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# Intro to square-root equations & extraneous solutions

CCSS.Math:

## 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 phenomenon that occurs when we do these equations so 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 B is equal to this squared is going to be 2x squared which is 4 x squared it's 2x squared the whole thing 4 x squared and then you multiply these two which is negative 12x and there's going to be twice that so minus 24x and the 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 two the product of those two is minus 12x or negative 12 X 2 times that is negative 24x and then that's squared so this is what our equation has I guess we could say simplify 2 and let's 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 0 Rho and the right hand side becomes four x squared minus 25 X plus 36 so this radical equation has simplified to just a standard quadratic equation and just for simplicity not having to worry how to factor it in group 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 let's get our calculator out so let's say so we have 625 - 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 over 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 over 8 8 goes into 18 2 times remainder 2 so this is equal to 2 and 2 eighths or two and one-fourth 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 four so this if X is equal to four works we get the principal root of four should be equal to two times four minus six the principal root of four 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 be let me make my erratic 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 should you might be able to figure out that the square root of 2.25 is 1.5 but let's 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 one point five 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 does this is this true this is telling us that 1.5 is equal to negative one point five this is not true two point five did not work for this radical equation we call this an extraneous solution so 2.25 is x trainee Asst 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 right when we substitute 2.25 we get one point five is equal to negative one point five so there's something here that 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 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 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 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 let me be very clear this statement we already established that most 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 so this is really an interesting conundrum and it 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 that you are to some degree you can either think of it as losing or gaining some information now this can be written as 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 we could 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 one the negative one will disappear right 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 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 assuming we're assuming that this was the only interpretation but this was the other one so we found two solutions to this but only only four satisfies this interpretation right here and I hope you get what I'm saying because we're kind of only taking you can imagine at 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 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 the solution 2.25 would have 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 point 5 minus 6 which is negative 1 point 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 if you square this you actually get this equation that both solutions are valid now that you might have found that a little bit confusing and all that I intention is not to confuse you the simple thing to think about when you 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 there are 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 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 prints Paul Drude