We’re now going to look at inequalities formulated in terms of quadratic polynomials. In Week 2, we studied carefully the geometry of these quadratic polynomials, and as we’re now going to see, our knowledge of that geometry will enable us to easily solve these inequalities. So consider a quadratic polynomial, ax squared plus bx plus c, the leading coefficient a being different from 0. The question is, what are the solutions to inequalities like the following, ax squared plus bx plus c less than or equal 0. In this discussion, I’m going to take the a to be strictly positive, but the case in which a is negative would be very similar.
While the discriminant– remember the discriminant capital delta b squared minus 4ac is going to play again a key role in the analysis. Why? Well, because we know, for example, that if the discriminant is strictly positive, then the quadratic polynomial has two distinct roots, r1, r2. To put this geometrically, it means that the graph of our quadratic function, which is a parabola pointing upwards because a is strictly positive– this graph will intersect the x-axis for two x values, the two roots r1 and r2. Now, if we want to answer the question where is this function negative, well, clearly it’s between the two roots because those are the points on the graph that are below the x-axis.
So the solution set to our inequality will simply be the closed interval r1, r2, closed because we have a non-strict inequality being considered. To continue, then, the possible cases, if the discriminant capital delta is strictly negative, then we know there are no roots. The parabola is always strictly above the x-axis. Therefore, the set of solutions of our inequality is empty. And finally in the remaining case, when delta, by coincidence, happens to be 0, then we know there’s exactly one root of the quadratic polynomial. We know what it is. It’s the point minus b over 2a. And at that point the quadratic is less than or equal 0. In fact, equal and at no other point.
And therefore, the solution set is the singleton set consisting of that point. How would the answer change if we were considering, for example, the strict inequality? Well, in the first case, where delta is strictly positive, that would simply change the solution set from the closed interval to the open interval because we need to exclude r1 and r2 for strict inequality. In the second case, the solution set would still be empty obviously. And in the third set– third case rather, it would go from being a singleton to being the empty set because the inequality is strict. Now, let’s apply these general observations on quadratic polynomials to the following example.
We have an inequality here to study. It’s expressed on the left-hand side by the sum of two rational functions. The natural domain for this inequality would be all of the real numbers except we exclude x equals 1 and x equals minus 2 to avoid denominators being 0. Perhaps you recall that we had solved the equals case of this inequality where the inequality is replaced by an equal sign. We had done it by initially multiplying across by x minus 1, x plus 2. This clears out the denominators on the left. Then we get a quadratic equation, which we had solved with the quadratic formula, and we had found the solution set two points, minus 5 and 2.
Now, in the case of an inequality, we have to be more careful. We can’t just multiply across by x minus 1, x plus 2 because that could either be positive or negative depending, and therefore, the sense of the inequality will be affected. So we adopt another procedure. I suggest that we use a common denominator to express the left-hand side, the sum on the left. When we do that, we get a certain expression, and now we have our inequality in a slightly more compact form. A certain function must be less than or equal 3. Let’s now study the inequality in that form. We see that the numerator here of the left-hand side is always strictly positive.
So it has no effect on the sign on the left. And we see that the denominator can either be positive or negative. In fact, we see easily enough that the denominator is positive when x is greater than 1 or less than minus 2. And in the other case, when it’s between minus 2 and 1, then the denominator is negative. So that’s two natural cases that arise. We’ll call the first one, logically enough, Case 1 and the second one Case 2. We have to look at these cases separately, find a number of solutions to our inequality in each case, and then lump them all together to get the final solution set.
Now, in Case 1, when the denominator is positive, we can multiply across by that denominator, and our inequality is not reversed and it becomes a quadratic inequality, which we now know how to solve from our previous discussion. And we find that the solution set is the union of two intervals, minus infinity up to the first root minus 5, union 2 infinity, 2 being the other root. Now, we have to ask ourselves a question, are all these points going to be retained in the final solution set? That amounts to saying, are they all admissible? Are they all in the domain? Are they all consistent with the defining inequalities for Case 1?
So we have to look at how Case 1 is defined, and we keep the points we’ve obtained but only those that are consistent with Case 1. As it turns out, they’re all consistent with Case 1. That is, all the points we’ve found have x either greater than 1 or less than minus 2. So we keep them all. That’s not always the case. In Case 2 now, the denominator is negative. So when we multiply across by the denominator, the sign of the inequality is reversed. We get the same actually quadratic inequality but with the sign reversed, and we easily find a solution set. It’s the interval minus 5 to 2. However, do we retain all those points? No.
We can only retain the points that are consistent with Case 2 under study; that is, we only retain those that are between minus 2 and 1. And that, therefore, will be the contribution of Case 2 to our ultimate solution set. Now, we lump these two parts together– the solutions retained from Case 1, the solutions retained from Case 2, and we have the answer to our problem. The solution set S is the union of three intervals. Notice the middle one– minus 2, 1– is an open interval. Why is 1 not in the solution set, for example? Because 1 is not in the domain of the original inequality.
Remark, suppose you wanted to consider the opposite inequality to the one we’ve been looking at. Here I’ve replaced less than or equal 3 by strictly greater than 3. We don’t have to do all the work over again. We merely need to take the complement of the set S that we have just found for the reverse inequality– the complement relative to the domain; that is, D minus S in the set difference sense. And we easily find the answer, the union of two intervals. Notice that the point 1 is still not in this answer because it’s not in the domain. So we can’t just say the points that are not in S.
We have to take the complement of S relative to the domain.