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Intro to square-root equations & extraneous solutions

Sal explains what square-root equations are, and shows an example of solving such an equation and checking for extraneous solutions. Created by Sal Khan and CK-12 Foundation.

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  • blobby green style avatar for user gna0905
    Hello, I have a question/concern . My question/concern is: At 5;53, one can see that the square root of 2.25 is 1.5. However, the square root of 2.25 can also be -1.5. If we put, " -1.5=4.5-6", x being equal to 2.25 would work. So why doesn't x=2.25 work?
    (31 votes)
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    • female robot grace style avatar for user Ethan Boggs
      The original equation is (sqrt)x=2x-6. When you see a radical with no + or - sign before it, we assume that we are only taking the principal root (the positive version). When he plugs that 2.25 into the equation, he can't use -1.5 because of that.
      (29 votes)
  • stelly yellow style avatar for user Kirstin Crump
    So why are there extraneous solutions?

    I know it's because a square root function (parent graph) has only positive y-values. But why can't we take both roots? Why can't there be 2 y-values for every x-value? A square root function is basically a 90 degree rotation of a quadratic function, so why does it only get half?

    I know it's because we want it to be a function. But why does it need to be a function? And who decides what a function is anyway? Who says a function can only have one y-value for every x-value?
    (12 votes)
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    • leafers ultimate style avatar for user tintintitli2008
      Interesting question. The thing is a function itself is defined as having one output for every input. If there are two or more outputs for any input, then you no longer have a function. Also, as to "who decides it gets to be a function", pretty much any equation with both variables and constants with non-ambiguous solutions (i.e. y = sqrt(x)) is a function. So let's see how that relates to your claim that a square root function is basically a 90 degree rotation of a quadratic equation.

      Sadly I can't put pictures here but if you go to Desmos and graph both y=sqrt(x) and y=-sqrt(x), you'd see you'd basically get the 90-degree rotated quadratic function. However, now we don't even have a function. There are two values per input. Also, this can't be represented in a single equation/function except for y=(plus or minus)sqrt(x). But that's really just the two equations we just graphed, since plus or minus is not an operator but rather the choice between addition or subtraction. The plus or minus operator cannot be used for functions because it just splits the whole equation into two (although it could be used to find solutions of a function like in the quadratic formula).

      So in short, the square-root equation is a function by nature because it returns one value per input. Keeping it as a function makes everything clearer. Using the non-function definition of square root really gives you two equations instead of one, so for simplicity, the square root in math returns a positive value by nature. Extraneous solutions from radical equations exist whenever it is assumed that the principal root can return two values in one function.
      (14 votes)
  • aqualine ultimate style avatar for user Cammy
    At or so, Sal multiplies 2x and -6 to get -12x which doubles to -24x. Why? I was understanding everything up till that point.
    (13 votes)
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    • starky sapling style avatar for user Nicky
      The reason why is that when you multiply 2x and -6 you have to do it twice have you ever heard of the F.O.I.L. method
      F.O.I.L. is for squaring a set of two numbers it stands for First Outer Inner Last
      For example if you do (3x+2)^2 you would rewrite it as (3x+2)(3x+2)
      Then you would do 3x^2+2^2+3x*2^2
      (12 votes)
  • leaf green style avatar for user Baukje Verwiel
    At , why did he take 25^2 and not -25^2 like he did with the -b part of the abc formula?
    (12 votes)
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  • aqualine ultimate style avatar for user Michelle Woodard
    What is the definition of "extraneous" and how do I know that an answer is extraneous?
    (5 votes)
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    • aqualine ultimate style avatar for user Jasmine Hansen
      “Extraneous” means “irrelevant or unrelated to the subject being dealt with”. In math, an extraneous solution is a solution that emerges during the process of solving a problem but is not actually a valid solution. You can only find out whether or not a solution is extraneous by plugging the solution back into the original equation.
      (17 votes)
  • leafers tree style avatar for user Oluchi Onyimah
    At , Sal uses the "principlal" root of 2.25. Is the principal root always the positive root?
    (7 votes)
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  • male robot johnny style avatar for user Rich Loper
    How'd you get 625-4 x 4 x 36/8?
    (10 votes)
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  • piceratops seed style avatar for user jjohnson
    what is 4 radical 3(2 radical 6)
    (8 votes)
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  • primosaur sapling style avatar for user Jess McC
    why does the 24x become 25x after subtracting the x from both sides around ?
    (6 votes)
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  • marcimus orange style avatar for user Viktor M.
    At Sal tells that "only 4 satisfies this interpretation".

    The interpretation he means is x = (2x-6)^2
    Opposing interpretation is x = (6-2x)^2

    We have two roots: 4 and 2.25 (extraneous)

    I plug both in to x = (2x-6)^2
    4 = (2*4 - 6)^2
    4 = (8 - 6)^2
    4 = (2)^2
    4 = 4 valid

    2.25 = (2*2.25-6)^2
    2.25 = (4.5-6)^2
    2.25 = (1.5)^2
    2.25 = 2.25 valid

    other interpretation
    x = (6-2x)^2

    4 = (6-8)^2
    4 = (-2)^2
    4 = 4 valid

    x = (6-2x)^2
    2.25 = (6-4.5)^2
    2.25 = (1.5)^2
    2.25 = 2.25 valid

    So what does Sal mean here, if we see that both roots fit?
    (5 votes)
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    • stelly blue style avatar for user Kim Seidel
      First, you need to check the solutions using the original equation (it has square roots). The process of squaring both sides of the equation to eliminate the square roots causes sign changes. We use it to help us solve the equation and find potential solutions. But, you can't use it again in the check as it will make the extraneous solution look correct (as you have demonstrated)

      Second, you have a sign error: 4.5-6 = -1.5, not 1.5

      Sal does the checks correctly (using the original equation and not squaring the equation). When he checks the solution 2.25, the check fails with 1.5 = -1.5

      Hope this helps.
      (6 votes)

Video transcript

In this video, we're going to start having some experience solving radical equations or equations that involve square roots or maybe even higher-power roots, but we're going to also try to understand an interesting phenomena that occurs when we do these equations. Let me show you what I'm talking about. Let's say I have the equation the square root of x is equal to 2 times x minus 6. Now, one of the things you're going to see whenever we do these radical equations is we want to isolate at least one of the radicals. There's only one of them in this equation. And when you isolate one of the radicals on one side of the equation, this one starts off like that, I have the square root of x isolated on the left-hand side, then we square both sides of the equation. So let us square both sides of the equation. So I'll just rewrite it again. We'll do this one slowly. I'm going to square that and that's going to be equal to 2x minus 6 squared. And squaring seems like a valid operation. If that is equal to that, then that squared should also be equal to that squared. So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to-- this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x. And then it's going to be twice that, so minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually, the special case where we square binomials. But the general view, it's this squared, which is that. And then you have minus 2 times the product of these 2. The product of those two is minus 12x or negative 12x. 2 times that is negative 24x, and then that squared. So this is what our equation has I guess we could say simplified to, and let's see what happens if we subtract x from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes zero and the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation his simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and grouping and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this, x can be negative b. Negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625, minus 4 times a, which is 4, times c, which is 36, all of that over 2 times 4, all of that over 8. So let's get our calculator out to figure out what this is over here. So let's say so we have 625 minus-- let's see., this is going to be 16 times 36. 16 times 36 is equal to 49. That's nice. It's a nice perfect square. We know what the square root of 49 is. It's 7. So let me go back to the problem. So this in here simplified to 49. So x is equal to 25 plus or minus the square root of 49, which is 7, all of that over 8. So our two solutions here, if we add 7, we get x is equal to 25 plus 7 is 32, 32/8, which is equal to 4. And then our other solution, let me do that in a different color. x is equal to 25 minus 7, which is 18/8. 8 goes into 18 two times, remainder 2, so this is equal to 2 and 2/8 or 2 and 1/4 or 2.25, just like that. Now, I'm going to show you an interesting phenomena that occurs. And maybe you might want to pause it after I show you this conundrum, although I'm going to tell you why this conundrum pops up. Let's try out to see if our solutions actually work. So let's try x is equal to 4. If x is equal to 4 works, we get the principal root of 4 should be equal to 2 times 4 minus 6. The principal root of 4 is positive 2. Positive 2 should be equal to 2 times 4, which is 8 minus 6, which it is. This is true. So 4 works. Now, let's try to do the same with 2.25. According to this, we should be able to take the square root, the principal root of 2.2-- let me make my radical a little bit bigger. The principal root of 2.25 should be equal to 2 times 2.25 minus 6. Now, you may or may not be able to do this in your head. You might know that the square root of 225 is 15. And then from that, you might be able to figure out the square root of 2.25 is 1.5. Let me just use the calculator to verify that for you. So 2.25, take the square root. It's 1.5. The principal root is 1.5. Another square root is negative 1.5. So it's 1.5. And then, according to this, this should be equal to 2 times 2.25 is 4.5 minus 6. Now, is this true? This is telling us that 1.5 is equal to negative 1.5. This is not true. 2.5 did not work for this radical equation. We call this an extraneous solution. So 2.25 is an extraneous solution. Now, here's the conundrum: Why did we get 2.25 as an answer? It looks like we did very valid things the whole way down, and we got a quadratic, and we got 2.25. And there's a hint here. When we substitute 2.25, we get 1.5 is equal to negative 1.5. So there's something here, something we did gave us this solution that doesn't quite apply over here. And I'll give you another hint. Let's try it at this step. If you look at this step, you're going to see that both solutions actually work. So you could try it out if you like. Actually, try it out on your own time. Put in 2.25 for x here. You're going to see that it works. Put in 4 for x here and you see that they both work here. So they're both valid solutions to that. So something happened when we squared that made the equation a little bit different. There's something slightly different about this equation than that equation. And the answer is there's two ways you could think about it. To go back from this equation to that equation, we take the square root. But to be more particular about it, we are taking the principal root of both sides. Now, you could take the negative square root as well. Notice, this is only taking the principal square root. Going from this right here-- let me be very clear. This statement, we already established that both of these solutions, both the valid solution and the extraneous solution to this radical equation, satisfy this right here. Only the valid one satisfies the original problem. So let me write the equation that both of them satisfy. Because this is really an interesting conundrum. And I think it gives you a little bit of a nuance and kind of tells you what's happening when we take principal roots of things. And why when you square both sides, you are, to some degree, you could either think of it as losing or gaining some information. Now, this could be written as x is equal to 2x minus 6 squared. This is one valid interpretation of this equation right here. But there's a completely other legitimate interpretation of this equation. This could also be x is equal to negative 1 times 2x minus 6 squared. And why are these equal interpretations? Because when you square the negative 1, the negative 1 will disappear. These are equivalent statements. And another way of writing this one, another way of writing this right here, is that x is equal to-- you multiply negative 1 times that. You get negative 2x plus 6 or 6 minus 2x squared. This and this are two ways of writing that. Now, when we took our square root or when we-- I guess there's two ways you can think about it. When we squared it, we're assuming that this was the only interpretation, but this was the other one. So we found two solutions to this, but only 4 satisfies this interpretation right here. I hope you get what I'm saying because we're kind of only taking-- you can imagine the positive square root. We're not considering the negative square root of this, because when you take the square root of both sides to get here, we're only taking the principal root. Another way to view it-- let me rewrite the original equation. We had the square root of x is equal to 2x minus 6. Now, we said 4 is a solution. 2.25 isn't a solution. 2.25 would've been a solution if we said both of the square roots of x is equal to 2x minus 6. Now you try it out and 2.25 will have a valid solution here. If you take the negative square root of 2.25, that is equal to 2 times 2.25, so that is equal to 4.5 minus 6, which is negative 1.5. That is true. The positive version is where you get x is equal to 4. So that's why we got two solutions. And if you square this-- maybe this is an easier way to remember it. If you square this, you actually get this equation that both solutions are valid. Now, you might have found that a little bit confusing and all of that. My intention is not to confuse you. The simple thing to think about when you are solving radical equations is, look, isolate radicals, square, keep on solving. You might get more than one answer. Plug your answers back in. Answers that don't work, they're extraneous solutions. But most of my explanation in this video is really why does that extraneous solution pop up? And hopefully, I gave you some intuition that our equation is the square root of x. The extraneous solution would be valid if we took the plus or minus square root of x, not just the principal root.