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Robert McKeown: This is a great question, we've

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been given a radical expression. And we're being

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asked to rationalize the denominator, basically

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remove all the square root symbols from the

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denominator, and then simplify it as much as

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possible. Now, this is not a function or an

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equation, there's no right hand side left hand

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side. So we're left just with what we have to

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try and follow these instructions. I'm going to

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use the difference of squares.

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And notice I've got a minus here, which means

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that if I multiply, explain the numerator in a

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moment, by multiply the denominator

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by the same expression, but with a plus, I'm

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going to be able to get rid of the square roots.

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And I'll show you why. And if you remember the

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difference of squares, you already know why,

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let's try this two times the square root of

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three. Well, I'll just write it like that with a

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square, there's two of them.

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And we'll do plus two square root of three times

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three. I'll just leave it like that, then we've

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got negative three, times two times the square

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root of three. And finally, we've got minus

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three, square three times three. Notice that the

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plus and the negative are going to sum to zero.

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So we're going to be left with

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now I have completely forgotten the numerator.

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So let me rewrite this out, we've got negative

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four times this thing right here. Right, as long

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as I multiply the numerator and denominator by

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the same value, I don't change the expression.

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And that's divided by

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this thing. Now I'm going to work through the

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exponents. So all right, the numerator as is,

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and I've got the square. So two squared is four,

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the square root of three squared is just three.

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And that's going to be minus nine.

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Now I work for the numerator as well. So I'm

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going to have negative eight times the square

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root of three minus 12. And that whole thing is

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going to be divided by 12 minus nine and

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over three, and I look at that the numerator, I

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guess, maybe I should not have brought the four

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in the brackets. So I'll take I can factor out a

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four, we're gonna have negative four, square

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root of three minus three, that whole thing is

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going to be divided by three. And I just

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realized that I made a little mistake here. I

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forgot to. I should have had a two in front of

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this and a four here, like so. And I'll just

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redraw that thusly. Now when it comes to ALEKS,

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ALEKS doesn't like these brackets. So this is

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the answer that ALEKS would like. Let's look at

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this next question, which is asking us to

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simplify a higher radical expression. So again,

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we've got no left hand side right hand side,

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there's no equality. It's not a function. We

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just have to work with what we've been given. So

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to get started, I well find the hardest part

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with these questions is what to do with numbers

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with numbers like 48. So why don't I break 48 up

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into the smallest factors possible. So I'll

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rewrite the cube root four, like so. And I know

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that two divided by 24, or excuse me, 48 divided

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by two is 24. So that would get me one, two. And

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then with 24, I could have six times four. And

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then maybe I'll just take a little break, I

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don't want to do too much in my head. Because if

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I try to do too much in my head, I'm most more

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most likely or more likely to make a silly

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mistake. Now let's keep going. Because we can

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break this up even further. I've got the two

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sixes, two times three, and four is two times

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two. So we've got s 19, here, and t 12. Now I've

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got a four up here, and I've got 1234 fours,

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excuse me, four twos. So I can sort of bring out

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those fours, or maybe I'll rewrite it like this

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as an intermediate step.

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I'm going to use our exponent rules to break up

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this fourth order root. And remember, the rule

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is something like well, if I've got x to the

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power of a, multiply by y to the power of B, and

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I've got this whole thing raised to say the

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power of C, that's equal to x A C, times why BC.

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And so that's the exponent rule. Or whatever one

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of them. Now looking over here, we got four

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squared, and I want to take the fourth root of

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that thing. So we're just going to be left with

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two on the outside. You know, it's kind of like

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having four for this thing to the power of one

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over four. And so we get four times one quarter,

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which is equal to one. Oh, I forgot that this

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three. So we should have a three in here, I'll

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add it to the end. And I've got s 91, I can just

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rewrite it like that and exponent form if I

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want, and I've got t to the 12 or four, and I

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still got the square root of three. Now

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rewriting this, so I've got to and outside the

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bracket, or maybe I'll just write with the whole

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number exponent, I'm going to write as to the

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power of four. Or maybe if to make it a little

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more clear for you, I'll write it as 16 over

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four. And you can see 12 divided by four, well,

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that's obviously to the power of three. Now,

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I've still got as to the power of three over

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four remaining, so I can put it back into the

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cube root like so. And then rewriting this kind

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of ugly thing I've got to ask to the power of

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four t to the power of three times the fourth

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root of three times s cubed.

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So here we are on ALEKS. The question is asking

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us to put the following expression in simplified

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radical form. So we're allowed to keep a square

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root, we just need to simplify it. Now I'll put

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in our answer. So we had two times s to the

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power of four times t to the power of three. And

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here, I'm going to use this expression here, the

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one that's highlighted in blue. And if I click

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on that, it's going to let me do an nth order

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root. So I've got a four there for the fourth

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root. And then I hit the right arrow key. Now

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I'm inside the bracket three times as to the

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power of three or s cubed. Now we're on the

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outside and I'll click the check button. And we

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had the we have the right answer. We're getting

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into an expression where we've got one radical

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subtracted from another radical. We have no

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equal sign here. We have no left hand side right

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hand side, it's not a function. It's not an

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equation. It's just an expression. And so we

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have to simply So as much as possible, why don't

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we start by doing what we did last time. And

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let's start with the numbers. So we've got y to

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the square root of 50. Well, I know that I could

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rewrite 50, like so. And I'll just leave the x

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cubed in there, minus nine x, and we've got the

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square root of y, I can't do anything with two.

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So I'll just rewrite the right hand term. Now

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going back to the next slide, we've got y, while

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y is just going to sit out there for now, that x

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cubed, I can do something with that I could

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bring out an x squared. And if I bring out the x

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squared, and take the square, I should say if I

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take the square root of the x squared, I'm just

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going to have x. And now I've got o and I now I

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can take out, I can take the square root of 25.

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So I've got five like that. And what's left in

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here is two X. Over here, we've got nine x, I

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can take the Y out. So now we've got the Y

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there. And what because of y squared, we just

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becomes the square root of that is just y. And

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we've got two x in there. And that's kind of

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handy. Because now I've got a square root of two

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x on both sides of the equation. So why don't I

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factor out the square root of two x. And I'm

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going to be left with five x y minus nine x y.

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And now I can factor out the x y. So we've got x

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y, square root of two x times five minus nine.

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And that gives us negative four, xy square root

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of two x, which looks like it could very well be

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the answer. Why don't we put it in ALEKS? And

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see if it is the answer. Here's ALEKS. I'm

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putting in the number or I'm putting in our

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answer negative for x, y. And here's the square

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root symbol that I want to use. And we're to x

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down there and click on the check button. And we

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got the right answer. We've got a square root of

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y here and we've got a y to the power of one. So

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I suspect two solutions exist. Just like if I

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have y squared is equal to four. That's like y

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is equal to plus minus two, right? It could be

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minus two, or could be plus two. So let's go

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ahead, I've got the square root on the entire

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left hand side. So one thing I could do, and

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I'll do this, so just again, to help you look at

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it, I could rewrite this whole thing, like so.

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Now, if I want to collect sort of like terms, I

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don't really know what to do with the square

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root of four y minus seven. So one thing I can

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do to try and make this easier to work with and

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get to the answer is I can square both sides of

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the equation, I can square both sides of the

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equation. And if I do that

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exponent rule says the one half is going to be

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multiplied by the square the two. And then I

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take the square this whole thing, remember,

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remember common student error, why minus three

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squared is not equal to y squared minus three

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squared. Right in general, that's not true. And

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specifically here, it's not true either. Now

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what have we got, we've got on the left hand

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side, we just have four y minus seven, that's

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nice. I can use I can split up the terms I can

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work with the four y and the negative seven

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separately. And on the right hand side, I've got

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this y minus three, that whole thing's squared.

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Want to open up the bracket what I call opening

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up the bracket? And if I do that, I'm going to

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have y squared. And I know I'll have six. Why?

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Because this is a special kind of equation that

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I've memorized the answer to most people

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memorize the answer two, plus nine, right three

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times three, you can work it out if you want to

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the FOIL method, as ALEKS explains to you, but I

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know what the answer is going to be that. Now,

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what do I want to do, I've got a quadratic, I've

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got a quadratic. So I want to set this thing

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equal to zero. And then I can solve it by

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factoring. Or I can solve it using the quadratic

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formula. And I want to, so I want to collect

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like terms here and set the whole expression

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equal to zero. If I do that, I'm going to get y

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squared minus 10 y plus 16 is equal to zero.

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Now, how can I have two numbers that sum to

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negative 10? And that multiply to each other is

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equal to 16? Well, if I've got y minus eight,

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and y minus two, I'm going to be able to factor

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that quadratic expression. And I can see here

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that if y is equal to eight, or y is equal to

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two, I should say the equation is true. Wait a

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second. Are we sure that this equation is solved

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when y is equal to eight, or y is equal to two?

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Well, let's give it a try if y is equal to two,

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and we get four times two minus seven is equal

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to two minus three.

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And we end up with one is equal to negative one,

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which is a contradiction.

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Y equals two is not a solution.

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How about if y is equal to eight? If y is equal

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to eight, eight times four minus seven, the

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square root of that whole thing is equal to

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eight minus three. We've got 32 minus seven is

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equal to five, which is the square root of 25 is

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equal to five and five equals five. So why is

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equal to two y? Well, y equals two is not a

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solution. We already said that y is equal to

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eight is a solution. So there's only one

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solution here, not two solutions.