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## Algebra (all content)

### Course: Algebra (all content) > Unit 11

Lesson 8: Radicals (miscellaneous videos)- Simplifying square-root expressions: no variables
- Simplifying square roots of fractions
- Simplifying rational exponent expressions: mixed exponents and radicals
- Simplifying square-root expressions: no variables (advanced)
- Intro to rationalizing the denominator
- Worked example: rationalizing the denominator
- Simplifying radical expressions (addition)
- Simplifying radical expressions (subtraction)
- Simplifying radical expressions: two variables
- Simplifying radical expressions: three variables
- Simplifying hairy expression with fractional exponents

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# Simplifying radical expressions (subtraction)

Sal simplifies 4∜(81x⁵)-2∜(81x⁵)-√(x³). Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- What is the difference between a square root and a principle square root?(51 votes)
- Since both (-x)^2 and x^2 equal x^2, we use the phrase principal square root to note that we are only interested in the positive value. Most 'real world' applications of radical expressions are not concerned with the negative square root (negative length, or a negative amount of time, for example) so we ignore them by focusing on the principal quantity.(66 votes)

- At the very end, isn't it possible to factor out a |x|?(9 votes)
- Yes you can factor it out. See https://www.wolframalpha.com/input/?i=is+6*|x|*x^%281%2F4%29-|x|*sqrt%28x%29+equal+to+|x|*%286*x^%281%2F4%29-sqrt%28x%29%29

So the fully simplified expression is |x|*(6*x^(1/4)-x^(1/2))(11 votes)

- I tried to solve this myself before I finished the video to see how Sal solved the expression, and I got a differently simplified answer. I turned the root of 4 into a fractional exponent and turned the square root into a fractional exponent as well. I presume that it's still technically the same answer, but just either not simplified completely or a different way of simplifying.

This is what I did:

4*4√(81x^5) - 2*4√(81x^5) - √(x^3)

2*4√(81x^5) - √(x^3)

2*(81x^5)^1/4 - (x^3)^1/2

2*3*x^5/4 - x^3/2

6x^5/4 - x^3/2

So if this is just not completely simplified, I guess the real question is, how do you simplify it further from here? Would you have to go back and do it the way Sal do it or is it possible to continue simplifying using this method?(4 votes) - Couldn't you get rid of the ∜x and √x by converting to x^1/4 and x^1/2, then putting them together as x^1/4 and x^2/4?... Or is this rulled out because it would result in a negative root?(4 votes)
- wouldnt 81 be squared into 9????? Or am i missing something?(0 votes)
- It doesn't ask for the square root, but the 4th.(19 votes)

- Why is 3 not an absolute value while x is?(3 votes)
- 3 is just a number, if you take the absolute value of 3, that's just 3. We need to take the absolute value of x because x is a variable, it can be positive or negative. 3 isn't a variable, it can't be positive or negative, it's always 3.

Although, as Sal noted, you might not need the absolute value signs anyway, since we will probably only be dealing with positive values of x in this particular expression, but it can't hurt to leave them in.(3 votes)

- How can I do this simplification/subtraction with fractions? For example say : Sqrt(6)/6 - Sqrt(2)/4?(3 votes)
- why couldn't he do 9x9 instead of 3^4 at the beginning(2 votes)
- Then you would not be able to take and cancel out the ^4. You would end up having to make it 3^4 anyways. So he just did it earlier.(1 vote)

- I have followed another path and it seems to offer a further simplification, I want to know if this does not work...

1. Undistribute the 4th root expression convert to a fraction exponent

(4-2)(3x^5/4)-x^3/2

No absolute value is required from this because both exponents have an odd numerator which would resolve a negative x into a negative radicant and it would not therefore be possible to take a principal 4th root.

This can be further simplified by creating a common denominator between the two fractional exponents.

(6x^5/4)-(x^5/4 * x^1/4)

And factor out the common expression...

x^5/4(6-x^1/4)

I would like to know if this line of reasoning is correct, and if so why we would even need to consider the absolute value of x if we always keep the exponent odd inside a radical?(2 votes) - What is the difference between a square root and a principle square root?(2 votes)

## Video transcript

We're asked to subtract all
of this craziness over here. And it looks daunting. But if we really just
focus, it actually should be pretty straightforward
to subtract and simplify this thing. Because right from the get-go,
I have 4 times the fourth root of 81x to the fifth. And from that, I
want to subtract 2 times the fourth root
of 81x to the fifth. And so you really can just say,
look, I have four of something. And then the something
I'll just circle in yellow. I have four of this--
it could be lemons. I have four of
these things, and I want to subtract
two of these things. These are the exact same things. They're the fourth root
of 81x to the fifth. So if I have four lemons and
I want to subtract two lemons, I'm going to have
two lemons left over. Or if I have four of this
thing, and I take away two of this thing,
I'm going to have two of these things left over. So these terms right over
here simplify to 2 times the fourth root of
81x to the fifth. And I got this 2 just by
subtracting the coefficients. Four of something
minus two of something is equal to two
of that something. And then that, of
course, we still have this minus the regular
principal square root of x to the third. Now I want to try to
simplify what's inside of these under the radical
sign so that we can, in this example, actually
take the fourth root, and over here actually take,
maybe, a principal square root. So first of all, let's
see if 81 is something to the fourth power,
or at least could be factored into
something that is a something to the fourth power. So 81, if we do prime
factorization, is 3 times 27. 27 is 3 times 9. And 9 is 3 times 3. So 81 is exactly 3
times 3 times 3 times 3. So 81 actually is 3 to
the fourth power, which is convenient
because we're going to be taking the
fourth root of that. And then x to the fifth
we can write as a product. Let me write it over here
so it doesn't get messy. So I'm going to write what's
under the radical as 3 to the fourth power times x
to the fourth power times x. x to the fourth times x
is x to the fifth power. And I'm taking the fourth
root of all of this. And taking the fourth root
of all of this-- that's the same thing as taking
the fourth root of this, as taking the
fourth root of this. And I'm going to
want to skip steps. So I'm taking the fourth root
of all of it right over there. And of course, I
have a 2 out front. And then x to the third could
be written as x squared times x. So it's minus the
principal square root of x squared times x. And I broke it up like this
because this right over here is a perfect square. Now, how can we simplify
this a little bit? And you're probably getting
used to the pattern. This is the same thing as the
fourth root of 3 to the fourth, times the fourth root
of x to the fourth, times the fourth root of x. So let's just skip
straight to that. So what is the fourth root? I could write it-- let
me write it explicitly, although you wouldn't have
to necessarily do this. This is the same thing as the
fourth root of 3 to the fourth, times the fourth root
of x to the fourth, times the fourth root of x. And 2 is being multiplied
times all of that. And then this over here is
minus the principal square root of x squared, times the
principal square root of x. And so if we try to
simplify it, the fourth root of 3 to the fourth
power is just 3. So we get a 3 there. The fourth root of x
to the fourth power is just going to be x. Actually, I just
reminded myself, we have to be careful there. It is not just x, because
what if x is negative? If x is negative, then
x to the fourth power is going to be a positive value. And when you take the
fourth-- remember, this is the fourth
principal root-- you're going to get the
positive version of x. Or really, you're going to
get the absolute value of x. So here, you're
going to be getting the absolute value of x. Although, well, you
could make an argument that x needs to be positive
if this thing is going to be well-defined
in the real numbers, because then what's under the
radical has to be positive. But let's just go with
this for right now. And then we have the
fourth root of x. And then over here, the
principal square root of x squared, by
the same logic, is going to be the
absolute value of x. And then this is just the
principal square root of x. So let's multiply
everything out. We have 2 times 3 times
the absolute value of x. So 2 times 3 is 6,
times the absolute value of x, times the principal
fourth root of x, I should say, minus we took out the
absolute value of x, times the principal root of x. And we can't do any
more subtracting. Just because you have to
realize this is a fourth root. This is a regular square
root, principal square root. If these were the
same root, then maybe we could simplify this
a little bit more. And so then we are all done. And we have fully simplified it. And if you make the
assumption that this is defined for real numbers. So that the domain
over here, what has to be under these
radicals, has to be positive, actually, in every
one of these cases. And if they need to
be positive, we're not going to be dealing
with imaginary numbers. All of these need
to be positive. Their domains are x has to be
greater than or equal to 0, then you could assume that
the absolute value of x is the same as x. But I'll just stick
it right here. If you restrict the
domain, you could get rid of the
absolute value signs.