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**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

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**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

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**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.

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**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.

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**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

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**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

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**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. ∴

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**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

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**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

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