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Number Systems and Inequalities Quadratic Inequalities Aim To demonstrate how to solve quadratic inequalities. Learning Outcomes At the end of this section you will be able to: • Distinguish between quadratic equations and inequalities, • Solve quadratic inequalities. Theinequalities that we have dealt with so far are called “linear” inequalities because they only deal with linear expressions like x − 2 > 0. An inequality where x2 is the 2 highest-degree term is called a quadratic inequality. In general, ax +bx+c = 0 is a 2 quadratic equation, while ax +bx+c ≤ 0 is an example of a quadratic inequality. Solving a quadratic inequality means that we are looking for the range of values of x where the inequality is greater than or less than zero. The first step is to factorise the quadratic inequality as if it was just a quadratic equation. Example 1 Find the set of values of x for which 1 + 2x − 3x2 > 0. Factorise the quadratic, 1+2x−3x2>0 ⇒ 3x2−2x−1<0, ⇒ (3x+1)(x−1)<0. What this result tells us is that the product of two values when multiplied together is negative. This can only occur when one of the values is negative and the other one positive. There are two possibilities that need to be looked at: 1. (3x+1) < 0 and (x−1) > 0, or 2. (3x+1) > 0 and (x−1) < 0. 1 Number Systems and Inequalities Looking at case 1 we have (3x+1)<0 and (x−1)>0, ⇒x<−1 and x>1. 3 and looking at case 2 reveals that (3x+1)>0 and (x−1)<0, ⇒x>−1 and x<1. 3 It can be seen by a quick inspection that the results from case 1 do not make any mathematical sense as there are no values of x that are less than −1 and at the same 3 time greater than 1. On the other hand, the conditions from case 2 are that x is greater than −1 and at the same time less than 1 which is possible. Therefore the final answer 3 is x>−1 and x<1. 3 This tells us the value of x where the graph is below the x axis (y < 0). y = 3x2 −2x−1 ✁✂✄ 2 ✁ Figure 1: The graph of y = 3x −2x−1 Example 2 Find the values of x for which 2x2 + 5x − 3 > 0. 2 2x +5x−3>0⇒(2x−1)(x+3)>0. In this case the product of (2x − 1) and (x + 3) must be greater than zero so either both of them must be positive or else both of them must be negative. Therefore 1. (2x−1) > 0 and (x+3) > 0 or 2. (2x−1) < 0 and (x+3) < 0. 2 Number Systems and Inequalities Looking at case 1: (2x−1)>0 and (x+3)>0, ⇒x>1 and x>−3. 2 and looking at case 2 reveals that (2x−1)<0 and (x+3)<0, ⇒x<1 and x<−3. 2 It is clear to see that in case 1 the only values of x which satisfy both conditions occur when x > 1. Similarly for case 2, the only values of x which satisfy both of these 2 conditions is x < −3. Therefore the final answer is x<−3 or x>1. 2 This tells us the value of x where the graph is above the x axis (y > 0). y = 2x2 +5x−3 ✁ ✂ 2 ✄☎ Figure 2: The graph of y = 2x +5x−3 Related Reading Morris, O.D. 1987. Text & Tests 1. The Celtic Press. 3
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