Solving Radical Equations 4

Uploaded by videosbyjulieharland on 25.05.2010

>>This is part four of solving equations containing
radical expressions.
We are just doing more examples,
so in this video we will be solving the following
three equations.
To solve this problem we have two radicals,
so we want to isolate one radical.
Now it doesn't matter which one you isolate,
I usually isolate the one that looks more complicated,
which will be the square root of "x" plus 9, so what I am going
to do is put that on the right side of the equation
so it's positive, so I will have a positive in front
of the square root sign by adding square root of "x" plus 9
to both sides, and then I will have to subtract 1
from both sides, so that I have square root of "x" minus 1
on the left hand side of the equation.
So I've isolated the square root of "x" plus 9
on the right side of the equation.
Alright, once we isolated it, we want to square both sides
so that we can then do that square root,
if it was a cube root you would of course cube it,
and fourth root, we would raise it to the fourth power.
Alright, so we are going to square both sides,
but on the left hand side, its binomial the binomial squared
so be careful if you want you can write that out as squared
of "x" minus 1 times squared of "x" minus 1,
and that will give you "x" -" t" square roots of "x" plus 1.
I'm not showing all these steps, but you can go out on the side
and do the foil method; squared
of "x" minus 1 times squared of "x" minus 1.
And on the right side, when we square the square root,
I get what's underneath the radical,
which will be the "x" plus 9.
All right, so there's still a square root In here,
so now we need to isolate that next square root.
So for the moment ill just leave this square root the
"t" negative "t" squared of "x" on the left hand side
of the equation, but let's get rid of the, subtract "x"
and subtract 1 from both sides, so well subtract "x"
and subtract 1 from both sides , so that I have negative
"t" squared of "x" equals 9 minus 1 is 8, okay,
and then we could divide both sides by negative "t"
so I have the squared of "x" all by itself on one side,
so I'm going to go ahead, and divide both sides
by the coefficient of the square root,
and that will give me square root of "x" equals negative 4.
And right there you should know that's impossible,
the square root of something is always non-negative,
so there is no solution of "x" so the answer
for this problem is the null set, right?
There's no solution , right,
because you can't have a square root equaling a negative number.
Remember you can also write the null set
as braces with nothing in them.
Alright here's a problem where we have a square root
on one side equaling to the square root of the other side,
so we just square both sides, because they are both isolated
in this case, and we get what's underneath the radical so that
"x" minus 5 equals 5 minus 7 "x".
Alright, so now to solve this we need to get the variables
on one side constants on the other because its linear, right,
there's no "x" squared terms, so let's say add 7x to both sides,
and add 5 to both sides, we get 14x equals 10, divide both sides
by 10, I'm sorry, divide both sides by 14,
so we get "x" equals a five-sevenths.
And then, of course, we want to check it.
So let's check it in the original.
So we're going to put in five-sevenths for "x",
so 7 times five-sevenths minus 5, so on the left side,
the sevenths cancel underneath here an d I end
up with 5 minus 5.
Everybody see that?
Which is zero, so you get the squared of zero, which is zero,
and now let's go ahead and simplify the right hand side
by plugging in five-sevenths for "x", so I get 5 minus,
again the sevenths cancel, I have 5 minus 5,
which is square root of zero, also zero.
So there we are, five-sevenths is the correct answer.
All right, here's a problem with fourth root.
Again, we want to begin by isolating the radical,
which is the fourth root, so we want to add 5 to both sides.
So, we've got the fourth root of "x" plus 3 is equal to 2.
Now, when you're dealing with the fourth root,
remember the fourth root can't equal a negative number,
but this has it equaling a positive number,
so we're okay here.
If it would have been, been equal to negative 3
or negative 2, you could have stopped right then,
and that would be an impossible situation,
and there would be no solution.
But in this case, since the fourth root is equal
to a non-negative number, you can go ahead and raise each side
to the fourth power, so we get "x" plus 3 on the left-hand side
and 2 to the fourth is 2 times 2 times 2 times 2, which is 16.
And then we simply subtract 3
from both sides to get "x" equals 13.
All right, now we want to check it.
So, we want to put the original problem,
fourth root of "x" plus 3 minus 5 equal negative 3,
and we're going to plug in 13 for "x".
All right, so that gives us the fourth root of,
I'm running out of space here, here we go.
This is the fourth root of 16 minus 5, fourth root is 16,
you have to think well, what number to the fourth is 16,
that's 2, so this becomes 2 minus 5
and 2 minus 5 is negative 3.
All right, so the left side simplifies to negative 3
and the right side is also simplified
to negative 3, so it checks.
So for this problem, "x" equals 13 is our answer,
and so there's our solution.
Checked and all.