Well we're not quite done yet, because we wanted to solve for X, and not for P. So either P equals zero, or P equals two. Well if P minus two is equal to zero, then that means P is equal to two. Needs to be equal to zero, so either P is equal to zero, or P minus two is equal to zero. ![]() Things and they equal to zero, at least one of them And we've seen this shown multiple times, if I have the product of two So we can write this as, if we subtract two P from both sides, we can get P squared minus two P is equal to zero, and we can factor out a P, so we get P times P minus two is equal to zero. Simplifies quite nicely, the left-hand side becomes P squared, P squared is equal to two times P, 'cause once again two X minus three is P, two times P. So let's say that P is equal to two X minus three. ![]() Replace two X minus three, we'll do a little bit of a substitution, let's replace that with P. So if we can solve for what the stuff in blue could be equal to, then we could solve for X, and I'll show you that right now. So if we can solve for the something, let me be very clear here, so the stuff in blue squared is equal to two times the stuff in blue. And so this is really interesting, we have something squared is equal to two times that something. That's just two times two X minus three, let me be clear there, so this is the same thing as two X minus three squared is equal to, four X minus six, if I factor out a two, that's two times two X minus three. Have two X minus three squared on the left-hand side, on the right-hand side Or a simpler way to do this if you really payĪttention to the structure of both sides of this equation. Of a classic quadratic form, but there might be a faster And I'll give you a little bit of a hint, you could do this in the traditional way of expanding this out, and then turning it into kind X minus three squared, and that is equal to four X minus six, and I encourage you to pause the video and give it a shot. Try to find the solutions to this equation right over here. I have ended up with 3 binomials two of which is what Sal got, (2x-5) and (2x-3) and an additional one: (2x+3) ![]() Use grouping, extracting a factor from both sides ![]() Multiply coefficient on 1st term with that of the last term to get a number where to look for sum of factorsĦ*10=60 and 6 -10= -4 <- Error, both of them need to be negative if I want their product to be a positive 60, this condition seems impossible, didn't notice at the time, counter-intuitively I still got both of Sal's results but also an additional one.Ĭan now split the middle term with my erroneous factors Get all of them on one side (-4x and +6 to both sides) I have several questions relating to the way I tried to solve it (which gave me a really counter-intuitive result):ġ) are there any factors of a (positive) 60 whose sum would be -4? (the 60 is positive, the 4 is negative, specifically)Ģ) I suspect the answer to previous is "no"? Why is that? I thought a trinomial can always be factored using sum-product method?ģ) also did I expand these parentheses correctly (2x -3)^2 -> 4x^2 +9 ETA: I found the mistake - the very first step of expanding the parentheses, after doing it correctly everything worked perfectly using my usual method :D
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