In this video we're going to learn how to solve a quadratic inequality. Here we have an example of a quadratic inequality. We know it's an inequality because of the less than sign here.
It could have also been a greater than sign, or even less than or equal to, or greater than or equal to. But that's why it's an inequality. This on the other hand is a quadratic equation.
It's an equation because it has an equals sign. Now we already know how to solve a quadratic equation like this. We would factorise it into two brackets.
This one would be x plus 5 and x minus 2 and that would give solutions x equals negative 5 and x equals positive 2. These numbers here, negative 5 and positive 2, are known as critical values. We didn't want to solve the equation though, we wanted to solve when it was an inequality. To do this we're going to draw the graph represented by the equation. So y equals x squared plus 3x minus 10. We draw some axes first of all, and we know that since negative 5 and 2 are critical values, solutions to the equation, it must cross through these on the x-axis.
We also know that quadratics like this make a nice u-type shape, so it would look something like this. Let's now consider our inequality again that we were trying to solve. If you substitute in the critical values, you'll see that they're solutions because they give 0. So if we start with 2, 2 squared plus 3 lots of 2 take away 10. gives 4 plus 6 and 4 add 6 take 10 is 0. The same works for the other critical value negative 5. Negative 5 squared plus 3 lots of negative 5 take away 10 gives 25 minus 15 and 25 take 15 take 10 again is 0. This shouldn't be surprising since they were solutions to the equation, but we're not interested in when it equals 0, we want to know when it's less than 0. So if we now look at the graph, you can see the graph is split into three sections. The first section here, where the graph is above the axis, we then have this middle section, where it's below the axis, and this final section where it's above the axis again. The two red sections are above the x-axis.
This means that for those x-values, the equation gave an answer that was greater than zero. The green section however is below the x-axis, so for those x-values, the equation gave an answer which was less than zero. Since our inequality asks for less than 0, we're interested in the green section of the graph. To solve this inequality then, we want to know the x values that give this green section.
So that's all of the values marked here by the pink arrow, so in between negative 5 and 2. We can write that down as an inequality x is in between negative 5 and 2. Notice both of the inequality signs point to the left. And that's the answer to this inequality. So what we did was instead of solving the inequality, we first solved the equation to find the critical values.
We used the critical values to draw a sketch of the graph, and then considered which section we were after. Once we had that section, we can write down the solution. Let's try a second one. For this example we have a greater than sign, but we'll still begin in the same way. We'll solve the equation instead.
To solve this equation we can factorise. The brackets this time would be x plus 3, and x minus 6, which gives solutions x equals negative 3 and x equals 6. Remember these are our critical values. So when we come to draw the graph, we know the graph will go through those critical values, so it must cross at negative 3 and it must cross at 6. And again this is a positive x squared so it'll make a u-type shape, something like this.
We can now look back to the question x squared minus 3x minus 18 is greater than zero. Notice this time it's a greater than sign. This means the section of the graph we're interested in is now above the x-axis, so the section we want to look at is this section, since it's greater than zero. We then want to write down the corresponding x values that give us this section of the graph. This is now in two distinct parts.
We have the part that is less than 3 here, and the part that is greater than 6 here. We can write these down as two separate inequalities. x is less than negative 3, or we have x is greater than 6. When we solve a quadratic equation we're used to writing the word AND in between because they're both acceptable solutions.
This time we're going to write OR since you can't have a number that's less than negative 3 and also greater than 6. So when you have inequalities like this we write an OR instead. In this final example there are two extra differences. Firstly the symbol is now a greater than or equal to.
This isn't going to change our approach, we just need to remember to write the extra little line. This will be the same if it was less than or equal to. The second though will change.
We have 3x-2 on the right hand side, where we are used to having 0, so we need to eliminate those terms before we begin. To do this, we will subtract 3x, and add 2 to the right hand side, and then do the same to the left hand side. On the right hand side this gives 0, since 3x take away 3x is 0, and negative 2 add 2 is 0. On the left hand side, the x squared will remain unchanged, but then negative 3x take away another 3x, gives negative 6x, and 3 add 2 gives 5. This inequality is now in the correct form.
We can then look at solving the equation. So to factorise this one, you'd get x minus 1 x minus 5, giving critical values, x equals 1, and x equals 5. These critical values can then be plotted. Ordinarily at this point we would draw the graph, and you can still do that, you'll still get to the solution, but I want to show you another method that some people prefer. This method is case analysis.
Instead of drawing the graph, we'll draw out just the x-axis, and we'll plot on the critical values, x is 1 and x is 5. These critical values split the x-axis into three distinct sections. The first section here on the left. is when x is less than or equal to 1. The next section in the middle is when x is in between 1 and 5, and the final section is when x is greater than or equal to 5. Notice I've used less than or equal to's and greater than or equal to's due to the symbol in the question. I would next write down the inequality we're trying to solve, and we'll go through each of these sections one at a time checking if they're valid. So we'll start with the section on the left, x is less than or equal to 1. We need to select a number from this section, substitute it into the inequality and check if it works.
I'm going to pick 0 since that's an easy number to deal with. So if I substitute 0 in for x I get 0 squared minus 6 lots of 0 plus 5. 0 squared is just 0 so that will disappear. Negative 6 times 0, that disappears too, so I'm just left with 5. And 5 is greater than or equal to 0. So this one works.
This means this section will form part of our solution. We can then try the next section. We need a number that's in between 1 and 5. I'm going to choose 2. So if we substitute 2 in, we get 2 squared, minus 6 lots of 2, plus 5. 2 squared is 4, negative 6 times 2 is negative 12, and then if we do 4 take 12, add 5, we get negative 3. Now is that greater than or equal to 0? No.
So this section will not form part of our answer. We move on to the final section, x is greater than or equal to 5, I'm going to pick the number 6, so we do 6 squared, take away 6 times 6, plus 5, 6 squared is 36, negative 6 times 6 is negative 36, so the 36's will cancel, so again we're just left with 5. Is 5 greater than or equal to 0? Yes, so this forms part of our solution.
What you now do is take any of the sections that have a tick on them. and write them down as part of your solution. Remember when we have more than one inequality we'll use the word OR.
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