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Current time:0:00Total duration:5:23

Square-root equations intro

CCSS.Math:

Video transcript

let's say we have the radical equation 2x minus one is equal to the square root of eight minus X so we already have the radical isolated on one side of the equation so we might say well let's just get rid of the radical let's square both sides of this equation so we might say that this is the same thing as 2x minus 1 squared is equal to the square root of 8 minus X 8 minus x squared and then we would get let's see 2x minus 1 squared is 4x squared minus 4x plus 1 is equal to 8 minus X now we have to be we have to be very very very careful here you might feel okay we did legitimate operations we did the same thing to both sides that these are equivalent equations but they aren't quite equivalent because when you're squaring something one way to think about it is when you're squaring it you're losing information so for example this would be true even if the original equation were 2x I'm just in a different color even if the original equation where 2x minus 1 is equal to the negative of the square root of 8 minus X because if you squared both sides of this you would also get you would also get that right over there because the negative squared would be equal to a positive so when we're finding a solution to this we need to test our solution to make sure it's truly the solution to the first yellow equation here and not the solution to this up here if it's a solution to this right-hand side or not the yellow one then we would call that an extraneous solution so let's see if we can solve this so let's let's write this as kind of a standard quadratic let's let's subtract 8 from both sides so let's subtract 8 from both sides to get rid of this 8 over here and let's add X to both sides so plus X plus X and we are going to get we are going to get 4x squared minus 3x minus 7 minus 7 is equal to is equal to zero and let's see it we would want to factor this right over here and let's see maybe I could do this by if I do it by well I'll just use the quadratic formula here so the solutions are going to be X is going to be equal to negative B so 3 plus or minus the square root of B squared so negative 3 squared is 9 minus 4 times a which is 4 times C which is negative 7 so I could just say x I'll just write I'll write a 7 here then that negative is going to make this a positive all of that over 2a so 2 times 4 is 8 so it's going to be 3 plus or minus the square root of let's see 4 times 4 is 16 times 7 16 times 7 is going to be 70 plus 42 let me make sure I'm doing this right so 16 times 7 - 4 so it's 112 plus 9 so 121 that worked out nicely so plus or minus the square root of 121 all of that over 8 well that is equal to 3 plus or minus 11 all of that over all of that over 8 so that is equal to if we add 11 that is 14 eighths or if we subtract 11 3 minus 11 is negative 8 negative 8 divided by 8 is negative 1 so we have to think about you might say okay I found two solutions to the radical equation remember one of these might be solutions to this alternate radical equation that got lost when we square both sides we have to make sure that they're legitimate or maybe one of these is an extraneous solution in fact one is very likely the solution to this radical equation which wasn't our original goal so let's see let's try out x equals negative 1 if x equals negative 1 we would have 2 times negative 1 minus 1 is equal to the square root of 8 minus negative 1 so that would be negative 2 - one is equal to the square root of is equal to square root of 9 and so we'd have negative 3 is equal to the square root of 9 the principal or nine is the positive square root this is not true so this right over here that is an extraneous solution extraneous extraneous solution it is a solution to this one right over here because notice for that one you if you substitute 2 times negative 1 minus 1 is equal to the negative of 8 minus negative 1 so this is negative 3 is equal to the negative of 3 so it checks out for this one so this one right over here is the extraneous solution this one right over here is going to be the actual solution for our original equation and you can test it out on your own I in fact I encourage you to do so