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CCSS.Math:

solve the equation x squared minus x squared minus 4 over X minus 2 is equal to 4 and they tell us that X won't or cannot be 2 because if it was 2 then this would be undefined to be dividing by 0 so let's see what X is let's see the X that actually satisfies this equation so you might be tempted to try to express this with the common denominator of X minus 2 and then add these two expressions but the thing that jumps out at me initially is that we have x squared minus 4 on the numerator which is a difference of squares or if you factor it out it's X plus 2 times X minus 2 so we should be able to factor this X minus 2 out so let's do that so if we were to rewrite it this is equivalent to x squared minus instead of writing x squared minus 4 we know that's a difference of squares that is X plus 2 times X minus 2 all of that over X minus 2 and that is equal to 4 and this whole time we're assuming that X won't be equal to 2 and because X does not equal to 2 X minus 2 divided by X minus 2 is going to be defined and it will be 1 so those two will cancel out and so we're left with x squared minus X plus 2 is equal to 4 we can distribute the negative sign and I'll just arbitrarily switch colors here we can distribute the negative sign so we get x squared minus X minus 2 is equal to 4 and what we want to do is put this in the form ax squared plus BX plus C is equal to 0 that allows us to either factor it or apply the quadratic equation or complete the square or any of the ways that we know how to solve quadratic so let's do that let's get a 0 on the right hand side the best way to do that is to subtract 4 from both sides of this equation subtract 4 and we are left with x squared minus X negative 2 minus 4 is negative 6 and then 4 minus 4 is 0 that was the whole point so we have x squared minus X minus 6 is equal to 0 let me write it up here x squared minus X minus 6 is equal to 0 and this looks factorable just have to think of two numbers and when we multiply them give us negative 6 so they're gonna have different signs when I add them now I'll get negative 1 so it looks like negative three and positive two works so if we do X minus three times X plus two it's a little bit of trial and error but six doesn't have that many factors to deal with and three and two or one apart they're different signs so that's how you can think of how we get to that conclusion negative three times two is negative six negative three plus two is negative one so that is equal to zero so we have two possible two possible ways to get 0 either X minus 3 is equal to 0 or X plus 2 is equal to 0 we have to take X minus 3 is equal to 0 if we add 3 to both sides of that equation we get X is equal to 3 or if we subtract 2 from both sides of this equation we get X is equal to negative 2 so both of these are solutions and let's apply them into this equation to make sure that they work because this is these are solutions to essentially the situation where we we got rid of the X minus 2 maybe it had some type of side effect so let's just make sure that both of these actually work in the original equation so let's try X is equal to 3 first so you get 3 squared minus 3 squared over 4 3 squared sorry 3 squared minus 4 over 3 minus 2 so this is equal to 9 minus 3 squared is 9 minus 4 which is 5 over 1 so 9 minus 5 which is equal to 4 which is exactly what we needed to equal now let's try it with negative 2 so if I take negative 2 squared negative 2 squared and I have a minus negative 2 squared minus 4 all of that over negative 2 minus 2 so negative 2 squared is 4 minus negative 2 squared which is 4 minus 4 all of that over negative 2 minus negative 2 over negative 4 well 4 minus 4 is 0 so this whole thing is just going to become 0 so this whole thing is going to equal 4 so both of these solutions work and that makes sense because when we actually cancel this out we never we actually didn't fundamentally change anything about the equation only if you had the situation where the X would have been equal to 2 that's the only thing that you're really changing so that's why it makes that both of these solutions work