I hope you've been notice that the theme here in our units are different types of expressions. We've been learning the skills into perform operations on polynomials, rational expressions, and now radicals. The last operation mean to learn radicals is division. We can think of the division of two radicals as fraction. We'll have root two divided by root five. But for radical expressions, we don't want a radical to appear in the denominator. There are few reasons why we want to change this. First, it's typically easier to approximate this division if we divide it by an integer here rather than a radical. Secondly, we can determine if the expressions are equivalent to other expressions by changing what this fraction looks like. Keep in mind that when we rationalize a denominator, we're going to change what the fraction looks like but we won't change it's value. What do you think we can multiply the numerator and denominator by so that we have a denominator without a radical? We want to multiply by a form of 1 here so whatever we multiply this denominator by, we'd also multiply this numerator by.
We want to multiply by root 5 divided by root 5. Now, that's great thinking if you figured it out. We want to multiply by a form of 1 so we don't change the value of this fraction. We do this because root 5 times root 5 equals root 25. Now, we can take the square root of our denominator, which equals 5. For the numerator, we'll have root 2 times root 5, which equals root 10. And that, we'll just leave as root 10. This means that root 2 divided by root 5 is really equal to root 10 divided by 5. Now, if you're in disbelief, I would plug this into a calculator, and this into a calculator, and see if you get the same result. I think this is pretty cool.
I mentioned before that we rationalize the denominator so we can make a better approximation for our answer. We can use the square root of 10 divided by 5 to help us approximate the answer for this expression. I don't know what it really means to divide by root 5. But I can approximate this by dividing by 5. We know the square root of 9 equals 3, so the square root of 10 is just a little bit more than 3. We're going to approximate it though and just use 3. So root 10 divided by 5 is about 3 5th or 6 10th. If we would had left this expression as it was, we would have had to divide by a non-terminating decimal. Dividing by the square root of 5 is much more challenging than doing this approximation. As we work through this unit, we'll be taking denominators that contain radicals and changing them into integers or other expressions that don't contain radicals. For square roots, this is pretty simple. We can just multiply the numerator and denominator. By the exact radical that appears in the denominator. Then we just multiply our radicals together and simplify if possible.
Try rationalizing these denominators. When we rationalize a denominator, we want to multiply by something so it eliminates the radical in the denominator. So what do you think we need to multiply each of these fractions by, and what would be the final result?
Here are the answers, and great thinking if you got all three correct. For the first one, we want to multiply by the square root of 2 divided by the square root of 2. We really multiply by a form of 1 so that way we can change our denominator to the square root of 2 squared. This allows us to take the square root, so we get 2 in our denominator. And this is what we mean when we rationalize a denominator. We multiply our fraction by a form of 1, so that way our radical does not appear here. For the second one, we want to get rid of the square root of y in this denominator. So we multiply the numerator and denominator by the square root of y. This gives us the root of xy here, and the root of y squared here. Our factor of y appears twice inside of our square root, so the root undoes our power and we're left with y. Notice, another fraction with no radical in the denominator. That's rationalizing. And finally, for this last one, we just want to multiply the numerator and denominator by the square root of 2xy. Now you might be wondering, how did I get this. Well there's an intermediate simplification step. In the numerator, we'll have 4 times the root of 2xy. In our denominator, we know if we multiply the same square roots together, we'll just get the radicand, which is just 2xy. Now we simply 4 divided by 2 to get 2 divided by 1. Here's our final result.
For higher order roots though, we need to do something a little bit different. If you remember we had a product rule that let us split up a root of a and b. If these were factors of a radicand we could split the root up into the nth root of a times the nth root of b. Well it turns out there's a similar rule for division. The quotient rule lets us take the in root of a fraction we can just split it up into m through it of a divided by the m through it of b. This means that the cube root of this fraction becomes the cube root of 27. Divided by the cubed root of 16. We're just using our quotient rule. We know the cubed root of exactly take the cubed root of 16, 16 is 2 to the 4th. We don't want this cubed root to appear in the denominator, so we need to rationalize this part. If we can get this power to be a multiple of 3, then we would be able to take the cube root of it. For example, I really want this denominator to become the cubed root of 2 to the 6. I can take the cubed root of this, it's just 2 squared. So when we reationalize the denominator of a higher order root. We want to raise the power of each factor of the radicand to be a multiple of the index of the root. In other words, we want to have powers inside the radical that are evenly divisible by the index of the root. So, what would be multiply this numerator and this denominator by to change our denominator into this form? Keep in mind that whatever we multiply the denominator by we need to multiply the numerator by since we're really just multiplying by a form of 1.
The key idea when rationalizing a denominator is to choose a way to multiply by a form of one, so this denominator ends up simplifying to something that doesn't contain a radical. So, if we want to get the power of 4 up to the power of 6. We need to multiply it by two more factors of 2 inside of a cubed root. We have to use an index of three here since this is also an index of 3. If the indexes match, then we can multiply these radicants together to get 2 to the
So we know we get the cube root of 2 to the 6 for a denominator, and we'll get the cube root of 3 cubed times 2 squared in the numerator. So from here, simplify the answer. What's the final result?
We get a final answer of 3 times the cube root of 4 all divided by 4. Nice thinking, if you found this answer. We know the cube root of 3 to the 3rd is just 3, so that comes outside of our radical. We can't take the cube root of 2 squared since this power isn't divisible by this root. 2 squared stays inside of our cubed root. For our denominator, we'll use power divided by root to get
So let's think about this again. What should the denominator look like in order to be able to take a cube root of this denominator? We know that each of these factors needs to be raised to a power that is divisible by 3. So we'll multiply this numerator by something. And this denominator by something to change this, to what we need. So we need to multiply this numerator and this denominator by some sort of expression. When we do that, we want to get an expression inside of our cubed root that can be simplified. So you tell me. What should this expression be? You want to use the factor of 3, x and y and indicate their powers. Don't include the cube root inside of this box, since I already have it written for you outside. Keep in mind that you want all of your power divisible by 3.
We want this expression inside of our cube root to be 3 to the 3rd, x to the factors. Nice work if you found this expression.
So, what are we need to multiply this numerator by, and this denominator by to change our fraction into this form?
It turns out we want to multiply each part by the cube root of 3 squared x squared y squared. We want to multiply by enough factors of 3 x and y. To turn each of these powers into these powers. So for 3 to the 1, we want to multiply by 3 square to get 3 cubed. For x to the 7th, we multiply by x squared to get x to the 9th and for y to the 4th we multiply by y squared to get y to the 6. Keep in mind that this expression needs to be inside of a cubed root for us to multiply these factors together. And since we multiplied this denominator by this amount we must also multiply our numerator by the same amount. We're multiplying by a form of 1.
To finish up our last problem, we simply simplify the denominator. The cube root of 3 cubed is 3, the cube root of x to the 9th is x cubed, and the cube root of y to the 6 is y squared. Remember, the whole idea here was to rationalize this denominator. We wanted a denominator that didn't contain a radical. So we multiplied by enough factors underneath of a cube root, in order to change our denominator to something that we could simplify. For our numerator, we just had 1 time this cube root, and then we simplified the cube root, 3 squared is 9, and then I have my variables x squared y squared. We know we can't simplify this cube root of all, since all the factors only appear twice. And it's okay that we can't simplify the numerator, because our ultimate goal was to rationalize this denominator. We wanted to get rid of this radical symbol here.
Try rationalizing this denominator, you can use the last problem as a guide, and think about what you would need to do to change this denominator so that it doesn't contain a cube root. When you think you have the final numerator, and the final denominator, enter them here. Also, be sure to simplify if possible.
We have 3 times the cubed root of 4 xy squared divided by xy cubed. Great work if you found these two answers.
In order to rationalize this denominator, we want our powers of 2, x and y to be divisible by 3. This means, we'll going to to have 2 cubed, x to the 6th, and y to the 9th. We need to up these powers, so that way they're divisible by squared. This means, we have to multiply by the cube root of 2 squared x to the the 1 and 2 squared, which is 2 to the 3rd. x to the 5th and x to the 1, which is x to the 6th. And y to the 7th times y squared, which y to the 9th. Since we multiply a denominator by this expression, we multiply our numerator by the same expression. Now, we want to simplify. Our numerator stays the same, since we can't pull out any of these factors, since they don't appear three times. This denominator simplifies to 2 x squared and y to the 3rd. Remember, we're just applying power divided by root to get each of these powers. And we can actually simplify further. The 6 and the 2 reduce to 3 divided by 1. And this power of x reduces with 1 of the powers in the denominator. In other words, we'll have 1 more power of x in the denominator, than we would in our numerator. This leaves us with our simplified answer. Now, the hardest part to this problem is understanding what to multiply by here. I think if you start by figuring out what your denominator should look like, you'll get there more quickly.
Try rationalizing this expression. Write your final numerator here, and your final denominator here. Good luck on this one.
The final solution is the cube root of 4xyz squared divided by 2x squared yz, that's some great algebra work, if you got that one correct. Again, let's start by thinking about what this denominator should look like in order for it to simplify. We know we need the factors 2, x, y, and z to appear a multiple of three times. This means, that we want to change to the 1 To 2 to the 3rd. We want to change x to the 5th to x to the 6th, y squared to y cubed, and z to z cubed. Notice how each of these powers is the next highest multiple of 3 for these powers. So to change this denominator we'll multiply by the cubed root of expression, we multiply the numerator by the same amount. Next we simplify the numerator and the denominator. Our numerator can't be simplified so we just rewrite 2 squared as 4. For our denominator, we take the cube root of each of these factors. We use power divided by root to get 2 x squared yz. And here's our final answer.
Here's our third practice problems for rationalizing higher order denominators. Now careful here, we don't have a cube root anymore. We have a six root. So think about what factors you need to multiply by or what powers these would become in order to rationalize your denominator. Good luck.
Here was the answer for this one. Amazing effort if you got both of these correct. Now this one was tricky since there was a simplification that you had to do at the end. Lets see if you got it. Lets start by thinking what we want our denominator to be in order to simplify. We want each of these powers to be a multiple of 6. So that way, they can be divided by 6. We'll want the powers of 2, x and y to be divisible by 6 since that's our index of the root. So we want to change to the 1 into 2 to the 6 x squared into x to the 6th and y to the 8th into y to the 12th. This means we have to multiply the denominator by the 6th root of 2 to the 5th x to the 4th and y to the 4th. These are the missing factors we need, to increase each of these powers to these powers. We want to multiply by a form of 1, so we multiply our numerator by the same expression as this denominator. Next we multiply 4 times this radical to get our new numerator. We simply our denominator by writing 2xy squared. It's the each of these powers is not divisible by 6. So we just rewrite 2 to the 5th as simplify. This leaves us with our final answer. Great work with your factors and radicals if you got this far.
Here's your fourth practical problem, on rationalizing denominators. Think carefully about your index, and your powers for each factor and then simplify in the end if you can. Good luck.
Here, the answer is 3 times the 6 root of a squared b to the 5th c to the 4th, all divided by b squared, c squared. That's amazing algebra work, if you got any of that correct. Just like in our last problem, we want to increase the power of each of these factors, so that way they're divisible by 6. This means we need to change a to the 4th and to a to the 6th, b to the 7th, and to b to the 12th, and c to the 8th, and c to the 12th, as well. This means we need to multiply it by 6th root of a squared b to the 5th, c to the 4th. We know this, since a to the 4th times a squared is a to the 6th, b to the 7th times b to the multiply our numerators together, leaving 3a in front of our radical. Next, we simplify our denominator as a, b squared, c squared. We just divide each of these powers by 6. And finally, we reduce a power of a, since a divided by a equals 1. This leaves us with our final answer. And notice that a denominator does not contain a 6 root anymore. Great work.
But what if we wanted to rationalize the denominator that looked like this. If we multiplied our original fraction by the root of 7 divided by the root of 7. We would still wind up with a radical in our denominator. So this wouldn't work. Keep in mind, when we're multiplying, we would just distribute this root multiply by the root of 5 divided by the root of 5. We would still get an expression that contained a radical in our denominator. And that's no good. So we need a different way to be able to simplify this denominator so it doesn't contain these radicals.
We solved the difference of squares pattern with polynomials. We knew that a minus b times a plus b would equal a squared minus b squared. And we even solved this for a reverse process when we factored. So let's use this to our advantage with radicals. If I want to multiply these two expressions together, what would I get for my answer? As a hint, you really just need to square a, and then square b, and subtract them. So see if you can figure these three out. Good luck.
We'll have positive 3 for the first one, negative 3 for the second one, and negative 51 for the third one. Great work, if you've done all three of those. Now if you weren't quite sure what to do, that's okay. Let's see how it's done. For the first problem, we know that the a value's really equal to 3. So, we're going to square this 3. Our b value is really squared to 6, so we'll square the square root of 6. 3 squared equals 9 and the square root of 6 squared is 6. Then we just subtract these two numbers to get positive 3. For our second problem, the a value is really equal to the square root of 2. And our b value is really equal to the square root of 5. Squaring this square root leaves us with 2. And squaring this square root leaves us with 5. So 2 minus 5 equals negative 3 and this last difference of squares, will have a as 2 root of 3 and b as 3 root of 7 and here's where we want to be careful. We want to square this so after we square there's still multiplication between this number and this number. We square 3 root 7 to get 9 times 7. Next we multiply 4 times 3 to get negative 51. Now I think that's pretty incredible that we can take something that contains radicals. and pop out just one number
We're going to use this difference of square pattern to help us simplify the binomial radicals we saw earlier, in the denominator. But before we do that, let's learn a new term, the term is conjugate. The conjugate of a binomial is the same exact binomial, with an opposite sign in between the two terms. So the conjugate of the root of 7 plus 2 would be the root of 7 minus 2. We just change the middle sign. What's so great about the conjugate is that if we multiply a binomial with it's conjugate, we just get a number. We won't have any more radicals. For example, if we multiply these two binomials together, we'll just have a squared minus b squared. We square the root of 7 to get 7, and we square the 2 to get 4, 7 minus 4 equals 3. So what do you think would be the conjugate of these binomials. List those conjugates here. And then I want you to multiply it, the binomial with it's conjugate, to get the product.
Here are the conjugates and here are the products. Great work if you found at least four of those. We know for the conjugate we just change the sign between the two terms so the conjugate of 1 minus root 11 is 1 plus root 11. Here we change the subtraction sign to a positive. And here we change the positive sign to a negative. For the products we want to use our pattern for the difference of two squares. We know the product of two binomials with differing signs will be a squared minus b squared. So we can square a which is 1 here And squared b, which is the root of 11. One squared equals 1, and root 11 squared equals 11. We find the difference of these squares, which is 1 minus 11, which equal negative 10. For the second one, when we square the a term, we'll have 9 times is four, and we square the square root of three, which equals three. We know nine times two is 18, and four times three is 12. This gives us a final answer of six. And for the last one, we have negative five squared for the a term and six root two squared for the b. Negative five squared is 25 and 6 root 2 squared is 36 times 2 or 72. 25 minus 72 equals negative 47.
So going back to our original problem, if we want to rationalize this denominator we can actually use the conjugate. We multiply the denominator and the numerator by the same amount. So that way we multiply by 1. We know we can multiply by the conjugate since if we multiply these two expressions together we'll wind up with a number in our denominator, with no radical. We'll simply have a squared minus b squared, which would be 7 minus 5. A squared is 7, and b squared is 5. We multiply each of these radicals by 4, to get this numerator. Next we simplify our denominator to get 2. If we leave the 4 outside of these two radicals then we can simplify a common factor of 2 in our numerator and denominator. So our denominator simplifies to 1, leaving us with 2 root 7 plus contained radicals to something that turned into 1. Now not every time our denominator will be so nice. Sometimes we'll just have a number down here. In this case our denmoinator was 1, but we know that's not always the case.
Try this first practice problem now. Rationalize this denominator by multiplying by the conjugate. When you think you have the final answer, put it here.
The correct answer is 5 times root 5 plus 10. Nice thinking if you got it. This would be an alternate form of the correct answer as well. You could have factored a 5 from each of these terms to have 5 times the square root of 5 plus you're distributing the 5 to each of hte terms. We start rationalizing this denominator by multiplying the numerator and the denominator by the conjugate. The conjugate of root 5 minus 2 is root 5 plus 2. We distribute the 5 to each of these terms to get 5 root 5 plus 10 in our numerator. For the denominator we'll use the difference of squares pattern to do a squared minus b squared. The A is route 5 and the B is positive 2. This leaves us with 5 minus 4 in our denominator. 5 minus 4 equals 1 and anything divided by 1 is itself. So we're left with our final answer.
Try rationalizing the second problem. You want to multiply by the conjugate divided by the conjugate first and then simplify your answer. Be careful when you simplify your answer, and when you think you've got it write the final result here.
To start our problem, we want to multiply the numerator and denominator by the conjugate of this denominator. The conjugate of root 2 plus root 5, is root 2 minus root 5. We distribute this 6 to each of these terms to get 6 root 2 minus squared minus b squared. We square root 2 to get 2 and we square root 5 to get here's where we want to be careful. When we simplify we can divide negative 3 into this term and into this term, 6 divided by negative 3 is negative 2, and negative 6 divided by negative 3 is positive 2. This is how we get our answer of negative 2 root 2, plus 2 root 5. Another way the think about it is to factor a negative 3 from each of these terms. If we factor a negative 3 from this first term, we'll be left with negative 2 times root 2. And we factor a negative 3 from this term we'll be left with 2 root 5. We can check this really quickly by distributing this negative 3 to each of these terms. We'll wind up with our numerator from before. Now that we have a common factor of negative 3 in the numerator and in the denominator we can simplify. These reduce to 1 which leaves us with just our numerator, negative 2 root 2 plus 2 root 5. And now there are a couple of ways to write this answer. You could have factored a negative 2 out to get this answer or you could have factored a positive 2 out. Either of these is correct. Nice algebraic thinking if you got any of these answers.
Try rationalizing this fraction. Write your final answer here, and be sure to simplify if you can. Good luck.
When we rationalize this fraction, we'll get 4 root 35 plus 2 root 15. Great effort if you found it. We start by multiplying the numerator and denominator by the conjugate of this expression. The conjugate of 2 root 5 minus root 7 is plus 26 root 7. For the denominator, we use our difference of squares to do a squared minus b squared. We can't simplify or combine terms in the numerator, since this is root 5 and this is root 7. So we just list the numerator here again, and then we simplify the denominator. 2 squared is 4, and the root of 5 squared is 5. We have our subtraction sign and then we have the square of root into each of the terms. 15 divided by 13 is 4, and 26 divided by 13 is 2. Another way to think about it, is we're really factoring a 13 from the 52 root plus 2 root 7. This allows us to simplify the factor of 13 that appears in the numerator, and denominator. This simplifies to 1. So we're left with this for a fraction, and anything divided by 1, is simply itself. So we can write our answer like this.
Here's our last problem for the lesson. Try rationalizing this denominator. Think about if you can simplify in the end, and think about if you can combine like radicals at all. Good luck.
The final answer is 2 root 35 plus root 15, all divided by 5. Now if you got this one correct, that's amazing. You've really mastered rationalizing denominators. We start rationalizing this denominator by multiplying by the conjugate divided by the conjugate. Here our conjugate is 2 root 7 plus 2 root A squared minus B squared. Next, we simplify our denominator. 2 root 7 squared is 4 times 7, and root 3 squared is 3. Next, we multiply here to get 28. And we really know 28 minus 3 equals 25. Now I notice that each of these terms has a common factor of 5. I can factor a 5 from each of these, to be left with 5 times the quantity, 2 root 35 plus one root 15. And for our denominator of 25, I'll split it up to factors of 5 times 5. These reduce to 1, leaving us with 2 root 35 plus root 15 in the numerator. And 1 times 5 or 5 in our denominator. Great work if you found some of the steps or most of the steps to rationalize this denominator.