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Evaluate the following.
In order to evaluate radicals all
that you need to remember is
In other words, when evaluating we are looking for the value, y, that we raise to the n to get x. So, for this problem
we’ve got the following.
Technically, the answer to this
problem is a complex number, but in most calculus classes, including mine,
complex numbers are not dealt with.
There is also the fact that it’s beyond the scope of this review to go
into the details of getting a complex answer to this problem.
Convert each of the following to exponential form.
To convert radicals to exponential
form you need to remember the following formula
For this problem we’ve got.
There are a couple of things to
note with this one. Remember and notice the parenthesis. These are required since both the 7 and the x was under the radical so both must
also be raised to the power. The biggest
mistake made here is to convert this as
however this is incorrect because
Again, be careful with the
Note that I combined exponents
here. You will always want to do this.
Simplify each of the following.
9. Assume that and for this problem.
The property to use here is
A similar property for quotients
Both of these properties require
that at least one of the following is true and/or . To see why this is the case consider the
If we try to use the property when
both are negative numbers we get an incorrect answer. If you don’t know or recall complex numbers
you can ignore this example.
The property will hold if one is
negative and the other is positive, but you can’t have both negative.
I’ll also need the following
property for this problem.
In the next example I’ll deal with
Now, on to the solution to this
example. I’ll first rewrite the stuff
under the radical a little then use both of the properties that I’ve given
So, all that I did was break up
everything into terms that are perfect cubes and terms that weren’t perfect
cubes. I then used the property that
allowed me to break up a product under the radical. Once this was done I simplified each perfect
cube and did a little combining.
I did not include the restriction
that and in this problem so we’re going to have to be a
little more careful here. To do this
problem we will need the following property.
To see why the absolute values are
required consider . When evaluating this we are really asking
what number did we square to get four?
The problem is there are in fact two answer to this : 2 and -2! When evaluating square roots (or any even
root for that matter) we want a predicable answer. We don’t want to have to sit down each and
every time and decide whether we want the positive or negative number. Therefore, by putting the absolute value bars
on the x we will guarantee that the
answer is always positive and hence predictable.
So, what do we do if we know that
we want the negative number?
Simple. We add a minus sign in
front of the square root as follows
This gives the negative number
that we wanted and doesn’t violate the rule that square root always return the
Okay, let’s finally do this
problem. For the most part it works the
same as the previous one did, we just have to be careful with the absolute
Note that I could drop the
absolute value on the term because the power of 2 will give a
positive answer for regardless of the sign of x. They do need to stay on
the y term however because of the