Intermediate Algebra
Answer/Discussion to Practice Problems
Tutorial 39: Simplifying Radical Expressions
Answer/Discussion
to 1a

Note that both radicals have an index number of 5, so we were able
to put their product together under one radical keeping the 5 as its index
number.
Since we cannot take the fifth root of and does not
have any factors we can take the fifth root of, this is as simplified as
it gets. 
Answer/Discussion
to 2a

Since we cannot take the square root of 5 and 5 does not have any factors
that we can take the square root of, this is as simplified as it gets. 
Answer/Discussion
to 3a

Even though 40 is not a perfect cube, it does have a factor that we
can take the cube root of.
Check it out: 
In this example, we are using the product rule of radicals in reverse
to help us simplify the cube root of 40. When you simplify a radical,
you want to take out as much as possible. The factor of 40 that we
can take the cube root of is 8. We can write 40 as (8)(5) and then
use the product rule of radicals to separate the 2 numbers. We can
take the cube root of 8, which is 2, but we will have to leave the 5 under
the cube root. 
Answer/Discussion
to 3b

Even though
is not a perfect square, it does have a factor that we can take the square
root of. Check it out: 
Answer/Discussion
to 4a


*Simplify fraction
*Take the square root

Note that both radicals have an index number of 2, so we are able to
put their quotient together under one radical keeping the 2 as its index
number. Since the radicand is a perfect square, we are able to take the
square root of the whole thing, which leaves us with nothing under the
radical sign. 
Last revised on July 19, 2011 by Kim Seward.
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