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Contents

- 1 Solution of a Quadratic Inequality Zeros
- 2 Solution of a Quadratic Inequality Zeros
- 3 Solution of a Quadratic Inequality Intervals
- 4 Solution of a Quadratic Inequality Intervals
- 5 Solution of a Quadratic Inequality on a Number Line
- 6 Solution of a Quadratic Inequality on a Number Line
- 7 Quadratic Inequality Check 1
- 8 Quadratic Inequality Check 1
- 9 Quadratic Inequality Solve for Zeros First
- 10 Quadratic Inequality Solve for Zeros First
- 11 Quadratic Inequality Testing Intervals
- 12 Quadratic Inequality Testing Intervals
- 13 Quadratic Inequality Solution in Interval Notation
- 14 Quadratic Inequality Solution in Interval Notation
- 15 Quadratic Inequality Check 2
- 16 Quadratic Inequality Check 2
- 17 Quadratic Inequalities Practice 1
- 18 Quadratic Inequalities Practice 1
- 19 Quadratic Inequalities Practice 2
- 20 Quadratic Inequalities Practice 2
- 21 Quadratic Inequalities Practice 3
- 22 Quadratic Inequalities Practice 3
- 23 Quadratic Inequalities Practice 4
- 24 Quadratic Inequalities Practice 4
- 25 Quadratic Inequalities Practice 5
- 26 Quadratic Inequalities Practice 5

Throughout this unit, we've been working with quadratic equations. Now, let's check out quadratic inequalities. A quadratic inequality looks a lot like a quadratic equation, except instead of an equal sign, we'll either have a greater than symbol, a less than symbol, a less than or equal to symbol, or a greater than or equal to sign. When we solve a quadratic inequality, we want to find all the values of x that make this statement true. In this case, we want any x value. So that way, when we evaluate this side, it's greater than 0. Now, we have an inequality here. So we know there's going to be more than one solution. In fact, there's going to be an infinite number of solutions. So to find the range of all possible values, let's start by finding which x values make this statement equal to 0. So, what two values of x make this statement equal 0. Write the smaller number here and the larger one here.

The smaller number is negative 3, and the larger number is positive 4. Nice work, if you found these two answers. To find these two answers, we simply want to factor this side of the equation, and then set each factor equal to 0. When we factor here, we want to find factors of negative 12 that sum to negative 1. Those factors are negative 4 and positive 3. Next, we set each factor equal to 0, and then solve for our two values of x.

We know the values of 4 and negative 3 would make this left-hand side equal to three separate intervals on a number line. These three intervals may contain the numbers that are solutions to our quadratic inequality. For example, all of the numbers to the left of negative 3 will either make this statement positive or negative. We can use a test point within this interval or a test value to find out. For example, I could use the value x equals negative 4 to determine whether or not this left side is positive or negative. We want this side to be greater than 0. So, when I plug in negative 4, I would want this to be a positive number. If that's the case, then every number from negative 3 and to the left will make this statement positive, which means we include it in our solution set. So, rewriting my original inequality here, I can plug in the value of negative 4 in for x. When we evaluate this, we'll get 8 is greater than 0. We know the statement is true, so we know all the values less than negative 3 are solutions to our original inequality. If you're in disbelief that all the other numbers in here work, then try plugging in negative 5 or something like negative 10 in for the value of x. You should wind up with a positive number on this side. Here, I've gone ahead and shown you another check for x equals negative 8. If we use this value at interval I, we'll wind up with a statement 60 greater than 0. We know this is true, so again, this is part of our solution set. We don't need to test every value in interval I, we only needed to test one of them. You try testing interval II and interval III. Pick one value for x in this region and in this region. Plug them into your original inequality, and see if you come out with a true statement. If true, then that interval is part of the solution. If false, then that interval will not be part of our solution. Also, as a hint, think of an easy value to plug in that's between negative 3 and positive 4. What do you think is the easiest value to use in between these two numbers?

It turns out that interval II is false and interval III is true. Let's see why. I think for the second interval, the easiest number to check would be x equals than 0. We know this is a false statement. So, this means that none of the values between negative 3 and positive 4 would satisfy our original inequality. We would exclude these values from our solution set. For interval III, I'm going to choose the number x equals 5. Now, you could have chosen any number, like six, eight, ten, even 14. You just want to choose one number to test in this interval to see if it makes our inequality statement true. So plug in x equal 5. I get the statement 8 is greater than 0. I know this is true, which means that all of these numbers satisfy our original condition. Great work if you found this to be false and this interval to be true.

For quadratic inequalities, we can show the solution in one of two ways. On a number line, we know that these values satisfy the inequality and these values do as well. We use parentheses at x equals negative 3 and x equals, 4 since we know these numbers make the equation equal to 0. These two numbers would not be part of the solution set. The other way we can write our solution set is in interval notation. Keep in mind, since we have two non-overlapping regions, we'll need to use the union symbol in between the two intervals. What I want you to do is to fill in each of these boxes using the correct interval notation. Be sure to use parenthesis or brackets and include the infinity symbol or the numbers as necessary. Good luck.

For the first interval, we would have a negative infinity to negative three. We use parentheses on each end, since we don't include these values. The right-hand side would be four comma infinity. This would be the second interval. Be sure to include parentheses here, because we can't include these two values. Nice work if you remembered interval notation. And, just to be even more certain that we're right, we can choose any number in either of these intervals, and they should check in our inequality. So, if I chose x equals 12, I know that's between 4 and positive infinity. So I could plug in 12 for x, and then I would get the value 120. We know 120 is greater than 0, which means that this statement is true. This quick check just reinforces that we have the correct intervals on either side. It's usually a good idea to test a couple numbers to make sure that you have the correct interval.

Now that you've seen one, let's see if you have the hang of it. Try finding the solution set for this inequality and write the solution in interval notation here. As a hint, this solution will be a union, just like the last problem. We'll have two different regions.

For this inequality, the interval would be negative infinity to negative 3, union 2 to positive infinity. Great solving if you found those two regions. When we find the solution set for a quadratic inequality, we first want to start by finding the zeros. These are the x values that make the equation equal to 0. We factor this quadratic using x plus 3 times x minus 2 as the factors, then we set each factor to equal to 0 to get x equals negative 3 and x equals positive 2. We use these x values to set up our three intervals, so we know the values of x might be less than negative 3, between negative 3 and 2, or greater than 2. These different regions are the possible solutions to our inequality. We test the value in the first interval to see if it makes the inequality true. I'm going to choose the value x equals negative 4. Plugging in negative 4 for x, we'll get 6 is greater than 0, which we know is a true statement. This means, any value less than negative 3 will make this inequality true. And notice, we've already got our first interval. Negative infinity to negative 3, here. We need to test the value now for interval II. I'll use x equals 0, since that's pretty easy to plug in and evaluate. If x is equal to 0, then these two terms would drop out. They would just be 0. And we'd be left with negative 6 is greater than 0. We know this statement is not true. So this region is false. We exclude it from our solution set. And finally, we try a different value of x for our third interval. I'll use x equals 3. When we plug in x equals 3, here and here, we'll get the statement 6 is greater than 0. This is true. This means any value greater than 2 will satisfy our original inequality. And notice we have our second region. From 2 all the way up to positive infinity.

If you're acing these problems, then great work. I don't think they're the easiest, so let's try one more problem broken down into parts just like the first one. For this quadratic inequality, I want you to tell me the 0s first. Find the values of m that would make this left side equal to zero. Write the lesser value for m here, and the greater value for m here.

Great work if you found negative 2 3rds and positive 5. You've really got your factoring down. We begin factoring this by finding the factors of negative 30 that sum to negative 13. We use these two factors to rewrite the middle term, negative 13m. Then, we use factoring by grouping to get 3m plus 2 times m minus So m could equal negative 2 3rds or m could equal positive 5.

Now that we know the 0s are negative 2 3rds and positive 5, we can set up our three intervals on our number line. Now, I want you to test each of these intervals to determine which ones are part of the solution set. Again, if you plug in a value in this region and the statement winds up being true, then we include this as part of our solution set. If the statement turns out to be false, then we exclude the interval from the solution set.

Here, the only values of m that make the inequality true are in interval II. The numbers between negative 2 3rds and positive 5. In the first interval, we can test the number that's less than negative 2 3rds. Well, negative 1 is less than negative 2 3rds. We know this since negative 1 can be rewritten as negative 3 3rds. We want to plug in negative 1 instead of negative 3 3rds, since it's easier to work with an integer than it is with a fraction. Substituting the value of negative 1, m for m. We get the inequality statement that 6 is less than or equal to 0. We know this isn't true, so interval I is false. Next, we want to test the value in interval II. I think the easiest value to test in this region would be 0. When we plug in 0 for m, we lose these two terms to get negative 10 is less than or equal to 0, which we know is true. So yes, these values can be included in our solution set. And finally, we test a number in interval III. I'm going to use the number 6. You could of course use other numbers, 10 would have been another great one to use. Plugging in 6 in for m will get the inequality 20 is less than or equal to 0. This isn't true, so we know that this part is not included in our solution set.

So now that we know that these x values make our original inequality true, what is the solution set written in interval notation? You'll want to write your answer here, and think about whether you need to use brackets or parenthesis.

Our interval would be negative 2 3rds, positive 5. We know we can use brackets on the ends here since negative 2 3rd and 5 make this left-hand side equal to values. So here's our answer written in interval notation And here's our answer written on a number line.

Now that we've done one together, I want you to try this one on your own. Notice how we have a constant term on this side of inequality. Well, we really don't want this over here. We want to set our quadratic equal to 0 first. What do you think you'll need to do first in order to get started? Once you've gotten that first step, I want you to try and solve for all the possible solutions. Write your answer as a solution set in interval notation here. And good luck.

Here, the possible solutions for p are between negative 2 3rds and positive 1 this problem, we want to subtract 2 from both sides of the inequality. This gives us a quadratic expression on the left and zero on the right. Now, we want to find the 0s or the values of p that make this quadratic expression equal to 0. We find factors of negative 18 that summed to positive 3. Those two factors are positive 6 and negative 3, and we use them to rewrite the middle term of plus 3p. Using factoring by grouping, we get the factors 3p plus 2 and values of p, so p could equal negative 2 3rds or p could equal positive 1 3rd. Next, we use our two zero ro values to create three different intervals. And now, it's just a matter of testing each one to see which ones are part of the solution. For the first interval, we'll plug in p is equal to negative 1. When we plug in this value for p, we'll get this statement, 4 is less than 0. We know this statement is not true, so we exclude this interval from our solution set. For the second interval, we'll test the value of zero. We know zeros in between these two fractions, since one of them is negative and the other one is positive. When we plug in p equals 0, we get the statement negative 2 is less than 0. This is definitely true, so this part is part of our solution set. For the third interval, we'll test p equals 1 and we'll get the statement 10 is less than 0. We know this is false, so this is not part of our solution, which leaves us with this interval only. Any value between negative 2 3rds and positive 1 3rd will make our original inequality true. We also use parentheses here and in our interval notation since this is just a less than than symbol. We can't be equal to zero.

Try finding the solution to this inequality. What do you think the values of p could be? As a hint, you'll need a union to find both the intervals that work for this inequality.

Here, the ranges were from negative infinity to negative 5, with the union of negative 1 to positive infinity. Great work if you found these two intervals. First, we want to find the values of p that make this equation true. We factor to find the series of p and we get p is equal to negative 5, and p is equal to negative 1. We use the values of negative 5 and negative 1 to create the three intervals on our number line. For the first interval, I'm going to test the value of negative 10. When I plug in at negative 10 for p, I'll get the statement 45 is greater than 0. I know this is true so I can include it, this part or this interval, in my solution set. For interval II, I'll plug in the value p equals negative 3. When we substitute this value in, we get the inequality negative 4 is greater than 0. We know this is not true, so we can include this interval in our solution set. And finally, for the third interval, we'll use the value p equals zero. If p equals zero, we lose these two terms and are left with a statement 5 is greater than zero, which we know is true. We'll use parentheses at negative 5 and negative 1, since we don't have the or equal to symbol here. So any value, from negative infinity to negative 5 or any value from negative 1 to positive infinity will satisfy our original inequality. Here's our answer on a number line, and here's our answer written in interval notation.

Here's a second practice problem on quadratic inequalities. When you think you have the answer, write it in interval notation here. Good luck.

Our interval would be from negative 4 to positive 2. x could actually equal negative 4 and positive 2 in this case, and any value in between them. Great solving if you found this interval. To begin solving this inequality, we want to subtract 8 from both sides. This allows us to set this quadratic expression less than or equal to 0. Now, we can rewrite this as an equation and find the 0s or the x values that make this equal to 0. We factor the quadratic expression, and then we set each factor equal to 0 to get x could equal negative 4 or x could equal positive 2. We use these to create the three separate intervals on our number line. Next, we want to test a value in each of the intervals. For the first interval, we'll test the value x equals negative is less than or equal to 0. Now, this of course, isn't true, so this interval is false. It's not part of our solution. For the second interval, we can test x equal 0. If x equals 0, we lose these two terms and we're left with negative 8 is less than or equal to 0. This, we know is true, so this part is included in our answer. And finally, for the third interval, if we test x is equal to 3, we'll get the statement 7 is less than or equal to 0. Again, we know this isn't true, so this region is out. So the values of x that satisfy our inequality are between negative 4 and 2. We include negative 4 and 2, since we have the or equal to symbol here. Since we know both of these values are 0s to the quadratic equation. I hope you're starting to see all sorts of patterns when you solve these types of problems.

Here's our third problem on solving quadratic inequalities. What do you think would be the solution set here? Again, for this problem here, we have a union. So you should have two intervals. When you think you have the values of y, that make this inequality true, type your intervals in here. Good luck.

Here, the solution was negative infinite to negative 3 or 1 half to positive infinity. Nice algebra skills if you got these two integrals. To begin to solve this quadratic inequality, we want to subtract 3 from both sides. So we have 0 on the left and the quadratic expression on the right. We can flip the entire inequality around, so we'll have 2y squared plus 5y minus 3 is greater than 0. These two statements here are exactly the same. They're just written in the reverse order. Once we have this inequality, we want to find the 0s. If we find the 0s, we can set up our number line with three intervals. So we'll factor the left-hand side of the equation to get 2y minus 1 times y plus 3. We set each factor equal to 0 and we solve for the values of y over here. In this case, y could equal 1 half, and in this case, y could equal negative 3. We use positive I, we'll try the number y equals negative 5. When we substitute negative 5 in for y, we'll get the statement 22 is greater than 0. We know this is true, so this is included for our solution set. For interval II, we'll try the number y equals 0. When we plug in 0 for y, we'll be left with negative 3 greater than part of our solution. And finally, for the third interval, we can try the number y equals 1. When we plug in 1 for y, we'll get 4 greater than 0. This is a true statement, so we can include these values as part of our solution set. We use parentheses at negative 3 and positive 1 half, since we need to be greater than 0, we can't be equal to it. So our final solution is negative infinity to negative 3 or 1 half to positive infinity.

Here's a fourth problem on quadratic inequalities. What do you think the solution would be here? Write the solution set in interval notation, and be sure to use the appropriate bracket or parenthesis.

The values of x that satisfy this inequality are from negative 5 halves to positive 3. Great work if you found this interval. As usual, we begin by finding the 0s for this quadratic equation. We set this quadratic expression equal to 0, and then factor it, and set each factor equal to 0. So we'll have see that x can equal negative 5 halves or x can equal 3. We'll use the values of negative 5 halves in positive 3 to set up our three intervals on our number line. These two values make this left side of our inequality equal to 0. So, we want to test other values, so that way, this left-hand side is less than 0. If we let x equal negative 5 in interval I, we get the statement 40 is less than in here satisfy our inequality. For the second interval, we'll test the value of 0, since that's in between negative 5 halves and positive 3. If x equals 0, we plug it in and we get the statement that negative 15 is less than 0. Well, this is true, so this part is part of the solution set. And finally, for our third interval, we just need to test a number greater than 3. If I let x equal is out of my solution set. These are the only values that make the inequality true. We use parentheses at negative 5 halves and positive 3, since we can't equal 0. We must be less than 0. Again, fantastic work if you got most of that solution correct. These are tough.

How about this fifth problem? What do you think would be the solutions for x?

Any value greater than or equal to negative 2 3rds, but less than or equal to 1 half would be correct. Fantastic algebra work for getting this one correct. We start solving this inequality by rewriting it as an equation. We factor this expression to get 3x plus 2 times 2x minus 1. Now we're ready to set each of these factors equal to 0 and solve for our values of x. Here, the 0s are the values that make this equation equal 0, our x equals negative 2 3rds and x equals 1 half. We use the values of x that make this statement equal to 0, instead of three intervals on our number line. Next, we want to test each interval and we'll start here. We'll let x equal negative 1 to get the statement 3 is less than or equal to 0. We know this isn't true, so our first interval is out. If we try a value between negative 2 3rds and 1 half, like 0, we wind up with a true statement. This means any number in this region will be a part of our solution. Finally, we test a number in the interval III, like 1, to get a false statement. We know this isn't true, so this interval is out of our solution set as well. We can use brackets at negative 2 3rds and positive 1 half, since we can be equal to 0. So these numbers on the number line would work for our inequality and this would be the set in interval notation.