## ← Rounding x intercepts - College Algebra

• 2 Followers
• 17 Lines

### Get Embed Code x Embed video Use the following code to embed this video. See our usage guide for more details on embedding. Paste this in your document somewhere (closest to the closing body tag is preferable): ```<script type="text/javascript" src='https://amara.org/embedder-iframe'></script> ``` Paste this inside your HTML body, where you want to include the widget: ```<div class="amara-embed" data-url="http://www.youtube.com/watch?v=H1oTifNkNy4" data-team="udacity"></div> ``` 2 Languages

Showing Revision 2 created 05/25/2016 by Udacity Robot.

1. In order to find out where this curve hits the x-axis, we need to set y equal to
2. 0. Then we can say that this will be satisfied if one of two conditions is true.
3. We could either have the first factor equaling 0, which gives us the equation x
4. plus 3 minus the root of 6 equals 0. Or we could have the second factor equals
5. 0, x plus 3 plus root of 6 equals 0. Now we just solve for x. This gives us x
6. equals negative 3 plus the root of 6, and x equals negative 3 minus the root of
7. 6. So if x equals either of these numbers, this expression will equal 0. These
8. are just the x-coordinates of our two x-intercepts, and of course the
9. y-coordinate of each is 0. So this is actually really incredible what we've
10. done. We figure out how to factor over not just the integers but over the real
11. numbers. In other words, we can use completing the square and understanding
12. difference of squares to find the x-intercepts of a ton of different equations
13. for parabolas, not just ones that we can factor in the traditional way. Oh, and
14. I almost forgot. The rounded version of these two answers are negative .55, 0
15. and negative 5.44, 0. We're going to keep working with the idea of finding
16. x-intercepts of any given parabola in the next lesson. Awesome job on a lot of
17. pretty complicated material.