WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today, we are going to be talking about graphing and solving quadratic inequalities.
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And we are going to apply some techniques learned from linear inequalities in our work.
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So, in order to graph quadratic inequalities, use the techniques learned for graphing linear inequalities.
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And recall: that involves graphing the related equation and using a test point,
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except this time, we are going to be using a quadratic equation.
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For example, just starting out with something simple: if you had an inequality y > x² - 2,
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the first step is to graph the related equation; and in this case, the related equation is going to be y = x² - 2.
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And that is going to give the boundary line for the solution set.
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So, looking at this form, this is actually in vertex form; and here, a is 1; h is 0; and k is -2.
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So, what this tells me is that the vertex of my graph is at (0,-2).
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So, I have the vertex at (0,-2), right here.
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And I also know that a is positive; a is greater than 0; so I know that the parabola opens upward.
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So, I have a general idea of the shape of this graph before I even get started.
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Now, to make a more exact graph, I am going to find some points.
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And the first point is going to be -2, squared is 4, minus 2 is 2.
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-1 squared is 1, minus 2 is -1; and then, I already have (0,-2); this is my vertex.
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1 squared is 1, minus 2 is -1; and now, 2 squared is 4, minus 2 is 2.
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This is enough for me to go ahead and graph.
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(-2,2), (-1,-1), the vertex I already have graphed...actually, this goes -1, -1 is right there; and then (1,-1); (2,2).
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So, I can see the shape of the graph.
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Now, since this is a strict inequality, the boundary line is going to be dashed; it is not going to be part of the solution set.
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And as expected, this parabola faces upward.
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So, just as with linear inequalities, my first step is to graph the related equation.
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My second step is to use a test point; and recall that a good test point to use is...I am going to use the test point (0,0),
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because it is not near the boundary line, and it is a very easy point to use.
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So, I am going to substitute (0,0) into this inequality, y > x² - 2; so 0 is greater than 0 squared minus 2.
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So, 0 is greater than -2; and that is true.
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Since this is true, this test point is part of the solution set.
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Now, the boundary line divides this into two sections: the area inside the parabola here, and the area outside the parabola.
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And what the test point tells me is that, since the test point is inside the parabola, the solution set for the inequality is also inside the parabola.
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If this had turned out to be not true, and the test point wasn't part of the solution set, then the solution set would have been out here.
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OK, so that was graphing quadratic inequalities.
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Now, we are going to talk about actually solving quadratic inequalities.
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In this case, we are going to again graph the related quadratic function.
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But then, we are going to find the roots of the related quadratic equation.
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So, we are looking for something slightly different: remember, what I just looked at was this,
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where I just had that y is greater than some function.
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Now, this time, I am actually going to say something like x² + x - 6 ≥ 0.
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So, I actually have a number here; I have a parameter; I am saying that this function value is greater than or equal to a specific value.
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So, I am going to start out by finding the roots, because by finding the roots, my graphing will be easier,
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because recall that the roots give me the x-intercept.
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Let's write the related quadratic--let's go ahead and find the roots by writing this as a quadratic equation.
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And looking at this, as you know, there are many different ways to solve quadratic equations.
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In this case, factoring would be an easy way; so I can see that I have a negative here, so I have x + something and x - something.
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And I also know that my middle term is x; so I have to look at factors of 6, which are 1 and 6, and 2 and 3,
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and look for a combination here that will add up to 1x.
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And it actually turns out to be (x + 3) (x - 2), because that would give me -2x + 3x, to give me x for the middle term.
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So, I did my factoring; and then, using the zero product property, I know that x + 3 = 0 and x - 2 = 0.
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So, this gives me x = -3 and x = 2, so those are my roots: x = -3, and it can also equal 2.
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That means that my x-intercepts are at -3 (this is -3) and 2; those are the roots.
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OK, the next thing I am going to do in order to have my graph is find the vertex.
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And recall that the vertex equals -b/2a; the x-value for the vertex equals -b/2a.
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Here, I have that b is 1, so this is -1; a is also 1; so this is going to give me -1/2.
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At the vertex, x equals -1/2; in order to find the y-value, I am going to substitute in.
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And I am going to find out that I am going to get (-1/2)² + 1/2 - 6.
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So, I need to solve this to get my y-value; and this is going to give me 1/2...
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this is actually -1/2 right here...(1/2)² is going to give me 1/4, minus 1/2,
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minus 6, which equals -1/4 - 6, which equals -6 and 1/4.
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OK, so now I have my vertex; my vertex is at (-1/2, -6 1/4).
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So, I can't graph that exactly; but 1, 2, 3, 4, 5, 6...-6 will be right here; and -1/2, -6 1/4 places it about there.
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Now, I also know that, since a is positive (a is greater than 0), this is going to open upward.
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So, I have my vertex here; and I know that this is going to open upward.
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And this is greater than or equal to, so I can use a solid line; OK.
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Now, what we are looking for, if we go back and think about this, is times when f(x), or y, is greater than or equal to 0.
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I can use a test point, or I can just use logic.
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I can look and see that here, at these x-intercepts, this is where y equals 0.
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Beyond that, up here...it was when y equals 0; y equals 0 right here at x = -3, and y equals 0 at x = -2.
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Beyond this point, for x-values over here, and for x-values over here, y is positive: y is greater than 0.
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So, I can actually just look at this graph and say, "OK, when x is less than or equal to -3, or when x is greater than or equal to 2, this inequality is satisfied."
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Now, you can always look at it and do it by logic; or you can use a test point.
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So, if I use a test point of (0,0), what I would do is go back here and say,
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"OK, I am going to let x equal 0; and I am going to see if that satisfies this inequality."
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So, I am using my test point, and then I put it in; now I have 0² + 0 - 6 ≥ 0.
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This gives me 0 - 6 ≥ 0; -6 ≥ 0.
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That is not true--no; therefore, this is not part of the solution set.
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These values in here are not part of the solution set; so that tells me that these are the correct values.
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And I knew that by logic; but you can always use a test point or verify using a test point.
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Therefore, my solutions for this inequality, for times when the function (or y) ends up
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being greater than or equal to 0, are at points when x is less than or equal to -3, or greater than or equal to -2.
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And you can see on the graph that that is true; when x is big over here, y is positive; when x is less than or equal to -3, y is also positive.
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OK, first we are just asked to graph this inequality.
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And we are going to begin by graphing the corresponding quadratic function in order to find the boundary line for the solution.
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The corresponding function is y = x² + 4.
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This gives me an equation with a vertex at (0,4).
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Now, let's find some points and then do the graphing.
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When x is -2, this gives me 4 plus 4, or 8; when x is -1, this is 1, plus 4 is 5.
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I already know that, when x is 0, y is 4, because that is the vertex.
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And I know that this is a minimum, because the parabola opens upward, because a is positive.
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So, a is greater than 0, so this opens upward.
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And when x is 1, this is 1 plus 4; that gives me 5.
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And when x is 2, that is 4 plus 4; that gives me 8.
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So, 2, 4, 6, 8, 2, 4, 6, 8; let's start graphing, starting with the vertex at (0,4).
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And I know this opens upward; and then, when x is -1, y is 5; when x is -2, y is 8.
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When x is 1, y is 5; and when x is 2, y is 8.
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So, because this is less than or equal to, I am going to use a solid line for this parabola.
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And it is opening upward, as I expected.
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Now, I graphed my corresponding quadratic function; I found this parabola; I graphed it;
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and now, I need to use a test point--I am going to use the test point of (0,0) once again, because that is an easy point to use.
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And I am going to go back here, and I am going to say y < x² + 4.
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0 is less than or equal to 0 squared plus 4; 0 is less than or equal to 0 plus 4; 0 ≤ 4.
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And this is true: since this is true, the test point is part of the solution set;
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and this parabola divides this graph into two sections--the area within the parabola and the area outside the parabola.
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And since this is true, then my solution set lies out here.
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So again, this is very similar to graphing linear inequalities, where you find the boundary line
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by graphing the corresponding equation, and then use a test point to determine where the solution set is.
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Now, I am asked to solve this quadratic inequality.
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I am going to start out by finding the roots of the corresponding quadratic equation, x² - x - 6 = 0.
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This is also one that is most easily solved through factoring.
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And since this term is negative, I am going to have x + something, and then x - something.
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And I know that my factors of 6 are 1 and 6, and 2 and 3, and that I want these to sum up to -1.
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1 and 6 are too far apart, so I look at 2 and 3.
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And if I added -3 + 2, I would get -1; therefore, this is (x + 2) (x - 3).
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And I can check that by saying, "OK, that would give me -3x + 2x, so my middle term would come out to -x."
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OK, and this equals 0; using the zero product property, I get x + 2 = 0, and x - 3 = 0.
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So, I solve and get x = -2 and x = 3.
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So, my roots are right here: -2 and 3.
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Something else I know is that this is going to open upward, because a is positive.
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And I am going to go ahead and find the vertex, and that is -b/2a (that is the x-value for the vertex).
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And that is going to give me negative...and b is -1/2, and a is 1.
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So, that equals 1/2; so the vertex is at 1/2.
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Now, to find the y-value, the x-value for the vertex (x = 1/2)--substitute that in to find y (the y point for the vertex).
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That gives me (1/2)² - 1/2 - 6.
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This equals 1/4 - 1/2 - 6, which equals -1/2 - 6...actually, -1/4...that would be minus 2/4, and that would give me -1/4, minus 6 equals -6 1/4.
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Therefore, the vertex lies at (1/2,-6 1/4).
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Now I have the vertex, which is going to be right about down here.
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And I know that this opens upward, and I know that it is going to be a solid line, because this is less than or equal to.
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Now, what I want to know is, "What are the situations when y, or this function, is less than or equal to 0?"
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And I can actually just look on here and say, "OK, y is less than or equal to 0 at all these values for x in here."
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So, when x is less than or equal to 3, I see that the y becomes negative--dips below 0.
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I could always use a test point of (0,0): substitute in (0,0) here; that would give me...
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if my test point is (0,0), I am going to get 0² - 0 - 6 ≤ 0, or -6 ≤ 0.
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And that is true; since that is true, this is part of the solution set.
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And the solution set is comprised of x ≥ -2 and x < 3.
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So, that is going to give me x ≥ -2 and ≤ 3.
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So, that is my solution for this quadratic inequality.
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OK, now graph the inequality: first graph the corresponding quadratic equation.
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And I have y = x² - x - 12, and I am going to start out by finding the vertex at -b/2a, which equals...
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b is -1, so it is negative -1, over 2...a is 1; that is going to give me 1/2.
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The x-value for the vertex is going to be 1/2; what is y going to be?
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y is going to be (1/2)² - 1/2 - 12, so y = -12 1/4.
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So, 2, 4, 6, 8, 10, 12; the vertex is going to be here at 1/2, which is going to be about here, all the way down here.
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And this parabola does open upward, because a is positive.
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Now, I am going to go ahead and find some more points (in addition to the vertex) to graph.
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So, when x is -2, if you figure this out, y ends up being -6.
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When x is -1, this is going to give me 1 + 1; that is 2, so this will come out to -10.
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When x is 0, that is going to give me -12.
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When x is 1, that is going to be 1 - 1 is 0, minus 12; so that is going to be -12.
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When x is 2, that is going to be 4 - 2 is 2, minus 12 is -10.
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OK, graphing these: this is -2 right here; when x is -2, y is -6.
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When x is -1, which is right about here, y is all the way down here at -10.
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When x is 1, y is down here at -12, right about here; when x is 2, we get -10 right here.
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So, I have a graph right here, and it is actually going to be a dashed line; it is going to give me a parabola that is a dashed line,
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because I have greater than; so, my boundary line is not part of the solution set.
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So now, I have this parabola; and I need to use a test point right here.
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Test point (0,0): substituting in here is going to give me 0 > 0² - 0 - 12.
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0 is greater than 0 - 12; 0 is greater than -12.
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OK, so is 0 greater than -12? Yes, therefore, since 0 is greater than -12, this point would be part of the solution set.
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So, the solution set is going to be in here.
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Again, we handled this much the way we would handle a linear inequality.
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And that is by graphing out the corresponding equation, y = x² - x - 12.
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And it gave me this parabola, with a vertex way down here that is a minimum that opens upward.
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Then, I went and found a test point, (0,0), and I substituted this in and came up with 0 > -12.
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And 0 actually is greater than -12, so this test point is part of the solution set, which would be right here within the parabola.
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Now, solve this quadratic inequality; first I am going to go ahead and find the roots,
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which are of the corresponding quadratic equation.
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And this is another one that I can do by factoring; I am going to have x + something, and I am going to have x - something.
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And I know that, because I have a negative 10; so I am going to have a positive and a negative.
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Now, the factors of 10 are 1 and 10, and 2 and 5; and I need something that is going to add up to 3x.
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And what would give me that is if I had a positive 5 and a negative 2, because then I am going to get -2x + 5x, to get 3x.
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Using the zero product property, I will have x + 5 = 0, or x = -5; x - 2 = 0 gives me x = 2.
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So, I have two roots that I found; and these are x = -5 and x = 2: right here and right here.
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Now, I could find the vertex, and I have been finding it; but you actually don't need it for these.
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All you need to know is that the zeroes are here, and this graph opens upward.
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So, I am just going to sketch this out--that this is a parabola that opens upward,
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and I don't know exactly where the vertex is; that is OK, because I am honestly not concerned with that.
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What I am concerned with are these points right here.
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And I am looking for values of x where this function would end up being greater than 0.
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So, I am going to look over here and see that, whenever x is less than -5, y is greater than 0.
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Whenever x is greater than 2, y is also positive.
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I could also use my test point; and again, I use the origin as a test point, as long as it is not too close to the boundary.
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So, my test point is (0,0); and now I am going to use the x-value and put it in here and find that 0² + 3 times 0 - 10 > 0.
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Let's see if that is true: 0 + 0 - 10 > 0--is -10 greater than 0? No.
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So, this test point is not part of the solution set; the solution set is not in here--it is out here, just as I thought.
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So, what this gives me is a solution set where x is less than -5, or x is greater than 2.
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So, my solution set are values of x that are less than -5 or greater than 2.
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That concludes this session of Educator.com on quadratic inequalities.
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And I will see you next lecture!