Integrated math 3
Extraneous solutions are values that we get when solving equations that aren't really solutions to the equation. In this video, we explain how and why we get extraneous solutions, by understanding the logic behind the process of solving equations.
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- At2:52, how is one extraneous if the square root of one equals plus or minus one?(14 votes)
- When we take the square root of one, or just the square root of 5x-4 in general, we're assuming it's the principle square root. If we were taking the plus or minus square root of 5x-4, this would work.
Another way of looking at this, if we square the square root of 5x-4, this allows for a negative value for a principle square root, because the negative value is squared to become positive. But the original equation allows only for positive square roots, because its the principle square root.
If this makes things more perplexing, you just need to know that the original equation takes the principle square root, and thus allows for no negative values.(9 votes)
- In the example where x=1, 1 - 2 = -1. Isn't -1 a square root of 1 and therefore there is no extraneous solution?(7 votes)
- Sal is looking at the principal square root. The principal square root of 1 is positive 1, not negative 1. Principal square roots are always a nonnegative number.(6 votes)
- in the question like this:
square root of 4x is equal to x-3
What are all values of x that satisfy the given equation?
I choose both because square root of 4 is equal to 2 and -2. But the answer was only II. Is that means on the SAT we should never consider extraneous solution as a true solution?(3 votes)
- That's correct. 'Extraneous' means not relevant to the problem, so we don't accept them as solutions.
Generally, the √x notation refers to the principal root of x when it's used in algebraic equations. We want √x to be a genuine function, which means it can't output two numbers on one input. So if you plug 1 into your equation, it becomes
which is untrue.(7 votes)
- How do I know if a solution is extraneous?(2 votes)
- Plug in your solution back into the original equation. If it shows a false meaning (e.g 2=3) or if the value is undefined (n/0), then it's extraneous.
hopefully that helps !(5 votes)
- Where the -4x + 4 come from? Wouldn't it have been x squared + 2 squared.(2 votes)
- When you square multiple terms, you can't just distribute the exponent like you do with multiplication(You can distribute exponents over multiple factors, however). Instead, you would have to properly multiply out the first binomial by an identical one. Let's say that we had (a + b)^2.
(a + b)^2 = (a + b)(a + b)
= a*a + b*a + a*b + b*b
= a^2 + 2*a*b + b^2
Sal uses a shortcut instead of multiplying everything out, as if you have any numbers a and b, the square of their sum will be a^2 + 2ab + b^2.(4 votes)
- At4:48he says sqrt(1) != -1 but i thought sqrt(1) = +1 or -1.
why does he only take the positive root in this scenario when in many other videos he takes the positive and negative root?(3 votes)
- Because he's talking about the principal sqrt which is always positive(2 votes)
- How would you solve this problem? 1 over 50 + 1 over x = 1 over 30(2 votes)
- 1/50 + 1/x = 1/30
1/x = 1/30 - 1/50
1/x = 50/50 * 1/30 - 30/30 * 1/50
1/x = 50/1500 - 30/1500
1/x = (50 - 30)/1500
1/x = 20/1500
x * 1/x = x * 20/1500
1 = x * 20/1500
1500 * 1 = x * 20/1500 * 1500
1500 = x * 20
1500 / 20 = x * 20 / 20
1500/20 = x
x = 75(4 votes)
- At2:48he says "Well, if a was two and b is three" but that is not possible because a=b?(3 votes)
- At2:58, Sal explains that 0 is an example of an x value for which xa is not equal to xb. Is there another number other than 0 for which that would be true?(2 votes)
- What Sal says is that for 𝑥 = 0,
𝑥𝑎 = 𝑥𝑏 is always true, regardless of whether 𝑎 is equal to 𝑏 or not.
There are no other numbers that do this.
For any 𝑥 ≠ 0, we can divide both sides of the equation by 𝑥:
𝑥𝑎 = 𝑥𝑏 ⇒ 𝑥𝑎∕𝑥 = 𝑥𝑏∕𝑥 ⇒ 𝑎 = 𝑏(2 votes)
- At3:45arent you supposed to square, and not foil, what sal did is foiling, why didnt he not just square both sides to get 5x-4=x^2+4(2 votes)
- No, you really have to put expressions in parentheses when you square, so (√(5x-4))^2 = (x+2)^2. Doing it your way gives x^2-5x+8 and determinant gives 5^2-4(1)(8)=25-40=-15 which give imaginary solutions. Sal finds real solutions where 1 is extraneous when substituted back in, but 8 works as a solution to the original.(2 votes)
- [Instructor] In this video, we're going to talk about extraneous solutions. If you've never heard the term before, I encourage you to review some videos on Khan Academy on extraneous solutions. But just as a bit of a refresher, it's the idea that you do a bunch of legitimate algebraic operations, you get a solution or some solutions at the end, but then when you test it in the original equation, it doesn't satisfy the original equation. And so the key of this video is why do extraneous solutions even occur? And it all is due to the notion of reversibility. There's certain operations in algebra that you can do in one direction, but you can't, and it'll always be true in one direction, but it isn't always true in the other direction. And I'll show you those two operations. One is squaring and the other is multiplying both sides by a variable expression. So let's just see the example of squaring, and then we're going to see it in an actual scenario where you're dealing with an extraneous solution. So we know, for example, that if a is equal to b, I could square both sides and then a squared is going to be equal to b squared. But the other way is not true. For example, if a squared is equal to b squared, it is not always the case that a is equal to b. What's a example that shows that this is not always the case? Actually, pause the video, try to think about it. Well, negative two squared is indeed equal to two squared, but negative two is not equal to two. So this shows that you can square both sides of an equation and deduce something that is true, but the other way around is not necessarily going to be true. Another non-reversible operation sometimes is multiplying both sides by a variable expression. So multiplying both sides, actually, like I said, that will make us confused, looks like an x. Multiply both sides by variable. I'll just write variable, but it could be a variable expression as well. For example, we know that if a is equal to b, that if we multiply both sides by a variable, that's still going to be true. xa is going to be equal to xb. But the other, the reverse, isn't always the case. If xa is equal to xb, is it always the case that a is equal to b? Well, the simple answer is no, and I always encourage you, pause this video and see if you can find an example where this doesn't work. Well, if a was two and b is three and the variable x just happened to take on the value zero. So we know that zero times two is indeed equal to zero times three, but two is not equal to three. Now, how does all this connect to the extraneous solutions you've seen when you were solving radical equations or when you were solving some rational, or equations with rational expressions on both sides? Well, let's look at an example. Let's solve a radical equation. If I wanted to solve the equation the square root of five x minus four is equal to x minus two, a typical first step is, hey, let's get rid of this radical by squaring both sides. So I'm going to square both sides, and then I'm going to get five x minus four is equal to x squared minus four x plus four. Once again, if this looks completely unfamiliar to you, we go into much more depth in other videos where we introduce the idea of radical equations. And let's see, we can subtract five x from both sides. We can add four to both sides. I'm just trying to get a zero on the left hand side. And so I'm going to be left with zero is equal to x squared minus nine x plus eight, or zero is equal to x minus eight times x minus one, or we could say that x minus one is equal to zero, or x minus eight is equal to zero. We get x equals one or x equals eight. So let's test these solutions. If x equals eight, we would get, and I'll color code it a little bit, for x equals eight, if I test it in the original equation, I get the square root of 36 is equal to six, which absolutely true, so that one works. But what about x equals one? I get the square root of five times one minus four is one is equal to one minus two, which is equal to negative one. That did not work. This right over here is an extraneous solution. If someone said, what are all the x values that satisfy this equation, you would not say x equals one, even though you got there with legitimate algebraic steps. And the reason that is true is, actually, pause this video, look back. For which of these steps does x equal one still work and what step does it not work? Well, you'll see that x equals one works for all of these equations below this purple line. It just doesn't work for the square root of five minus four x is equal to x minus two. In fact, you could start with x minus one and then you could deduce all the way up to this line here. But the issue here is that squaring is not a reversible operation. This is analogous to saying, hey, we know that a squared is equal to b squared. We know that this is equal to this. But then that doesn't mean that a is necessarily equal to b for x equals one. And we could do the same thing with a rational, or an equation that deals with rational expressions. So, for example, we might have to deal with, and let me make sure I have some space here. If I had to solve x squared over x minus one is equal to one over x minus one, the first thing I might wanna do is multiply both sides by x minus one. So multiply by x minus one. Now, notice I'm multiplying both sides by a variable expression, so we have to be a little bit contentious now, but if I'd multiply both sides by x minus one, I'm going to get x squared is equal to one, or I could say that x equals one or x is equal to negative one. Well, we could test these. For x equals one, if I go up here, I'm dividing by zero on both sides. So this is an extraneous solution. They key here is that we multiplied both sides by a variable expression. In this case, we multiplied both sides by x minus one. You can do that. You can multiply both sides by a variable expression, and it is a legitimate algebraic operation. It's completely analogous to what we saw right over here. Just because zero times two is equal to zero times three does not mean that two is equal to three. It's completely analogous because we multiplied by a variable expression that actually takes on the value zero when x is equal to one. So the big takeaway here is, hopefully you understand why extraneous solutions happen a little bit more. When you square, when you multiply both sides by a variable expression, completely legitimate as long as you do it properly, but it's not always the case that the reverse is true. You could add or subtract anything from both sides of an equation, and that's always going to be reversible. And so that's not going to lead to extraneous solutions. You could multiply or divide by a non-zero constant value. That's also not going to lead to anything shady, but if you're squaring both sides or multiplying both sides by a variable expression, you should be a little bit careful.