Int Algebra: Solving Equations Containing Radicals

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Roots and Radicals
Equations with Radicals
Solving an Equation Containing a Radical Page [1 of 3]
Okay, so now I actually want to look at real honest to goodness radical equations. So let’s take a look at one right
now.
Suppose I wanted to solve the following: x = 15 - 2x. Now what makes this a radical equation is that there’s the
square root here. Now how would I solve this? Well, to get rid of a square root one way to do that is that to square
both sides. Now I have to square both sides even though the square root is only on this side because, of course, if I
just squared this side and do nothing here, that’s going to change the equality, it will no longer be equal, it won’t be
the same in equality. So to keep everything balanced perfectly if I square one side, I also square the other side. So, of
course, here I’m not going to have an x2, but that’s a small price to pay for lifting that whole radical, so this looks good.
So what do I do? If I square both sides, I would see on this side x2 and on this side if I square a square root, the
radical just lifts up and goes away, they cancel each other. So I’m just left with the inside which would be 15 - 2x.
Okay, great. Well, now you’ll notice that now we have squares in it, before we didn’t have any squares in it, but the
happy thing is there are no more square roots. And let’s face it, if you had a choice between facing a square or a
square root in a dark ally you always go for the square. So we got rid of the square root and now we can just solve
this. Well, how would you solve this? Well, this is just now a quadratic equation so what do I do? I push everything
over to one side, in my case my left, and have everything equal zero and then try to factor and solve.
So let’s try that technique right now. So if I bring everything over, what’s going to happen? Well, this - 2x we brought
over as a + 2x, so I would see x2 + 2x and then this 15 would brought over as a negative 15. Now let me say a little
thing here. Sometimes when students are working through these problems, they’re getting so involved and so forth as
you can see, sometimes they start to work and factor this thing. Don’t do that. We’re dealing with an equation so
always remember to right equals and then whatever’s on the other side. In this case since I brought everything over,
subtracted everything over, added everything over, I’ve just left it as zero here that way it’s always clear where you
are. I want to solve this equation.
Okay, well now there’s only one game in town, let’s hope this can factor and let’s see what happens. Well, I put an x
here, x here, negative tells me that I’m going to have opposite signs, since these are the same I can put those signs
wherever I want. I’ll put a plus and a minus here and I need two numbers that multiply to get 15, but then subtract
somehow to give 2. So let’s see, 3 and 5 sound pretty good. Should I put 3 and 5 here? If I put the big number 5 with
the negative sign when I combine them I’ll get a negative 2, so it’s probably a bad idea. I better put the 5 here and the
3 here. Notice the inside term is 5x, the outside term is -3x, their net gain is a + 2x and the last times the last is a -15.
Again, I’m factoring really fast, but I hope that as you practice these things you’ll be able to factor fast, too, but even
more important than factoring fast is to factor accurately. So after you factor just take a second to check and make
sure that everything’s looking okay.
Anyway, this looks good to me and I’ve got a product of two things that multiply to give zero. So either this thing is
zero or that thing is zero. So either x + 5 = 0, that’s one possibility or the other is that x - 3 = 0. Well solving this is
easy, that means that x = -5 and solving this is pretty easy, x = 3. So there are two solutions to this quadratic.
However, the mission was not to solve the quadratic, the mission originally was to solve this radical equation. So what
I have to do now is check and make sure that, in fact, both of these things are solutions. Let’s check the easy one first,
the smaller number 3. So let’s plug in 3 here and see what happens. If I plug in 3 for x what I’m going to do is where
ever I see an x I’m going to plug in 3 and I’m going to see if the left-hand side equals the right-hand side. So let’s do
that right now, in fact, I’ll just it right on top of this and I’ll even use a different color. So if I put in an x = 3 everywhere
on this side I would have a 3 and the question is does that equal, I don’t know, I’m going to check. Does that equal
when I put in the 3 on this side? Well, if I put in the 3 on this side I see 15 - and if I put in a 3 here I’d see 2 x 3 = 6. So
this side here would equal the square root of 9 and you’ll notice the square root of 9 = 3 and so I see that 3, the righthand
side, does equal 3, the left -hand side. So, in fact, this checks. So this is really one answer.
What about the -5? Well, let’s try that. So let’s check the -5. If I plug in -5 for x, I see -5 = and now I’m going to try and
see what happens on the right-hand side, but I’ll tell you right now, I’ll tell you what I’m thinking, I know this is bad
news because this square root is positive. If it were negative, there would have to be a negative sign in front of it. So
Roots and Radicals
Equations with Radicals
Solving an Equation Containing a Radical Page [2 of 3]
already I know this is not going to work. But let’s just keep going and see what value this equals. I have 15 - and if I
plug in a -5 here that would give me a + 10. So I would see a + 10 and this thing, and this is a question mark because
it’s equal, and this thing would equal 25 , but the square root of 25 is 5, and you’ll notice that 5 is not equal to -5. Do
you see that? So, in fact, this is an extraneous root. Even though it’s the root to the quadratic, it’s not a root to the
original thing so we cross that out and the only answer is x = 3. See the importance of checking your answers?
Okay. Let’s try one more. This one’s going to be even a little bit more eclectic one because this one looks like this, x =
3 + 3- x . Okay, now let me tell you why this is eclectic. You’re saying, it looks like the exact same thing as the
other one. Well, let’s just think through this. You see a square root so you’re initial reaction should be, I’m going to
square both sides. That’s a great reaction but let’s just think through this together and reason together. When I square
this side this is actually a binomial, there are two terms here. So when I square that out, I’m going to actually have to
do a little bit of foiling and when I start foiling there’s going to be inside terms and outside terms, and notice that those
inside terms and outside terms are going to have square roots in there. So actually by squaring this side I am not
going to, in fact, clean off that radical because remember, if I have something A and B and I square that, I do not get
A2 + B2. Right? This is the number one classic mistake on my top 10 list, the squaring mistake. This is not true.
Remember, don’t forget to foil, so there’s middle terms, there’s inside terms and outside terms and that’s going to
actually contribute more square roots. So the important thing to remember here is that when I see a square root
equation, what I have to do is get all the square roots by themselves. I have to isolate it so there’s no foiling. So if that
3 weren’t there, then if I squared that I would just lift the radical. So how do I get rid of the 3? I’ll bring that 3 over to
the other side. So what I’m going to do is bring the 3 over and I see x - 3 = 3- x . And I want you to notice what just
happened there. By bringing this over now I’ve isolated the square root so when I square both sides, this will just lift,
the radical will go away. You see that? So now I’m going to square both sides. On the left I see x - 32, and on the right
the squaring just lifts the radical. So I see 3 - x and now I’m returning to a quadratic, which I can solve. Again, this is
not just x + 9. Remember the number one classic mistake is that you have to foil out. If I foil this out I’ll see x 2, the
outside term is - 3x and so what I see is a -6x and the last times the last would be a + 9, I just foiled. I multiplied x -3
by itself in my head and this equals 3 - x. What do I do? I see it’s quadratic; I pull everything over to the left so it
equals a zero on the right and see what happens. If I move this -x over, it becomes the +x and I see x2 - 6x + x = -5x
and then if I bring this 3 over, I could subtract it and so I see 9 - 3 = +6 = 0. So now I want to factor this and see what
happens. Well, I’ll put an x, an x, the positive sign here tells me that the they’re same sign and this tells me they must
both be negative, two numbers that multiply to get 6, but combined to get 5, it looks like 2 and 3. Right? Minus 2x - 3x
combined to give -5x and -2 to -3, great.
Okay, so now what I want to do is say okay, here we have this product and it equals zero. So what does that mean? It
means that either this equals zero or that equals zero. And so if this equals zero that means that x - 2 = 0 or the other
possibility is that x - 3 = 0. Okay, well if x - 2 = 0, that means that x has to be 2. So in this case we see that x would
have to = 2. So there’s one solution and what’s the other solution? The other solution is, if x - 3 = 0, which means that
x would have to equal 3. So I get two solutions to the quadratic, but what about to the original question? Well, to the
original question what I have to do now is plug back and see.
So let me actually do that, let me lower this for you a little bit here. So let’s see, if I plug back and check, I have to see
if these are going to work or not, let’s check the first one. So I take my check font, I’m going to plug in a 2 here. If I
plug in a 2 what happens? On the left-hand side I see just 2 and the question is does that equal? What I get when I
plug in 2 on the right. So that would be 3 + 3-2, but what does that equal? That’s 3 + 1 = 4, 4 equals 2? I don’t
think so. So, in fact, this is false. So, in fact, this is an extraneous root. It’s a root to the quadratic, but not a root to the
original answer. By checking I saw that, in fact, this is not a root.
Now maybe, in fact, the other one’s not a root, too. Just because you get one thing not to work doesn’t mean the other
one automatically works, maybe this thing has no solutions, you always have to check. So let’s plug 3 in here, 3 on
the left for x would be 3. Does that equal? 3 + 3-3? Well, 3 - 3 = 0, so this is 3 + 0 = 3, 3 = 3 checks, so in fact,
there’s one answer to this radical equation and that’s the answer, x = 3.
Roots and Radicals
Equations with Radicals
Solving an Equation Containing a Radical Page [3 of 3]
Okay, up next what I’ll do is one more example of this where we’ll have two square roots in the same problem. I’ll see
you there.