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Current time:0:00Total duration:3:59

CCSS.Math:

let's say that we have the radical equation the square root of 3x minus seven plus the square root of 2x minus one is equal to zero I encourage you to pause the video and see if you can solve for X before we work through it together alright so one thing we could do is we could try to isolate each of the radicals on either side of the equation so let's subtract two let's subtract this one from both sides so I can get it onto the right-hand side so or a version of it on the right-hand side so I'm subtracting it from the left-hand side and from the right and from the right-hand side and so this is going to get us that is going to get us on the left-hand side I just have square root these cancel out so I'm just left with the square root of 3x minus seven is going to be equal to this the negative of the square root of 2x minus one so now we can square both sides and we always have to be careful when we're doing that because whether we're squaring the positive or the negative square root here we're going to get the same value so the solution we might get might be the version when we when we're solving for the positive square root not when we take the negative of it so we have to test our solutions at the end to make sure that they're actually valid for our original equation but if you square both sides on the left hand side we are going to get 3x minus seven and on the right hand side a negative square is just a positive and the square root of 2x minus 1 squared is going to be 2x minus 1 and I'll see we could subtract 2x from both sides to get all of our X's on one side so I'm trying to get rid of this and we can add 7 to both sides because I'm trying to get rid of the negative 7 so add 7 to both sides and we are going to get we are going to get 3x minus 2x is X is equal to negative 1 plus 7x is equal to 6 now let's verify that this actually works so if we look at our original equation the square root of 3 times six minus seven minus seven needs two plus plus the square root of two times six minus one needs to be equal to zero so does this actually work out three times six minus seven so this is going to be the square root of eleven plus the square root of eleven needs to be equal to zero which clearly is not going to be equal to zero this is two square roots of 11 which does not equal zero so this does not work and you might say wait how did this happen I did all of this nice neat algebra I didn't make any mistakes but I got something that doesn't work well this right here is an extraneous solution why is it an extraneous solution because it's actually the solution to the equation it's a solution to the equation the square root of three X minus seven minus the square root of two X minus one is equal to zero and you might say well if it's a solution to that if it's the solution to this thing right over here how did I get the answer while I'm trying to do algebraic steps there well the key is if when we added when we took this onto the right hand side and squared it well it all boiled down to this regardless of what starting point you started with if you did the exact same thing you would've gotten to that say this point right over here so the solution to this ended up being the solution to this starting point versus the one that we originally started with so this one interestingly has no solutions and it actually be fun to think about why it has no solutions we've we've shown to a certain degree the only solution you got by taking reasonable algebraic steps is an extraneous one it's a solution to a different equation that gets that it has a common intermediate step but it's also fun to think about why why this right over here is impossible