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Hi Class, Today we are going to solve radical equations. A radical equation is going to look like this where you have some kind of radical within
the equation. The first thing that you are going to want to do when you solve these is you are going to want to isolate the radical just like you would
want to isolate a variable when solving an equation, notice this one has a 3 in front of it being multiplied by it so that is going to have
some how get moved. We also have a minus 5 here on the end and that will have to be moved some how.
and when your solving a regular equation you always start with what is being added or subtracted so we are going to start right here
with this minus 5 and since it is a minus 5 the opposite of
subtracting five it is adding five so we'er going to go ahead and add
5 to both sides
that's going to make this go away so we are left with 3 square root of x + 5 equals
15 now that we've moved what' being added or subtracted we need to take care of the 3 which is being multiplied by it just like that would
be the next thing that you would do if this is a normal equation, and this radical was something y instead of the radical
and the 3 is being multiplied by the radical, the opposite of that is to divide so we are going to go ahead and divide both sides by 3
When we divided both sides by 3 the 3's end up cancelling which leaves us with the radical all by itself.
over here 15 divided by 3 is 5
now that the radical it is by itself what we need to do is somehow get rid of that square root and the opposite of square rooting something is to square
something so what we do is we raise both sides t0 whenever power would make
the radical go away so if this had been a cube root we would cube both sides, if it was a 4th root we would raise both sides to the 4th power, if it
was a 5th root we would raise both sides to the 5th power and so on. Since this is a square root the opposite of square rooting is squaring.
so we will square both sides
When we do that since the square and the square root are opposite operations they cancel each other out.
and your just left with x + 5
on the other side 5 squared turns into 25
now we have a regular equation so once you've gotten rid of your radical you go ahead and
solve equation the same way you would normally do so in this case are going to
subtract 5 from both sides and 25 -5 ends up giving us
20 so 20 would be what we would believe is the solution however sometimes when you are dealing with these radicals you'll do everything
correctly, and when you get to the end you'll end up with what is called an extraneous solution so when you plug it back in to check it actually
doesn't work so you want to make sure you always check your solution on these, so we are going to check our solution of x = 20
by going ahead and going back to this original equation
and
plugging in 20 where we see our x and our x is right here, so what that is going to give us for our check
is 3 times the square root of, the x is going to be replaced by our 20
and then we have our +5 minus 5 equals
10
so now we just follow order of operations and the first thing I'll do is the 20 plus 5 which is 25
and then we have to take care of the square root and the square root of 25 is 5 and remember there is an implied multiply
right here between the 3 and square root of 25 so this becomes 3 times 5
-5 equals 10
and 3 times 5 is 15
and when you subtract 5 from that you do get 10 so this one does check out so we would say the solution
set
is
and then in set brackets we put
our 20 and that would be how we solve this. So the first thing that you want to do when solving these equations is get the square root by itself,
then you want to square both sides, or if it is a cube root cube both sides and so on, then just solve your equation normally
Now let's take a look that's one more of these
this one here is going to be slightly different than what we just saw so what would have here is something where you actually have a radical on
both sides of the equation so for instance we might have 5y
-1 underneath our radical equaling
this square root of the y + 1 notice that there's two square roots but both square roots are actually by themselves already so
we can actually go straight into squaring both sides in order to make the radicals go way
so if we square both sides
the square and the square root cancel here to leave us with 5y-1
and they also cancel over on this side to leave us y + 1 now we have a regular equation where we can go ahead and solve
so we are going to start by getting our y's on one side and that's going to happen by subtracting y
from both sides. That leaves us with 4y - 1
equals 1 and now we need to move the -1 here and the opposite of subtracting 1
is adding 1 so we're going to add 1 to both sides those cancel and we get 4y = 2
and then finally divide the sides by our 4
and reduce our answer so y ends up being 1/2
now just like the last one you do want to make sure you check this so you make sure that it works so we're gonna go ahead and
plug it into our original equation
which
is right up here so to check this
we are actually going to plug the ½ in for y
right here and right here and that is going to give us the square root of 5 times 1/2
-1 equaling the square root of 1/2
plus 1 now on the right hand side that's gonna be the easier one because 1 is the same thing as
2 over 2 which would give us common denominators so this becomes square root of, if we add our numberators
3 over 2 to what we hope is that on the other side we get the square root of 3/2 also
so when we multiply the 5 by 1/2 you can remember that the 5 is over 1 so you multiply straight across and this becomes a square root
of 5/2 and if we want common denominators this 1 here will become 2 over 2
and now you can subtract your numerators and 5 - 2 gives us 3/2
and again checks out so we would say that
the solution
set is
½
now if we had checked this hypothetically and instead of getting square root of 3/2 equals square root of 3/2 let's say we had gotten
the square root of -3/2 we didn't but what if we did get square root of -3/2 equals square root of -3/2, they are equal but when checking these and
negative underneath the radical is not OK so if you get something like this were you get negatives under the radical when you are
checking then you would say that it is an extraneous solution and instead of having the solution set as ½
, we would say that there is no solution. In this case we didn't get the negative underneath so it did check out just fine
but if wehad, we would not accept that as a solution
so, keep that in mind that you don't want the negatives under the radical when checking. Let's take a look at one more of these
this next one is going to have