Determining the number of solutions of a quadratic graphically
The second equation is graphed and read aloud as y = x^2 + 4x, but the narrator later calls it x^2 + 4x + 5 and says it has no solutions; x^2 + 4x would cross the x-axis at 0 and -4. Interventions avoid relying on that equation's exact form.
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The video looks at three quadratic equations written in the form (expression) = 0 and solves them graphically using Desmos. By graphing y = the left-hand side and looking for x-intercepts, it shows one equation with two solutions, one with no solutions, and one, (x + 5)^2 = 0, with exactly one solution at the vertex.
You'll be able to
- solve a quadratic equation set equal to zero by graphing y = the left-hand side and finding where it crosses the x-axis.
- tell from a graph whether a quadratic equation has two, one or no solutions.
- explain why a parabola that only touches the x-axis at its vertex gives exactly one solution.
Key words
- quadratic equation
- An equation whose highest power of x is 2, like x^2 + 2x - 3 = 0.
- solution
- An x value that makes the equation true; here, where the left-hand side equals zero.
- x-axis
- The horizontal line on a graph where y = 0.
- parabola
- The U-shaped curve you get when you graph a quadratic.
- vertex
- The turning point of a parabola, its lowest or highest point.
Transcript
- [Instructor] So we have three different equations here. And what I wanna do is think about how many solutions there are to each of these equations, and you might notice something in common. They all will have a second degree term. This one over here, if you expand out x + 5 squared, you would get an x squared term as well.
So these are all quadratic equations. Now we'll learn there's many ways to solve these, but I'm gonna first think about this graphically. So how can we think about it graphically in order to solve it? Well, I got out a graphing calculator right over here. This is the Desmos calculator.
And instead of... I can't graph this equation per se, but I could graph the equation of y equals the left-hand side of this equation. So y = x squared + 2x - 3. Whoops, not equals three, minus three. Now why is this useful? Well, if this is the graph of y = x squared + 2x - 3, when this graph equals zero would be when we would satisfy this equation.
And we could see it equals zero in two places, at x = 1 and x = -3. So it has two solutions which is always set out to figure out, but just by looking at it here, it looks pretty clear that those actually are the two solutions. But let's actually compare it to these other ones.
So let me graph that second equation, but I'm gonna graph it as not the equation exactly as it is, but y is equal to the left half of that equation, y = x squared + 4x. And this is useful because all of these equations have already been written in the form where you have all of the constant and x terms on one side and it equals zero.
So if you graph y equals the left-hand side here, you just have to see where it equals zero. Now this one is interesting because when I graph it, it does not intersect the x-axis. At no point does y = 0 here. So this equation actually has no solutions. There are no x values where the left-hand side is equal to zero.
We can see that graphically. It does not intersect the x-axis. Now last but not least, let's graph x + 5 squared is equal... Whoops, y = x + 5 squared, I should say. I'm not graphing those and it kind of just already graphed it that way. I guess it just assumes y equals that, but y = x + 5 squared.
Notice, this intersects the x-axis exactly one place when x = -5. And you can validate that when x = -5, y does indeed equal zero, but that's the only solution. It only intersected the x-axis exactly one place, right at the vertex. So now you see visually the three scenarios that you might run into when you're trying to solve quadratic equations.
You could have two solutions because the parabola that you graph for the left-hand side, y equals the left-hand side, intersects the x-axis twice. You could have no solutions if y = x squared + 4x + 5 does not intersect the x-axis and you could have one solution.
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