Hi, and welcome. My name is Anthony Varela. And today, we're going to talk about quadratic inequalities. So we'll talk about solutions to quadratic inequalities, how to solve quadratic inequalities, and this is all going to involve creating intervals on a number line.
So first, let's review inequalities versus equations. So here's a quadratic equation. We have a quadratic expressions set equal to something. So we have our equal sign.
But with inequalities, we have a different type of symbol. We don't have an equal sign. We have an inequality symbol. So here's an example of one of our inequality symbols, greater than. And this is one of our strict inequality symbols.
In a number lines, we use open circles or curved parentheses when we're dealing with strict inequalities. Now we also have non-strict inequalities. So this would be less than or equal to and greater than or equal to.
In our number lines, we used filled in circles or square brackets when dealing with non strict inequalities. So now what are solutions then to quadratic inequalities? Well, this would be a range of x values that satisfy our inequality statement. And non-solutions then would be a range of x values that don't satisfy the inequality.
So here's the process for solving quadratic inequalities. What we're going to do first is express it as an equation set equal to 0. That's going to be important, because we know how to solve quadratic equations that are set equal to 0. We can factor or use the quadratic formula.
So once we find solutions, that's going to help us create intervals on a number line. And then what we do with our intervals is we choose an x value that fits within all of our intervals. And we substitute it into our inequality. And then we're looking for true statements or false statements.
If it's a true statement, that entire interval is part of our solution. If it's a false statement, that entire interval is not part of our solution. So let's go ahead and work with our first example 2x squared plus 3x minus 2 is greater than 0.
So our first step is to express as an equation set equal to 0. So we already have a 0 right here. So we just need to swap out our inequality symbol for an equal sign. So now let's solve this equation. And I'm going to solve by factoring.
So factoring this I have 2x minus 1 as one factor. And x plus 2 as another. Now you can go ahead and pause and FOIL this out to make sure that this does indeed equal 2x squared plus 3x minus 2.
Well, when we have in factors now, what we can do is set each factor equal to 0 and solve for x. So solving for x and 2x minus 1, we see that x equals 0.5, 1/2. And solving for x when x plus 2 equals 0, we get x equals negative 2.
So let's go ahead and represent this on a number line. So we have an infinite number line. And I have it markdown to negative 2 and 1/2. And notice that this creates then 3 intervals.
We've broken up our infinite number line into three sections. So what we're going to do now is pick a number that fits within each of our three intervals. So what I like to do whenever we're jumping from a negative to a positive, I like to use 0, because we're eventually going to be plugging this in. And it's really easy to multiply and add or subtract 0.
So I always recommend doing that. And then we don't have to go too far here. I'm just using 1 on this side and then negative 3 on this side. So we've chosen our test values. Now what we do then is substitute each of our test values into our inequality and look for true statements.
So when x equals negative 3, 2x squared is 18. Plus 3x is negative 9. And then we have our minus 2. And we see then that 7 is greater than 0.
So what this means then is that entire interval is part of our solution. When x equals 0-- so this is within our middle interval now, 2x squared is 0. 3x is 0. And then we have minus 2.
Now negative 2 is not greater than 0. So that means are middle interval is not part of our solution. Finally, when x equals 1, 2x squared is 2, plus 3x is 3, and then we subtract 2. And we see that 3 is greater than 0.
So that entire interval then is part of our solution. So we can write our solution a couple of different ways. In set notation, this would be x, all x, such that x is less than negative 2 or x is greater than 1/2. We can also write this in interval notation.
So our first interval is from negative infinity to negative 2. And we have 1/2 to positive infinity. And we combine those two intervals with this u symbol here for union. Let's go through another example.
Here we have a non-zero on one side of our inequality. And it's important that we express our equation set equal to 0. So the first thing I'm going to do is subtract 12 from both sides. So we have a 0 here. And then this makes a minus 12 here. So now we can set this equal to 0. And now we need to solve this equation.
So you could use the quadratic formula. I'm going to factor again. The first thing that I'm going to do is factor out that common factor of 2 that I see in all of my terms. And I can factor x squared minus x minus 6 as x plus 2 times x minus 3.
And here we can pretty easily see then that our solutions are x equals negative 2 and x equals positive 3. So that's going to help us create intervals on our number line. And now let's choose values that fall within each of these intervals.
Once again, I see a jump from negative to positive. So I'm going to use a 0. And then I have positive 4 over here and negative 3 over here. So now we substitute each of these values in for x and see if we get a true statement.
So when x equals negative 3, 2x squared is 18. Minus 2x is plus 6. And then we have our minus 12. And simplifying this, we see that 12 is not less than 0. So this interval here is not part of our solution.
When x equals 0, 2x squared is 0 minus 2x is 0. And then we have a minus 12. And we see that negative 12 is less than 0. So that entire interval then is part of our solution.
And lastly, when x equals 4, 2x squared is 32. Minus 2x is minus 8. And then we have minus 12. And we can see that 12 is not to less than 0.
So this interval is not part of our solution. So let's write our solution then as a set. This would be all x values, such that x is between negative 2 and positive 3. As an interval, this would be from but not including negative 2, 2 but not including positive 3.
So let's review quadratic inequalities. When we talked about our strict inequality, which use open circles or curved parentheses and number lines and our non-strict inequalities, which have closed circles or square brackets are number lines. To solve a quadratic inequality, expressed as an equation set equal to 0, solve that equation, our solutions create intervals on a number line.
So choose test values within each of those intervals. Substitute back to your inequality to see where your solutions are or what your solutions are. So thanks for watching this tutorial on quadratic inequalities. Hope to see you next time.