Today, we're going to talk about quadratic inequalities. Remember, an inequality is just a statement that says that two quantities are unequal in value, and we use inequality symbols to show that one quantity is less than or greater than the other. So we'll start by reviewing how to graph an inequality on a number line. We'll talk about what it means to be the solution to a quadratic inequality, and then, we'll do some examples solving and graphing quadratic inequalities.
So let's start by reviewing how to graph an inequality on a number line. The inequality symbols, less than or equal to, or greater than or equal to, can be graphed on a number line in two different ways, one using a square brace, and the second way is to use a closed, or a filled-in circle. Inequality symbols that are strictly less than or strictly greater than, so not equal to, can be graphed either using a curved parentheses, or an open circle.
So from our first example, I've got the inequality x squared plus 6x plus 5 is less than or equal to 0. Since it's already set to equal 0, I'm going to go ahead and rewrite this as an equation and solve it for x. I'm going to solve this by factoring, so I know I'm looking for two numbers that multiply to 5 and add to 6.
Integers that multiply to 5 would either be positive 1 and positive 5 or negative 1 and negative 5. Since they have to add to be a positive 6, I know my two integers are positive 1 and positive 5. So I can write this in factored form as x plus 1 times x plus 5 equals 0, and using my zero-product property of multiplication, I know that x plus 1 equals 0 and x plus 5 equals 0. Solving each of these for x, I get x is equal to negative 1, and x is equal to negative 5.
So I'm going to put these two values on my number line. Since the inequality symbol is less than or equal to, I know I'm going to use a filled-in circle above the negative 1 and negative 5. So a filled-in circle. Now, I'm looking for the intervals that are bounded by these two points. And so my first interval, I'm going to choose a test value of negative 6. In my middle interval, I'm going to use negative 3. In my last interval, I'm going to use 0. So because the numbers, or the values, in each interval all represent solutions or non-solutions to the inequality, I only need to pick one value within each interval to test.
So I'm going to start by substituting negative 6 into my inequality. So that would give me negative 6 squared plus 6 times negative 6 plus 5, and I want to know if that's less than or equal to 0. So simplifying, I have 36 plus negative 36, which will give me 0, and 0 plus 5 is 5, which is not less than 0. So my first test value is not a part of the solution.
My second test value is negative 3. So substituting that into my inequality and simplifying, 9 plus negative 18 gives me negative 9. Plus 5 would give me negative 4, which is less than 0, so I know that my inner interval is part of the solution region. And my last test value is 0. 0 squared is 0, plus 0 plus 5. 0 plus 0 plus 5 is not less than 0. So I know that my last interval is also not a part of my solution region. So in between negative 5 and negative 1, also including negative 5 and negative 1, represents the solution region to my quadratic inequality.
So for the second example, I've got x squared plus 4x is greater than negative 3. So I'm going to solve this as an equation, but first, I need to set this equal to 0. So to do that, I'm going to add 3 on both sides, making this x squared plus 4x plus 3, and I'm going to just go ahead and write it as an equation is equal to 0.
I'm going to solve this by factoring, so I'm looking for two numbers that multiply to positive 3. I know that that is going to be 1 and 3 or negative 1 and negative 3. Since they also need to add to 4, I know that my two values are going to be positive 1 and positive 3. So I can write this in factored form as x plus 1 times x plus 3 is equal to 0. Using my zero-product property of multiplication, I can write two equations and solve them for x. So I find that x is equal to negative 1, and x is equal to negative 3.
So I'm going to use these two values on my number line to create some intervals. So I know that because I have a strictly greater than inequality symbol, I'm going to use open circles over negative 1 and negative 3. So I'm going to pick a test value within each interval and substitute it into my inequality to see which will yield true statements. So for this first interval, I'm going to pick negative 4. Then, I'll use negative 2 and then, finally, 0.
So substituting negative 4, negative 4 squared plus 4 times negative 4, is that greater than negative 3? Simplifying, I have 16 minus 16, which is 0. And 0 is greater than negative 3, so I know that my first interval is a part of my solution region. Then, I'm going to test the point, negative 2, by substituting negative 2 into my inequality. Negative 2 squared is 4. This will give me minus 8. 4 minus 8 is negative 4, which is not greater than negative 3, so I know that my second point, negative 2, is not part of my solution.
And finally, my last value 0-- 0 squared is zero plus 0. 0 is greater than negative 3, so I know that my last test value works. So negative 2 was not in my solution region, but negative 4 and 0 were. And so I know that this interval that is greater than negative 1 and this interval that is less than negative 3 represent my solution region.
So finally, let's review how we can write our answer in either set notation or interval notation. Our first example is represented by this shaded region on the number line. And so in set notation, we would write that as all x values, such that x is greater than or equal to negative 5 and less than or equal to negative 1. And in interval notation, we would use square braces with the values of negative 5 to negative 1 included in the interval.
Our second example had a number line with these two shaded regions for our solution. So in set notation, we would write this as all x values, such that x is less than negative 3 or greater than negative 1. And in interval notation, we would write it as the union of two intervals, using curved parentheses, from negative infinity to negative 3 and from negative 1 to positive infinity.
So let's go over our key points from today. The solution to a quadratic inequality is a range of x values that make the inequality statement true. And the process for solving a quadratic inequality is one, solve as an equation set equal to 0; two, use solutions to the equation to create intervals on a number line; three, choose a test value that falls within each interval on the number line; and four, plug each test value into the inequality to identify the solution regions.
So I hope that these key points and examples helped you understand a little bit more about quadratic inequalities. Keep using your notes and keep on practicing, and soon, you'll be a pro. Thanks for watching.