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# Quadratic systems: algebraic solution

CCSS Math: HSA.REI.C.7

## 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.