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Sal solves a system of two quadratic equations that are solved for y by substituting the expression from one equation for y in the other equation. Created by Sal Khan and Monterey Institute for Technology and Education.
Video transcript
Solve the system of equations by substitution. Check your solution by graphing the equation. Let's do it by substitution. We know that y is equal to this thing over here, and we also know that y is equal to this thing over here. So they're going to intersect when they both equal the same y, or when this thing is equal to this thing, or when negative x squared is equal to 2x squared plus 3x minus 6. The x value, which these two are equal, is going to be the x values where these y's are equal, so that's going to be their point of intersection. That will be an x and y pair that satisfies both of these equations. Let's solve for x here. A good starting point-- let's just add x squared to both sides, and we end up with 0 is equal to 3x squared plus 3x minus 6. Then, just to simplify this on the right hand side, we could divide both sides of the equation by 3-- we can divide everything by 3-- and we're left with 0 is equal to x squared plus x minus 2. We could do the quadratic equation, or complete the square, or all sorts of crazy things, but this is actually very factorable-- 0 is equal to two numbers, which are 2 and negative 1. When you multiply them, you get negative 2, and when you add them, you get positive 1. So this is x plus 2 times x minus 1. That tells us that x plus 2 is equal to 0, or x minus 1 could be equal to 0. Subtract 2 from both sides of this. You could get x is equal to negative 2. Or add 1 to both sides of this, x is equal to 1. These are our two solutions-- x is equal to negative 2 or x is equal to 1. Let's verify it. When I put x is equal to negative 2-- let's do it over here-- what do I get? Negative 2 squared is positive 4. And then you put a negative there, so it's negative 4-- the point here is negative 2, and then negative 4. That's what happens when you put negative 2 there. And then what happens when I put 1? 1 squared is 1, and then if you put a negative there, it's negative 1. So it's 1, negative 1. And so these are both points on this equation right here, on this function. If you look at this one over here, if you put negative 2 over here-- 2 times negative 2 squared. Negative 2 squared is positive 4. 2 times positive 4 is 8. 3 times negative 2 is negative 6, so 8 minus 6. 8 minus 6 is 2, and 2 minus 6 is equal to negative 4. The point negative 2, negative 4 is on this function. They both share that point in common, so they intersect there. If I put 1 into this, I get 2 plus 3 minus 6. 2 plus 3 is 5, minus 6, when x is 1, which is negative 1. The point 1, negative 1 is also on this top graph. We can plot these points: negative 2 comma negative 4. Negative 2-- 1, 2, 3, 4-- is right there. That's going to be a point of intersection. Then we have the point 1, negative 1. 1, negative 1 is also going to be a point of intersection there. Let's graph this second one and just verify-- they say, check your solution by graphing the equation. Let's graph the first one. This is pretty easy to graph-- y is equal to negative x squared. It's going to intersect the point 0, 0. It's going to be a downward slope, downward opening parabola. When x is positive or negative 1, y is going to be negative 1, because you square them and then you take the negative. When x is positive or negative 2, y is going to be negative 4. So the graph will look something like this-- 4, 5, 6, 7, 8, 9. The graph will look something like that. That is that equation right here. This is going to be an upward opening parabola, and a quick way-- we could do all of that completing the square, but if you think about it, let's just apply the quadratic equation right here. I want to show you something, a quick way of where the vertex formula comes from when you apply the quadratic formula. If you take this, and you wanted to find its zeros, you would say x is equal to negative b-- which is negative 3-- plus or minus the square root of b squared-- which is 9-- minus 4ac. Minus 4 times 2 times negative 6. All of that over 2 times a, or 2 times 2. Now, we can evaluate this and figure out the zeros of this, but the real point why I wanted to show you this is because we always have two solutions. Let me actually write down the quadratic formula here for you. The quadratic formula is negative b plus or minus the square root of b squared minus 4ac over 2a. You always end up-- or if this is a positive number-- with two solutions, and they're equal distant. They're this far away, this far over 2a away from negative b over 2a-- we could write this as negative b over 2a plus or minus the square root of b squared minus 4ac over 2a. You have at most two solutions that are equal distant from this x value right here. We've seen in multiple videos: what is that point that is equidistant from the two solutions, that is right in between? That's going to be the line of symmetry, or the x value of the vertex. This is where the x value of the vertex formula comes from: x val of vertex is negative b over 2a. If we wanted to find the vertex, the x value of this guy right here, we just take negative b-- which is negative 3-- over 2 times a, 2 times 2, which is 4. So, x is equal to negative 3/4 is the vertex of this parabola. And when x is equal to negative 3/4, what is y? We could actually-- that's a little bit more complicated right here, but I'll just go through it. It's going to be 2 times 9 over 16 minus 9 over 4 minus 6. Let me actually get the calculator out for this one. It's going to be 18 over 60. Let me just get the calculator out, because it will be simpler, and I don't want to waste your time doing arithmetic. It's going to be 18 divided by 16 minus 9 divided by 4 minus 6 is equal to negative 7.125. This is equal to negative 7.125. So the vertex occurs when x is negative 3/4-- when x is right there-- and y is negative 7. This is essentially 7 1/8, so 1, 2, 3, 4, 5, 6, 7.125. So it's a little over 7. The vertex of this top graph is right over there. It's symmetric around the vertex. That is the line of symmetry, and so this top graph is going to look something like this. Just like that and we're done. We've found our two points of intersection, right over there and right over there, and when you graph it it looks pretty good.