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hi this is george woodbury from College of the Sequoias in Visalia California
and in the short video I'm going to go over how to simplify radical expressions
we begin with an example
we want to simplify the square root of x to the ninth power
and we're going to assume x is non-negative
now one way to think of square roots
we're looking for pairs of factors
that can come out as one
although i wouldn't recommend you do this
in every problem
let's start by writing x to the ninth as x times x
times x etcetera where we are using x as a factor nine different times
here's a pair of
factors that are the same
and we know from before that the square root of x squared
is equal to x when we know x is not non-negative
so this allows us to take one factor of x to the outside
here's another group of two
that will give us our second factor that's outside the square root
here's a third pair
and a fourth pair
and notice we're going to have one factor of x that is going to remain unpaired
taking all four of these out to the front of the square root
this is the same as x of the fourth power
times
the square root of the single factor of x which remains inside the square root
and basically the idea here is that every two on the inside
is equivalent to one on the outside
this radical is considered simplified but let's talk about a
quicker way to get from the problem to our results here
so if i divide index of the radical : 2
into nine
it goes in four times
with a remainder of one
the quotient tells me
how many factors are coming out
the remainder
tells me how many factors are staying in
let's use that in a few examples
let's simplify the square root of a to the twenty third power again assuming a is not
negative
so when i divide
twenty three
by 2
again you should use long division for this but that's going to be eleven
with a remainder of one
that tells me that
a to the eleventh is the expression in front of the square root
and a to the first power
or simply a
is the expression
that stays inside the square root
now a radical is considered simplified whenever the power
of any base inside of it is less than
the index of the radical which here is 2
so since this power of one
which we don't usually right
is less than the index of two which again we do not right
then this is considered simplified
let's try it with a root other than the square root
uh... the third root of b to the forty fourth power
now here because it's an odd root it doesn't even matter if b is non-negative
or not but we'll leave that assumption in
i divide forty four by three - let me do this out one time
three goes into four once
bring down the four
three goes into fourteen four times
with a remainder of two
that tells me that
i have b to the fourteenth power
outside of the radical
and then inside of the radical
i'm going to have b to the second power
by the way be sure to write this index of three if you drop the index your
radical turns into a square root which is not what you started with
one last problem to work through here what's different about this
problem well we have two variables but we handle this the same way
i divide five into seventeen
to determine what comes out and what stays in for x
and then i'll divide five into thirty four to determine the same for y
let's begin with the variable x first
five goes into seventeen three times
with a remainder of two
now five goes into thirty-four six times so you get y to the the sixth on the
outside
with a remainder of four
y to the fourth on the inside
let's move over to problems that don't contain variables but instead
we're just taking the square root of a number here
now we don't know the square root of twenty four we know the square root of
twenty five is five
so this answer is close to five but it's not exactly five
when we are taking a root of a number that we don't know the exact square root
of
the first thing we want to do is find the prime factorization of twenty-four that
value
lots of different ways to break down twenty-four you can use two and twelve
three and eight, four and six
i'm going to begin with four and six
i know that four is equal to two times two
i know that six is equal to two times three and those are all prime
so the square root of twenty four
is equal to the square root of two times two times two times three
and because it's the square root we're looking for pairs
and here we have
a pair of twos
that will come to be outside as a two
but the remaining factors are unpaired
and so i just multiply those together : two times three is six
so the square root of twenty four simplifies to be two times the square root of
six
by the way on this line
where i had four ten six
this is a pretty convenient factoring
because i know that the square root of four
is
two
and i don't know the square root of six
there's no perfect square that divides into six evenly other than one
so the six is going to stay inside
so either you can factor at all the way through
or you can look for convenient factorings for that type of
radical you're working with
let's try one together
square root of twenty eight
let's go to a factor tree first twenty eight is the same as two times fourteen
two is prime
fourteen is two time seven and both of those are prime
so the square root of twenty eight is the same as the square root of two times two
times seven
again because it's a square root we're looking for factors that repeat in pairs
we have a pair of twos that comes out as a two
the seven stays behind
now another way to think of this
is twenty eight is the same as four times seven
i know the square root of four is two
and the square of seven i don't know
let's try one with a third route
here i'm looking for the third root of five hundred forty will this is gonna take a
little while to factor
but i know that five hundred forty is the same is ten times fifty four
whenever a number ends in zero we know that ten is a factor of that number
and ten can easily be factored is to two times five
both of those are prime
fifty four
if i think about my multiplication facts i know that's equal to six times nine
and i can easily break down six to be two times three
both prime
and nine to be three times three
both prime
so when i rewrite this i have the cube root
of two times two
times three times three times three times five
and because the index is three i'm looking for repeated factors in groups
of three
and i only have one group of three
that's the three threes
so that comes out
and inside the cube root i'm going to multiply two
times two
times five
which is twenty
so the cube route of five hundred forty simplifies to be three times
the cube root of twenty
one last example mixing the two ideas together
i have a square root of a product of a constant
and a variable term
i'm going to begin by factoring one hundred fifty
using the same approach as last time i know the ten is a factor
ten is two times five
fifteen is three times five and those are all prime
so that's the square root of two
times three
times five times five writing those in ascending order
and then x to the nineteenth
because it's a square root for the constants I'm taking out groups of two
so i can take out a five
and left in the square root i'm going to have two times three
or six
now i turn my attention to the variable
two
goes into nineteen nine times
with a remainder of one
the nine tells me that i'm going to have x to the ninth
outside the square root
the remainder of one tells me that i'm going to have x to the first or just x
like inside the square root
notice how we treated
coefficients differently than we handled the exponents
prime factor the coefficient and remove pairs if it's a square root
or groups of n for whatever root you're taking
but the
exponent needs to be divided by the index of the radical
if you have any questions or comments or if you have a request for a video that i
can put on you tube for you
you can reach me through the contact page on my website
which is george woodbury dot com
thanks for watching and good luck with this