Exact vs. approximate solutions to quadratic equations
At about 50s the narrator says '8x squared plus 8x' when rewriting x^2 + 8x (uncorrected misspeak); he also briefly says x = 4 + sqrt(11) before correcting to -4 + sqrt(11).
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The video solves x^2 + 8x + 5 = 0 by completing the square, since it does not factor nicely, to get the exact solutions x = -4 + sqrt(11) and x = -4 - sqrt(11). It uses a calculator to approximate these as -0.683 and -7.317, then graphs y = x^2 + 8x + 5 in Desmos to see the same x-intercepts. The key point is that completing the square gives exact answers, while calculators and graphs give approximations.
You'll be able to
- solve a quadratic that does not factor easily by completing the square.
- write exact solutions using square roots and find decimal approximations with a calculator.
- explain the difference between exact and approximate solutions and connect them to x-intercepts on a graph.
Key words
- completing the square
- Adding (half the x coefficient)^2 to make a perfect square like (x + 4)^2.
- exact solution
- A precise answer, often written with a square root, like -4 + sqrt(11).
- approximate solution
- A rounded decimal close to the exact value, like -0.683.
- x-intercept
- A point where a graph crosses the x-axis, where y = 0.
Transcript
- [Instructor] What we're going to do in this video is try to solve the equation x squared plus 8x plus 5 is equal to 0. And then we're also going to see if we can think about it graphically. So like always, pause this video. Have a go at this before we do this together.
Okay, whenever I see something like this where I have a quadratic expression on the left-hand side, and I have it equaling zero, my instinct is to see, okay, can I factor this in any way? The first thing I might say is, "Are there two numbers that if I multiply I get to 5, and if I were to add them, I would get 8?"
And I really can't think of easy numbers there, integers there, so what I'm going to do is complete the square. And to do that, I am going to just rewrite this, 8x squared plus 8x, and I'm just going to think about, what could I add here so that I have a complete square and actually, or I should say, so I have a perfect square?
So notice, all I've done is I've rewritten what I have up there, but I've written it slightly different spacing. Now if we take half of this value, half of 8 is 4, and if we were to square it, we put a 16 here. Well, if I add 16 to one side of the equation, I have to add it to both in order for the equation to be true.
Notice I still haven't changed this equation. I just added 16 to both sides of it. I guess I've changed it, but I haven't fundamentally changed it. And the reason why I did that is now what I have in the left in this bubble gum color, which I like, I haven't historically used bubblegum, is we can rewrite this as x plus 4 squared.
This is what completing the square is all about. And then that is going to be... Or plus 5 is equal to, we get this 16 here. Now we could subtract 5 from both sides, and I'll switch back to one color. So we get x plus 4 squared. If we subtract 5 from both sides, we get 11.
And then we could get x plus 4 is equal to, it could be the positive or negative square root of 11. Notice, if it's the positive square root of 11, well, if you square it, you're going to get 11. If it's a negative square root of 11, and you square it, you're also going to get 11.
So now to solve for x, we just subtract 4 from both sides of this equation. So if we take the positive square root of 11, we would get x is equal to 4 plus the square root of 11. Oh, let me be careful. We're subtracting 4 from both sides. So, negative 4 plus the square root 11, or we could have x is equal to, we subtract 4 from both sides.
But if we use the negative square root of 11, so we have negative 4 minus the square root of 11. Now we could also get a calculator out to figure out approximately what these are. So if I have -4 plus the square root of 11, I guess I'll close that parentheses there, that gets me to, if I round to three decimal places, this is approximately equal to
- 0.683. And then over here, if I evaluate the other one, if I have -4 minus the square root of 11, close those parentheses, that is approximately equal to
- 7.317. Now let's see what happens when we graph this same equation and see if we can approximate or if we see these same values. So I'm going to get Desmos out right over here, and actually let me just type in our equation. And what I'm going to type in is y is equal to x squared plus 8x plus 5, and I'm going to see at what x values is y equal to zero.
So let me zoom out a little bit. All right, so all the interesting things are right over here. So let me now zoom in a little bit right over there. I want to see this behavior. And so we see we get pretty close to zero if we approximate, and this is why we are still approximating.
We get about negative... Let's see as close as I can get. Let me zoom in a little bit more. So right over here, we get... Actually, an even cleaner way of trying to figure this out is I could say, let me draw a line, y is... Let me show you what I'm doing. Let me draw another line, y is equal to zero, and then it might show me the intersection points a little bit clearer.
So that point right over there is, yes, -7.317 if I were to round, which is what we got with our calculator. And similarly, this one right over here is -0.683, which is what we also got on our calculator. But it's important to realize that these are approximations.
We're really dealing with expressions that involve the square root of 11. And so that's why you can see completing the square actually gets you the exact answer, while, if you use your calculator or a graphing calculator like we just did, you're really getting an getting an approximation.
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