##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- Attend and watch FREE live webinar on useful topics

# Solving a Quadratic Inequality using a Sign Chart - Concept

###### Alissa Fong

###### Alissa Fong

**MA, Stanford University**

Teaching in the San Francisco Bay Area

Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts

Similar to when we solve quadratic equations, there are several ways we can solve quadratic inequalities. **Solving quadratic inequalities** can be done in two ways: by graphing the quadratic inequality or by using a sign chart. One of the advantages of using a sign chart to graph quadratic inequalities is that we aren't required to graph the parabola in order to solve the inequality, which saves time.

When we're looking at quadratic inequalities, one of the things you want to keep in mind is what you know about graphs of quadratics. I thought up an example here. Think about the equation y equals the quantity x-3 times x+2. This is already in factored form right? Check out the graph here. This has intercepts of -2 and +3 on the x axis and then the parabola shape opens upward. What you're going to be looking at is equations that look a lot like this only instead of y equals you're going to have some kind of inequality business. Something like this. Let me rewrite that with a better marker, like that, okay.

So we're looking for where is this graph less than zero and if you are to look at this picture, looking at this picture you can tell pretty easily the places where the graph is less than zero is this whole region down here or for x values that are in between -2 and +3. Notice I'm using strict less than signs, I'm not using equal than or equal to because of this guy here. So that's one way to do it if you have the graph. But a lot of times you guys I know graphing quadratics is a real drag. You don't want to have to draw the picture. So what we're going to be working on and what you guys will be doing in your homework is how to do these problems without a graph. How would you do this if all you had was this original problem statement?

Well the first thing you're going to be looking for is the places where you have x intercepts. We already know that the x intercepts will be at -2 and +3. I'm going to stick those guys on a number line. -2 and +3. Okay, what happens all around them is what's going to be important. I'm going to have to think about open circles and closed circles as well as how to shade my solution region. Well I'm going to go ahead and stick open circles on those guys based on this inequality sign. Then what I'm going to do is pick test points, a lot like what you do when you're graphing linear inequalities on the xy plane, this is the same idea. Like what would happen if I were to pick an x value out here? Like let's say I chose x is equal to -10. I don't know. I just picked that. Any value less than -2, I'm going to test this region.

Well, if my x number were equal to -10 and I substituted in there, I want to see whether or not I get a true inequality. -13 times -8 is a positive answer, right? So no this would not be part of my solution region. I'm not going to shade out there. But I need to test this region as well in between my two open circles. My two x intercepts. What happened if I pick x equals zero. I just chose zero because zeroes for me are pretty easy to work with.

When I substitute zero in here I want to see if I get a true inequality. Is it true that zero is greater than -6. Yes, that means this middle part is a solution. I'm going to be shading at least that piece and perhaps even more I got to test this last chunk of my line graph.

I'm going to pick another x number out here. I don't know I'm going to pick x=5. If 5 works, and so should every single number that's bigger than 3. I'm going if I can use any value I want to and if there's a solution, then every value out here would be a solution. Let's see, if I stick in 5, is it true that zero is greater than 2 times 7? No. That means not a solution. None of these points work. Don't shade.

So this is just my final graph of representation. This is my number line that shows the possible x values that would be solutions to this inequality statement. So if you want to, you could use a graph to to look at quadratic inequalities. But sometimes that takes takes a long time especially if you don't have a graphing calculator handy. These testing points and graphing your solutions on the number line is often faster.

Please enter your name.

Are you sure you want to delete this comment?

###### Alissa Fong

M.A. in Secondary Mathematics, Stanford University

B.S., Stanford University

Alissa has a quirky sense of humor and a relatable personality that make it easy for students to pay attention and understand the material. She has all the math tips and tricks students are looking for.

Your tutorials are good and you have a personality as well. I hope you have more advanced college level stuff, because I like the way you teach.”

Thanks alot for such great lectures... I never found learning this easier ever before... keep up the great work.... :)”

You seem so kind, it's awesome. Easier to learn from people who seem to be rooting for ya!' thanks”

###### Get Peer Support on User Forum

Peer helping is a great way to learn. Join your peers to ask & answer questions and share ideas.

##### Sample Problems (2)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete