WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today, we will be covering operations with radical expressions.
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And in earlier lessons in Algebra I, we talked about square roots and properties when working with square roots.
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And now, we are going to go on to talk about other roots.
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I will be starting out with reviewing properties of radicals and applying these properties
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to roots other than square roots: cube roots, fourth roots, sixth roots, and on.
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It is the same properties, such as the quotient property.
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And the **quotient property** says that, if you have the root (it could be a square root;
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it could be a cube root) of a/b, this is equivalent to the root of a divided by the root of b,
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with the restriction that b cannot be equal to 0, because if b did equal 0,
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the square root of that would be 0 (or cube root, or whichever root).
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And then, you would end up with 0 in the denominator, which would result in an undefined expression.
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OK, so first looking at this quotient property with an example: the fifth root of 17x^10, all of that divided by 12y⁴.
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This is the fifth root of 17x^10, divided by 12y⁴.
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This can be split up into this: the fifth root of 17x^10 over the fifth root of 12y⁴.
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And the quotient property, along with the product property, allows us to simplify radical expressions,
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which we are going to talk about in a few minutes.
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Here, I have the **product property**: this would be something along the lines of
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the fourth root of 4x⁶ equals the fourth root of 4, times the fourth root of x⁶.
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And again, this helps us to simplify--the same idea as working with square roots, only now the index is a different number.
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There is also the restriction that, if n is even, a and b are greater than or equal to 0.
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And that would be to avoid this situation: look at something like the square root of 10.
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Using the product property, I could say, "OK, this equals the square root of 5 times 2, which equals the square root of 5, times the square root of 2."
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So, I have an even index, and I end up with this--not a problem.
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However, I could also say, "Well, the square root of 10...10 also factors out to -5 times -2."
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So, I follow the product property, and I end up with the square root of -5, times the square root of -2.
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And this is not what we are looking for when we talk about a radical; we are staying with real numbers.
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And so, when we have an even index (it could be 2, 4, 6, 8, something higher), then we define a and b as greater than or equal to 0, not as negative numbers.
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Again, reviewing some concepts that were learned in Algebra I and applied to square roots,
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but applying them to other roots this time: simplifying: a radical expression is simplified if the index is as small as possible,
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the radicand contains no nth powers and no fractions, and no radicals are in the denominator.
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Let's go ahead and look at this part: the radicand contains no nth powers.
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Starting out with square roots, just to illustrate this: if you have something like the square root of 18,
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this can be rewritten, using the product property, as 9 (which is 3²) times 2.
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Well, since the root here, the index, is 2, here I have n = 2.
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I look in the radicand, and I have an nth power; I have 2.
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So, this could be further simplified to say "the square root of 3², times the square root of 2, equals 3√2."
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We did this earlier on; but we just restricted this discussion to square roots.
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I could look at a more complicated example, using an index of 3, looking for the cube root.
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If I had something like the cube root of 27, times x⁷: this is not in simplest form,
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because I have an index of 3, and I could rewrite this as 3³...well, x⁷ is equal to x⁶ times x.
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And x⁶ is equal to (x²)³; so if I rewrite this as x⁶(x),
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I could then go on and write this as the cube root of 3³,
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times...since x⁶ is equal to...(x²)³, times x.
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OK, so now, I can see that this was not in simplest form, because I have elements here in the radicand that were raised to the n power.
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So now, what I can do is say, "OK, this 3 essentially cancels out that 3; and I end up with just a 3."
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The cube root of 3³, of 27, is 3; the cube root of x², cubed, is x².
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And that just leaves an x behind, like this.
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OK, so simplest form means that the radicand contains no nth powers.
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Look at the index and make sure that you can't factor out something that would be raised to that same power.
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Also, the radicand cannot contain fractions--no fractions if it is in simplest form.
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Something such as the fourth root of x, over 2z, is not in simplest form,
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because there is a fraction under this fourth root sign.
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And we can use the quotient property to simplify this; and we will talk in a few minutes
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about how you go about getting rid of fractions that are under the radical sign.
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OK, the other thing is: no radicals in the denominator.
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So, if I have something such as 3y divided by the cube root of 2y, this is also not in simplest form.
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And again, we are going to talk about how to get rid of radicals in the denominator, how to go about simplifying those, in just a second.
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So, the index is as small as possible; that is what we are going to be working with (this is usually not an issue;
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usually the index is as small as it can be); the radicand contains no nth powers (so if there is an
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nth power inside, as part of that radicand, we need to take the root of that and remove it from under the radical;
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the same here with the cube root; in addition, the radicand contains no fractions (so something like this
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is not in simplest form); and there are no radicals in the denominator (so this is also not in simplest form).
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You can go through this checklist in your mind, when you think you are done simplifying,
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and make sure that the expression you are working with meets these conditions.
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OK, so I mentioned that you cannot have radicals in the denominator if something is going to be in simplest form.
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So, getting rid of radicals in the denominator is known as rationalizing the denominator.
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So, now we are going to talk about rationalizing denominators.
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And we are going to start out just talking about when we are working with a monomial--when we have a radical in the denominator that is a monomial.
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So, to eliminate radicals in the denominator, multiply the numerator and the denominator by a quantity, so that the radicand is an nth power.
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What does this mean? 2 divided by √5xy: what I want to do is make this radicand (5xy)²,
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because here, n equals 2; so I want to get 5xy, squared, as the radicand.
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OK, so in order to do that, what I need to do is multiply the numerator and the denominator by the square root of 5xy.
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This is going to give me...2 divided by the square root of 5xy, times the square root of 5xy, divided by the square root of 5xy.
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Now, I am allowed to do that, because this is just 1; these would cancel out and give me 1, and I am allowed to multiply this by 1.
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So, the numerator is 2 times √(5xy), divided by...looking at the product property:
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the product property tells me that, if I multiply these two,
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I am going to end up with √(5xy times 5xy), which equals √(5xy²).
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So, when I have the index the same as the power that the radicand is raised to, I can just eliminate that radical sign, and then eliminate this power.
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So, I have 2 here, really, although it is not written out, and a 2 here; so I can get rid of both of those and get rid of the radical sign.
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So, this is going to equal 2√5xy, divided by 5xy.
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Now, this is in simplest form, because I no longer have a radical in the denominator.
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And I achieved that by multiplying both the numerator and the denominator by the square root of 5xy, so that I ended up with √(5xy²).
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And I could take the square root of that to just get 5xy.
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But you need to make sure that you multiply both the numerator and the denominator by the same term,
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so that you are actually just really multiplying this entire thing by 1.
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OK, if the expression in the denominator is a radical, but it is a binomial, you need to use a different technique.
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We just talked about rationalizing a denominator when the radical was the monomial.
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But if you have a binomial, then what you need to do is work with a conjugate.
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So, first let's just review what a conjugate is.
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Conjugates are the sum and difference of two terms--not even worrying about the radical sign right now.
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It is something like a + b and a - b; but here, we are working with radicals, so it could be something like √2 + 1 and √2 - 1.
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So, these two are conjugates--the same numbers; the same radical sign; the only difference is that this one is positive; this one is negative.
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OK, so these radical expressions, a√b and c√d and a√b - c√d--the only difference here is in the sign.
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These are called conjugates; so they can be used to rationalize denominators that are binomials.
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For example, if I have 5 divided by the square root of 7 minus 3, well, √7 - 3...the conjugate of that would be √7 + 3.
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So, these two are conjugates; this is a conjugate pair.
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So, what I need to do is multiply 5, divided by the square root of 7, minus 3, times √7 + 3, over √7 + 3.
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Now, since the numerator and the denominator are the same, I am really just multiplying by 1; so again, this is allowed.
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All right, so in the numerator, this gives me...using the distributive property...5√7, plus 5 times 3, which is 15.
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The denominator: recall that here, if you look at what I am doing, it is multiplying a sum and a difference.
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So, if I had a + b times a - b, this is going to end up giving me a²...the outer term is -ab;
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the inner term is positive ab; so that is going to cancel out, and then I am going to get -b².
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b times b is b², but I have a negative sign in front of it.
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So, I am going to end up with a² - b²; and in this case, multiplying √7 - 3 and √7 + 3,
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a equals √7 in this situation, and b equals 3; so this is going to give me
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(√7)² - (I am using this format) 3², which equals 5√7 + 15.
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OK, I have (√7)², √7 times √7; it is just going to give me 7.
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Minus 3² (which is 9) is going to give me 5√7 + 15; 7 - 9 is -2, and I can put that negative out in front.
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So, this is going to give me 5√7 + 15, divided by 2.
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And now, the radical is gone from the denominator.
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So, I have rationalized the denominator when I had a radical, and I had a situation where it was part of a binomial.
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I had a denominator that was a binomial.
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And I did that by multiplying both the numerator and the denominator by the conjugate of the denominator.
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Another review of a concept you may have learned earlier on, which is adding and subtracting radicals:
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radicals are **like radicals** if they have the same index and the same radicand.
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For example, 4 times the cube root of 5x, plus 2 times the cube root of 5x:
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recall that this is the index; here, the index is 3 and the radicand is 5x; here the index is 3 and the radicand is 5x.
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So, I am going to use the distributive property, and I am going to say, "OK, I have the same; I can pull out this cube root of 5x."
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And that leaves behind 4 + 2; so this becomes the cube root of 5x times 6, or 6 times the cube root of 5x.
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All I did is add 4 and 2; and you can really just look at this as a variable, almost, with the radical.
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If I had given you something like 4y + 2y, that equals 6y; and here, we are going to let y equal the cube root of 5x.
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You could just look at it this way: that this whole thing is like a variable.
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And you can add these two together, but the variables are the same,
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because it is just saying 4 y's and 2 y's equal 6 y's, and that is the same idea here.
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OK, subtraction: the same thing--you just have to be careful (as always, when you are working with subtraction) with the signs.
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So, if I am subtracting something like 5 times the fourth root of 7yz, minus 3 times the fourth root of 7yz,
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I check and see that I have the same index (which is 4) and the same radicand (7yz).
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So, this becomes pulling out the same factor, which is the fourth root of 7yz, leaving behind 5 - 3.
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This is going to give me the fourth root of 7yz times 2, or I am rewriting it as 2 times the fourth root of 7yz.
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So, adding and subtracting like radicals is pretty straightforward.
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Just make sure that you check and make sure that the index numbers are the same, and the radicands are the same, before you try to combine radicals.
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Multiplying: with multiplying radicals, we are going to use the product property.
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And if we are going to multiply sums or differences of radicals, we will be using the distributive property.
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So, let's just start out with multiplying two monomials that involve radicals,
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the fifth root of 2x³ times the fifth root of x².
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Well, the product property, recall, tells me that the square root of ab equals the square root of a, times the square root of b.
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So, what I am doing down here, instead of going from left to right--I am going from right to left.
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I already have these two split up, but the product property tells me I can combine them.
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So, this would actually be the fifth root of 2x³, times x².
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Recall that, if you are multiplying exponents with a like base, then here, I can just add these exponents.
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So, this is going to give me 2 times x⁵.
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Now, I see that what I have (using the product property again) is the fifth root of 2, times the fifth root of x⁵,
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which equals...well, the fifth root of x⁵ is just x, times the fifth root of 2.
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So, you can see how multiplication and using the product property allowed me to actually simplify this.
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First, I used the product property to multiply these two together.
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Then, I used my property of exponents that says I add the exponents, since there are like bases here.
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That gave me 2x⁵; and I saw that this is not in simplest form,
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because I have a radicand that contains the nth power, the fifth power.
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So then, I further simplified.
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OK, so that is if you have monomials; now, for multiplying sums or differences of radicals, we need to use the distributive property.
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For example, if I am multiplying 2 times the square root of 3x, plus the square root of 2, times the square root of x,
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minus 2, times the square root of 5, we are going to use FOIL.
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The first terms (multiplying the first terms, because I am multiplying two binomials--FOIL--First terms):
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that is 2 times the square root of 3x, times the square root of x.
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Plus the outer terms--that is 2√3x, times -2√5.
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Inner terms are √2 times √x.
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And finally, the last terms are √2 times -2√5.
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OK, using the product property right here tells me that √a times √b is √ab.
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So, I am going to apply that here to get 2 times 3x times x, plus 2 times -2...that is actually going to give me a -4;
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so I am going to rewrite this as -4; √3x times 5...3 times 5x, plus the square root of 2x.
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And then here, I have a -2 out in front; and then, that is the square root of 2 times 5.
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I am doing some simplification: this equals 2 times 3x², minus 4 times √15x, plus √2x, minus 2√10.
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And I see here that I am not quite done yet, because I have an index of 2, and my radicand contains something to the second power.
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So, I can pull this out, and I need to remember to use absolute value bars, because this was an even index,
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and when I took that root of this, I ended up with an odd power, 1, so I need to use absolute values.
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And I can't simplify any further, because I can't combine these, since they are not like radicals.
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They are to the same powers, but none of them have the same radicands, so I can't add or subtract them.
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So again, multiplication with sums or differences of radicals--you just use the distributive property,
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like we have earlier on, when working with numbers or variables.
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OK, in this first example, we are going to simplify this expression; and it is the fifth root of x⁸y⁹z⁵, divided by 243.
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So, I am going to use the quotient property, because I know that this equals, according to the quotient property,
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the fifth root of x⁸y⁹z⁵, all divided by the fifth root of 243.
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So, recall that, in order to be in simplest form, a radical expression needs to have an index that is as small as possible;
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no nth powers; no fractions under the radical; and no radicals in the denominator.
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So, you should be familiar with these rules.
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What we have is: when we started out, we knew it wasn't in simplest form, because I did have a fraction under that radical.
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I then took care of that by using the quotient property.
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I no longer have a big radical sign where I have this fraction under it; I split it up.
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The only problem is that I now have a radical in the denominator.
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So, this is still not in simplest form.
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In addition, I also have some nth powers under here; when I work on this,
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I will see that there are some terms here that are to the fifth power.
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So, I can rewrite this as x⁵ times x³, because these have like bases, so I add the exponents.
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That would give me x⁸ back, so I can see here that I have n = 5, and the radicand contains some fifth powers.
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y⁹ would be y⁵ times y⁴, because I would add these to get 9 back; and then leave z as it is, z⁵.
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243 is not totally obvious, but it turns out, if you work this out, that 3 to the fifth power is 243.
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So, I also have a fifth power as part of the radicand in the denominator.
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OK, I am going to use the product property to rewrite this with my fifth powers all together here:
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x⁵y⁵z⁵ times what is left over (that is the fifth root of x³y⁵).
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OK, what this gives me is the fifth root of these; well, these are all to the fifth power.
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So, I simply remove the radical, get rid of the n, and get rid of this power to get x; the same with y and z.
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Since this is odd, I don't have to worry about absolute value bars.
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We only worry about that when the index is even.
00:27:47.500 --> 00:27:55.000
And this is times the fifth root of x³; this should actually be y⁴ right here,
00:27:55.000 --> 00:28:05.900
and y⁵ here and y⁴ here, divided by...well, this is the fifth root, and this is raised to the fifth power.
00:28:05.900 --> 00:28:22.100
So, that just becomes 3; or I could rewrite this as xyz divided by 3, all that times the fifth root of xy⁴.
00:28:22.100 --> 00:28:24.700
So, I double-check: is this in simplest form?
00:28:24.700 --> 00:28:33.200
There are no nth powers in the radicand; what I have left is xy⁴; there are no fifth powers here.
00:28:33.200 --> 00:28:39.700
There are no fractions under this radical sign; there are no radicals in the denominator.
00:28:39.700 --> 00:28:45.000
So, this is in simplest form.
00:28:45.000 --> 00:28:53.800
Here, we are asked to add and subtract some square roots; recall, though, that you can only add or subtract radicals
00:28:53.800 --> 00:28:59.400
if they have the same index (which these all do) and the same radicand (which they don't).
00:28:59.400 --> 00:29:04.400
However, they are not in simplest form yet; it is important to always simplify first.
00:29:04.400 --> 00:29:11.700
The problem is that I have some perfect squares left here as part of the radicand that I could pull out.
00:29:11.700 --> 00:29:16.300
So, I am going to rewrite this with the perfect squares factored out.
00:29:16.300 --> 00:29:28.900
And I am going to use the product property; 24 is 4 times 6; 48 is 16 times 3, so I have a perfect square;
00:29:28.900 --> 00:29:37.600
54 is 9 times 6, so that is another perfect square; and then, 75 is 25 times 3.
00:29:37.600 --> 00:29:46.000
According to the product property, the square root of ab equals the square root of a times the square root of b.
00:29:46.000 --> 00:30:02.500
So, I can rewrite this as √4 times √6, minus √16 times √3, plus √9 times √6, minus √25 times √3.
00:30:02.500 --> 00:30:06.000
And as you get better at this, you might not need to write out every step.
00:30:06.000 --> 00:30:08.800
But for now, it is a good idea, just to keep track.
00:30:08.800 --> 00:30:25.400
This is going to give me 2√6, minus √16, which is 4, √3; plus √9, which is 3, √6; minus √25, which is 5, √3.
00:30:25.400 --> 00:30:32.500
Now, I am looking, and I actually do have some like radicals that I can add, because they all have the same index;
00:30:32.500 --> 00:30:47.400
and now I see that these two (let's rewrite it like this): 2√3 + 3√6, have the same radicand.
00:30:47.400 --> 00:30:55.200
And then, I have -4√3 - 5√3.
00:30:55.200 --> 00:31:05.100
In this situation, what I am going to do is add the 2 and the 3, and this is going to give me 5√6.
00:31:05.100 --> 00:31:13.400
You are looking at this the same way that you would a variable: if I had 2y and 3y, it would become 5y.
00:31:13.400 --> 00:31:29.300
Plus...this is going to be -4 and -5; that is going to be -9, and then √3, which is 5√6 - 9√3.
00:31:29.300 --> 00:31:36.800
So, this is now in simplest form; and when I looked at this, it first looked like I could not combine these.
00:31:36.800 --> 00:31:42.400
But when I went about my checklist of how to simplify, I didn't have to worry about radicals in the denominator
00:31:42.400 --> 00:31:49.200
or fractions under the radical sign, but I did have to get rid of the perfect squares that were part of the radicand.
00:31:49.200 --> 00:31:52.600
And I used the product property to achieve that.
00:31:52.600 --> 00:32:01.800
Once I did that, I saw that I actually could combine some of these radicals to get this simplest form.
00:32:01.800 --> 00:32:06.700
OK, simplify: this time, I am asked to multiply two binomials.
00:32:06.700 --> 00:32:12.100
And I am going to use FOIL, just like I normally would.
00:32:12.100 --> 00:32:22.300
I am rewriting this here; use FOIL just as though you are multiplying any other two binomials.
00:32:22.300 --> 00:32:30.200
So, this is going to give me 2√3 times the other first term, which is 6√3,
00:32:30.200 --> 00:32:38.400
plus 2√3 (the outer terms) times 2√5.
00:32:38.400 --> 00:32:54.700
Now the inner terms are -4√5 times 6√3; and then the last terms are -4√5 times 2√5.
00:32:54.700 --> 00:32:59.000
OK, I am making sure that I have everything correct...inner, and then last.
00:32:59.000 --> 00:33:12.400
OK, we can use the product property; and the product property tells me that √ab = √a times √b.
00:33:12.400 --> 00:33:18.200
And I am actually moving from right to left here, because these are separated, and I want to put them together.
00:33:18.200 --> 00:33:34.900
So, this is going to give me 2 times 6, √3 times 3, plus 2 times 2, and then this is √3 times 5,
00:33:34.900 --> 00:33:49.000
plus -4 times 6, and this is √5 times 3, plus -4 times 2, times √5 times 5.
00:33:49.000 --> 00:33:57.900
This gives me 12; and I could rewrite this as 3² + 4; this is 15.
00:33:57.900 --> 00:34:09.100
-4 times 6 is going to give me -24√15; -4 and 2 is going to give me -8; and I can write this as 5².
00:34:09.100 --> 00:34:14.900
OK, I am not done simplifying yet, because recall that I look at the index; it is 2;
00:34:14.900 --> 00:34:21.700
and I see that I do have a term here and here that are to the power of 2.
00:34:21.700 --> 00:34:29.000
I can take 3²; that square root is just going to be 3; so this gives me 12 times 3.
00:34:29.000 --> 00:34:36.000
This does not have any perfect squares within it as factors, so I leave it alone.
00:34:36.000 --> 00:34:44.700
The same with this term; here, I do have 5², so the square root of 5 squared is just 5.
00:34:44.700 --> 00:34:56.300
Continuing to simplify: 12 times 3 is 36; -8 times 5 is -40.
00:34:56.300 --> 00:35:05.900
I can combine these two, 36 - 40; that is going to give me -4.
00:35:05.900 --> 00:35:16.600
I also can combine these two, because these are the same index, 2, and they are the same radicand.
00:35:16.600 --> 00:35:28.500
So, this would be the same as 4 - 24 times √15,
00:35:28.500 --> 00:35:39.700
which is going to give me -4; and then 4 - 24 is just going to give me -20√15.
00:35:39.700 --> 00:35:50.900
So, the simplified expression here is this; and I know it is in simplest form, because I don't have any fractions under the radical sign.
00:35:50.900 --> 00:35:59.800
There are no radicals in the denominator; and I don't have any nth powers here; I don't have any perfect squares under here.
00:35:59.800 --> 00:36:06.700
So, I first multiplied these two binomials out, using the distributive property.
00:36:06.700 --> 00:36:16.200
I got down to here; then I took my perfect squares; I took the square root of this number, 9, which is 3.
00:36:16.200 --> 00:36:20.900
I took this square root of 5², which is 5.
00:36:20.900 --> 00:36:34.500
I did some more simplifying, and then combined these two radicals that had like radicands and had the same index.
00:36:34.500 --> 00:36:40.000
OK, Example 4: Thinking about my rules, is this is simplest form?
00:36:40.000 --> 00:36:48.400
No; it doesn't have any fractions under the radical; however, there is a radical in the denominator.
00:36:48.400 --> 00:36:54.200
So, a radical expression is not in simplest form if there is a radical in the denominator.
00:36:54.200 --> 00:37:03.400
Recall that, to simplify a radical binomial expression, you multiply both the numerator and the denominator by the conjugate of the denominator.
00:37:03.400 --> 00:37:12.200
So here, I have 2 + √3; the conjugate of that is going to be 2 - √3.
00:37:12.200 --> 00:37:18.600
So, these are conjugates; this is a conjugate pair.
00:37:18.600 --> 00:37:35.100
I am going to take 4 - √3, divided by 2 + √3; and I am going to multiply that times this conjugate, 2 - √3.
00:37:35.100 --> 00:37:39.400
This is just the same as multiplying this by 1.
00:37:39.400 --> 00:37:46.500
I am going to have to use the distributive property, because I am multiplying these binomials.
00:37:46.500 --> 00:37:49.500
So here, I am just going to have to use FOIL, as usual.
00:37:49.500 --> 00:37:59.100
The first two terms are going to give me 4 times 2; the outer is going to give me 4 times -√3.
00:37:59.100 --> 00:38:14.400
The inner two terms--that is -√3 times 2; and then, the last two terms are -√3 times -√3.
00:38:14.400 --> 00:38:23.500
The denominator is a little bit easier, because this denominator is in the form (a + b) (a - b).
00:38:23.500 --> 00:38:27.300
It is the product of a sum and a difference, which gives me a² - b².
00:38:27.300 --> 00:38:45.100
Here, a = 2, and b = √3; so that is going to give me a², which is 2², minus (√3)².
00:38:45.100 --> 00:39:06.400
OK, simplifying: 4 times 2 is 8; this is 4 times -1, so that is -4√3; this is -1, essentially, in front of here, times 2 is -2√3.
00:39:06.400 --> 00:39:16.500
Here, I have a negative and a negative; that is going to give me a positive, so it is going to be + √3; and that is squared.
00:39:16.500 --> 00:39:28.200
OK, all divided by...2² is 4; minus...well, the square root of 3 squared is just 3.
00:39:28.200 --> 00:39:43.200
OK, so this gives me 8 - 4√3 - 2√3.
00:39:43.200 --> 00:39:51.200
Well, this √3 squared is also 3, so I am going to change that to a 3...divided by 4 - 3, which is 1...
00:39:51.200 --> 00:40:06.600
so I can just not write that 1; and here I have 8 + 3; that is 11; I also see that I have -4√3 and -2√3.
00:40:06.600 --> 00:40:13.800
Since these have the same index and the same radicand, I can combine these two to get -6√3.
00:40:13.800 --> 00:40:19.600
So, we started out with something not in simplest form, because it had a radical in the denominator.
00:40:19.600 --> 00:40:26.900
And since that was a binomial, I multiplied both the numerator and the denominator by the conjugate of the denominator.
00:40:26.900 --> 00:40:34.100
I got this whole thing; I then just continued to simplify.
00:40:34.100 --> 00:40:43.900
And when I got to here, I saw that I had two radicals that could be combined to give me -6√3.
00:40:43.900 --> 00:40:45.900
And then, I combined my constants.
00:40:45.900 --> 00:40:50.300
And this is now in simplest form, because there are no radicals in the denominator;
00:40:50.300 --> 00:40:57.800
there are no nth powers in this radicand (no perfect squares, in this case);
00:40:57.800 --> 00:41:06.200
and no fractions under the radical sign; and the index is the smallest power that it can be.
00:41:06.200 --> 00:41:11.000
That concludes this session of Educator.com; thanks for visiting!