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# Solving quadratics using structure

CCSS Math: HSA.REI.B.4, HSA.REI.B.4b, HSA.SSE.A.2, HSA.SSE.B.3, HSA.SSE.B.3a, HSF.IF.C.8, HSF.IF.C.8a

## Video transcript

- [Voiceover] So let's
try to find the solutions to this equation right over here. We have the quantity two
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. 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
of a classic quadratic form, but there might be a faster
or a simpler way to do this if you really pay
attention to the structure of both sides of this equation. Well let's look at this, we
have two X minus three squared on the left-hand side, on the right-hand side
we have four X minus six. Well four X minus six,
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. And so this is really interesting, we have something squared is equal to two times that something. 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. 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 let's say, let's just
replace two X minus three, we'll do a little bit of a substitution, let's replace that with P. So let's say that P is equal to two X minus three. Well then this equation
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. And now we just have to solve for P. And I'll switch to just one color now. 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. And we've seen this shown multiple times, if I have the product of two
things and they equal to zero, at least one of them
needs to be equal to zero, so either P is equal to zero, or P minus two is equal to zero. Well if P minus two is equal to zero, then that means P is equal to two. So either P equals zero, or P equals two. Well we're not quite done yet, because we wanted to solve for X, and not for P. But luckily we know that two X minus three is equal to P. So now we could say either two X minus three is going
to be equal to this P value, is going to be equal to zero, or two X minus three is going
to be equal to this P value, is going to be equal to two. And so this is pretty
straightforward to solve, add three to both sides, you get two X is equal to three. Divide both sides by two, and we get X is equal to 3/2, or over here if we add
three to both sides, we get two X is equal to five. Divide both sides by two, and you get X is equal to 5/2. So these are the possible solutions, and this is pretty neat. This one right over here, you could almost do this in your head, it was nice and simple. Well if you were to expand this out, and then subtract this, it would have been a much more complex set of operations that you would have done. You still would have hopefully
gotten to the right answer, but it would have just
taken a lot more steps. But here we could appreciate some patterns that we saw in our equations, and namely, we have
this thing being squared and then we have two
times that same thing, two times two X minus three.