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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.

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  • leaf grey style avatar for user Xander2000
    What is the difference between a square root and a principle square root?
    (50 votes)
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    • leaf grey style avatar for user CasualJames
      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.
      (65 votes)
  • piceratops ultimate style avatar for user Parthik Patel
    At the very end, isn't it possible to factor out a |x|?
    (9 votes)
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  • leafers ultimate style avatar for user Elaine Nagahara
    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)
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  • leafers seed style avatar for user JasonWilson.DLBA
    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)
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  • blobby green style avatar for user Gabriella Newt M
    wouldnt 81 be squared into 9????? Or am i missing something?
    (0 votes)
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  • starky ultimate style avatar for user Daniel
    Why is 3 not an absolute value while x is?
    (3 votes)
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    • leafers ultimate style avatar for user Alex
      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)
  • piceratops ultimate style avatar for user Ricardo Marques
    How can I do this simplification/subtraction with fractions? For example say : Sqrt(6)/6 - Sqrt(2)/4?
    (3 votes)
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  • spunky sam blue style avatar for user Daniel Mathew
    why couldn't he do 9x9 instead of 3^4 at the beginning
    (2 votes)
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  • aqualine ultimate style avatar for user Lindsay MacVean
    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)
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  • female robot grace style avatar for user Nishtha Dani
    what is the difference between a radical and a surd?
    (2 votes)
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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.