Video: Finding the Solution Set of an Inequality of the Second Degree

Determine the solution set of the inequality (π‘₯ + 3)Β²< (5π‘₯ βˆ’ 9)Β².

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Video Transcript

Determine the solution set of the inequality π‘₯ plus three squared is less than five π‘₯ minus nine squared.

We write down the inequality again. And you might think that the first step in solving this inequality is somehow to take square roots on both sides. π‘₯ plus three squared is less than five π‘₯ minus nine squared. So can we say that π‘₯ plus three is less than five π‘₯ minus nine? Unfortunately, this isn’t necessarily true. For example, one squared is less than negative two squared. But one is certainly not less than negative two.

Let’s go back to the inequality that we want to solve. We’ve seen that we can’t just take square roots on both sides. So what’s the first step? The answer is that we expand both sides. π‘₯ plus three squared becomes π‘₯ squared plus six π‘₯ plus nine. And five π‘₯ minus nine squared becomes 25π‘₯ squared minus 90π‘₯ plus 81. We then subtract the left-hand side from both sides to get a zero on one side. We could have also done this term by term.

And then we can simplify the right-hand side by combining like terms. We subtract the π‘₯ squared terms to get 24π‘₯ squared, the π‘₯ terms to get minus 96π‘₯, and the constant terms to get plus 72. Now, we have something that looks slightly more like a normal quadratic inequality. We swap sides so that the quadratic expression is on the left-hand side as is convention. Remembering of course that as we swap sides, we also have to swap the direction of the inequality sign.

We can notice that all the coefficients on the left-hand side are divisible by 24. So we divide both sides by 24 to simplify. And because we’re dividing by a positive quantity, we don’t have to flip the inequality sign. We’ve rearranged and simplified our inequality as far as possible. And so now, we just have to solve it.

How do we solve this inequality? Well, there are various methods, but we’re going to graph the quadratic expression on the left-hand side. Let 𝑓 of π‘₯ equal π‘₯ squared minus four π‘₯ plus three. We’re going to graph this function, paying particular attention to two features of the graph: the π‘₯-intercepts, that is where the graph touches or crosses the π‘₯-axis, and the direction that the curve passes through these π‘₯-intercepts.

Okay, so what are the π‘₯-intercepts? These are points on the π‘₯-axis for which 𝑓 of π‘₯ is equal to zero. We use our definition of 𝑓 of π‘₯. We notice that the quadratic can be affected by inspection. For the product of the two factors to be zero, one of the factors has to be zero. And so π‘₯ is three or π‘₯ is one. And we mark these π‘₯-intercepts on the graph.

Having found these π‘₯-intercepts, we move on to the second important feature. This is a graph of a quadratic function. And so it’s a parabola in shape. But is it an upward facing parabola or a downward facing one? We look to the coefficient of π‘₯ squared. This is one, which is positive. And so we have an upward facing parabola.

Now that we’ve determined our second important graph feature. Our graph is good enough to solve the inequality. We’re looking for values of π‘₯ for which 𝑓 of π‘₯ is greater than zero that corresponds to values of π‘₯ for which the graph of 𝑓 of π‘₯ is above the π‘₯-axis. The graph is above the π‘₯-axis here when π‘₯ is less than the π‘₯-intercept one and also here when π‘₯ is greater than the π‘₯-intercept three.

The inequality doesn’t hold it either π‘₯-intercept itself. At these values, 𝑓 of π‘₯ is equal to zero. And remember we want 𝑓 of π‘₯ to be greater than zero. This is why we’ve used less than and greater than signs and not less than or equal to and greater than or equal to signs because the end points one and three are not solutions of the inequality and so should be excluded.

Between the π‘₯-intercepts, so when π‘₯ is between one and three, the graph of 𝑓 of π‘₯ is below the π‘₯-axis. And so 𝑓 of π‘₯ is less than zero and not greater than zero as we want. Okay, so what’s the solution? We have seen that the inequality holds if π‘₯ is less than one or π‘₯ is greater than three. We’re looking for the solution set of this inequality. And so rather than a constraint on π‘₯, we want the set of values of π‘₯, which satisfy it.

The set of values of π‘₯ for which π‘₯ is less than one is the interval from negative infinity to one, which is written like so. And the set of values of π‘₯ for which π‘₯ is greater than three is the interval from three to infinity written like so. π‘₯ satisfies either one of these constraints and so lies in either one of these sets or intervals. And the solution set is there for the union of these intervals.

There is another way of writing this solution set. The set of values which satisfy the inequality are all real values apart from those which don’t satisfy the inequality. Stay with me here. Remember that we found the values of π‘₯, which don’t satisfy the inequality. They were the values of π‘₯ between and including the π‘₯- intercepts. So the set of values of π‘₯ which don’t satisfy the inequality is the set of values of π‘₯ for which π‘₯ is between one and three inclusive.

And if we write this set in implementation, our answer becomes the set of real numbers minus the interval from one to three inclusive. Remember that these square brackets mean that the end points one and three are included in the interval, whereas the other intervals have parentheses to show that their end points are not included.

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