2-3x -10 > 0. Consider x2-3x -10 = 0, y x y = x2-3x+10 O 5 -2 (x+2)(x-5) = 0 x = -2 or x = 5 x < -2 x > 5 Graphical ∵ a = 1, i.e. a > 0 ∴ According to the above result, we can sketch the graph. The graph above the x-axis stand for y > 0. ∴ or

12. Inequalities and Linear Programming (b) For algebraic method, how to use the graphs or tables to find the solutions of the quadratic inequalities in one unknown? E.g. (i) Solve the inequality x2-3x -10 > 0. Consider x2-3x -10 = 0, (x+2)(x-5) = 0 x = -2 or x = 5 Tabular x < -2 -2 < x < 5 x > 5 x+2 x-5 (x+2)(x-5) -2<x<5 -2+2<x+2<5+2 0<x+2<7 x+2>0 -2<x<5 -2-5<x-5<5-5 -7<x-5<0 x-5<0 x>5 x+2>5+2 x+2>7 x+2>0 x<-2 x-5<-2-5 x-5<0 x<-2 x+2<-2+2 x+2<0 - + + x>5 x-5>5-5 x-5>0 x+2>0 x-5>0 (x+2)(x-5)>0 x+2>0 x-5<0 (x+2)(x-5)<0 x+2<0 x-5<0 (x+2)(x-5)>0 - - + + - + ∴ x < -2 or x > 5

12. Inequalities and Linear Programming (b) For algebraic method, how to use the graphs or tables to find the solutions of the quadratic inequalities in one unknown? E.g. (ii) Solve the inequality x2+5x-6 < 0. Consider x2+5x-6 = 0, y x y = x2+5x-6 1 -6 O (x+6)(x-1) = 0 x = -6 or x = 1 Graphical ∵ a = 1, i.e. a > 0 ∴ According to the above result, we can sketch the graph. -6 < x < 1 The graph below the x-axis stand for y < 0. ∴

12. Inequalities and Linear Programming (b) For algebraic method, how to use the graphs or tables to find the solutions of the quadratic inequalities in one unknown? E.g. (ii) Solve the inequality x2+5x-6 < 0. Consider x2+5x-6 = 0, (x+6)(x-1) = 0 x = -6 or x = 1 Tabular x < -6 -6 < x < 1 x > 1 x+6 x-1 (x+6)(x-1) -6<x<1 -6+6<x+6<1+6 0<x+6<7 x+6>0 -6<x<1 -6-1<x-1<1-1 -7<x-1<0 x-1<0 x<-1 x-1<-6-1 x-1<-7 x-1<0 x>1 x+6>1+6 x+6>7 x+6>0 x<-6 x+6<-6+6 x+6<0 - + + x>1 x-1>1-1 x-1>0 x+6>0 x-1<0 (x+6)(x-1)<0 x+6<0 x-1<0 (x+6)(x-1)>0 x+6>0 x-1>0 (x+6)(x-1)>0 - - + + - + ∴ -6 < x < 1

12. Inequalities and Linear Programming Solving Quadratic Inequalities Easy Memory Tips： ax2+bx+c > 0 a > 0 When the quadratic function and a (the coefficient of ax2+bx+c < 0 a < 0 x2) are both larger than zero or smaller than zero, the solution of the inequality is x <  or x >β. ax2+bx+c > 0 a < 0 Otherwise, the solution of the inequality is  < x <β. ax2+bx+c < 0 a > 0