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## Imperfect Squares

In the first lesson, we are able to simplify radicals because they were perfect squares or perfect cubes. But not everything we come across is going to be so perfect. What about this square? If this square had area of two units squared, what do we think of the side length of the side link would be? Choose one of these answers for the side length, is this one of these numbers or it is something else?

## Imperfect Squares

This side length, or this side, would have to be something else. We can't have negative side lengths, so this first three choices would not make sense. Zero times 0 equals 0, so that also wouldn't make sense, since that doesn't give us an area of 2. One time 1 equals 1, so that also would not give me an area of 2, so this one's out. We know 2 times 2 or 2, or 2 squared is 4. So this one won't work either. And 4 times 4, or 4 squared equals 16. So, this one's out as well. The answer is something else, lets see what it is.

## Irrational Numbers

It turns out that that something else is the square root of 2. When we multiply the square root of 2 by the square root of 2, we get the number 2. We know this is true since we can rewrite this as the square root of 2 squared. We use our knowledge of fractional exponents to rewrite this radical as 2 to the 1 half. We're raising a power to a power so we multiply these exponents together to get simply 2. Now, we're not so much interested in multiplication of radicals right now, we'll run that later. But what we are interested in is the square root 2. This is an irrational number. And the irrational number is a number that doesn't terminate. It means that whatever is after the decimal never ends. The square root of 2 is about 1.4, but that's just an approximation. This decimal actually has an infinite number of places following the decimal point, it goes on forever. So, when we think about irrational numbers, we're just thinking about numbers that can't be written as a fraction. It's not important to plug a number like the square root of 2 into your calculator since we really only get an approximation of that answer. Throughout the rest of this unit, we won't be working with these crazy decimals. Instead, we'll just write the radical symbol and the number.

## Simplify Square Roots 1

This one would have been 5 times the square root of 3. Great work if you figured that one out. We know 75 is the same thing as 25 times 3. We want to make sure that we split this up into factors where one of the factors is a perfect square. This allows us to use our product rule and then to take the square root of this one. So we use the product rule in this step, and then we take the square root of 25 which is just 5. We're left with 5 times the square root of 3, which is just equal to 5 root 3. And notice that we make another assumption. If we write a number in front of a radical sign, we assume there's multiplication in between them. So here's our answer.

## Simplify Square Roots 2

This would simplify to 10 times the square root of 2, or 10 root 2. The key ideal is to remember that multiplication is between the numbers and the radical sign. We can rewrite the square root 50 as the square root of 25 times 2. Then we use our product rule to break this radical up into two separate radicals. We know the square root of 25 is 5, so we have 2 times 5 which is 10, times the square root of 2. We don't want to add these numbers together, since multiplication was really between the radicals and the numbers to begin with. So that's why the answer wasn't 7 root 2. It might be easy to think that way. But just remember what's actually going on between the numbers.

## Simplify Square Roots 3

Try simplifying each of these square roots.

## Simplify Square Roots 4

Alright, let's see you try these. Try finding the square root of 98, 150, and the square root of 72.

## Simplify Square Roots 4

The square root of 98 equals 7 root 2, the square root of 150 equal 5 root 6. And the square root of 72 equals 6 root 2. Great work if you got each of these correct. You're really mastering your square roots. We can split up 98 into 2 times 49 and then take the square root of 49 which equals 7. The 2 states inside of the radical are the square root. Because we can't simplify it. We use a similar approach for the square root of 150. We rewrite 150 as 25 times 6. And take the square root of 25 to get 5 root 6. And finally, for the square root of 72, we rewrite 72 as 36 times 2. And then take the square root of 36. Which leaves us with 6 times the square root of 2 or 6 root 2. I hope you're starting to see some patterns with square roots. And, if anything, you can always start simplifying these even numbers by dividing by 2. I've shown you the quickest way to get to the final answers but there are, of course, some other ways you could have gotten there as well. Be sure to compare your work to these answers to see if you could have gotten to the answer more quickly.

## Product of Roots

We've seen how we can use the product rule to simplify a radical by splitting it up into the multiplication of two different radicals. We can also use this product rule to multiply two radicals together, but note that we can only do that when the radicals have the same root. In this case, they're both square roots, so we can multiply them together. If we had two third roots or two fourth roots, then we could also multiply them together. This number, which is our index, just needs to match. So use our product rule here to simplify each of these multiplication problems. And don't worry about simplifying in the end, because you actually can't in these cases.

## Product of Roots

Here are each of the answers. Great work if you figured those out. We want to make sure that when we multiply, we just multiply the numbers inside of the radical, and we leave the index alone. So we have the square root of 5 times the square root of 7, which equals the square root of 35. Here, we just multiply the radicands of 3 and 12 to get the cubed root of 36. Be sure you listed the 3 as the index for your radical. For the third one, we have the 5th root of 100, since 5 times 20 equals 100. And for the last one, we have the 4th root of 21, since 3 times 7 equals 21.

## Simplify Higher Order Roots

We've simplified square roots and cube roots, so let's try some more higher order roots. When simplifying higher order roots, we want to look for a factor that has an exponent or power, equal to the index of the root. So this means that we need to find a factor of 40, that appears three times, or has a power of 3. If we have a 4th root, then we need to find a factor that appears four times. Or has the power of 4 in order to remove that factor from the radical. So you try simplifying each of these. Write your answer in each of these boxes and be sure to use the apropriate index for the roots.

## Simplify Higher Order Roots

For the cube root of 40, we have 2 times this cube root of 5. For the fourth root of 48, we have 2 times the fourth root of 3, and for the fourth root of we'll have 5 times the cube root of 20. Fantastic work for getting at least of those correct. These were pretty tough. For the first one, we can rewrite 40 as eight times five. Eight is the same thing as two cube so now we have a perfect cube or a factor that appears three times within our radical. We can use the product rule to split this radical up into the product of two radicals. We know that a cubed root undoes a cubed power. So we're really left with just 2 here. This also goes in line with our thinking, a power divided by root. 3 divided 3 equals 1 so we just have 2 to the 1. We can't simplify the 5 so that stays inside of its cubed root and we have that multiplied on the end with our answer. 2 times the cube root of 5. For the fourth root of 48, we split up 48 into 16 times 3. We rewrite 16 as 2 to the 4th and then we can take the fourth root up to the fourth. The root undoes the fourth power. We leave the 3 inside of the fourth root since we don't have four factors of 3, we just have one of them. So our final answer is 2 times the fourth root of 3. We use a similar method for finding the fourth root of 64. We rewrite 64 as 16 times 4. Take the our last one we split up 2,500 into 25 times 100, 25 equals 5 squared and 100 equals 5 times 20. We do this step so that way we can get three factors of 5 together. We know that the cube root of phi cubed is just 5. 20 is not a prefect cube since the factors of 20 are 2, 2, and 5. There's not three factors that repeat. This means 20 stays inside of our cubed root, and we are left with our final answer. 5 times the cube root of 20.

Simplifying variables underneath a radical works the same way as it does for factors, or just numbers. We'll look at simplifying the cubed root of x to the fourth times y to the sixth. Keep in mind that these variables actually represent numbers, like x could equal 2 and y could be 3. Our rules for these won't change, we'll still apply the product rule if we can, and we can only remove factors if they repeat three times. For example, we can split this up into the cube root of x times x times x times x. These four xs are the same as x to the fourth. Similarly, our second radical will have the cube root of y to the sixth. Since our x variable repeats three times, we you can pull one of them out, we can also pull out one y from these three, and one y from these three. This x however is a little bit lonely, it doesn't have three of itself, so we leave it inside the cube root. So our final answer is x times y squared time the cube root of x. So in order to simplify any radical we want to take the radican and split it up into its factors. If a factor appears the same number of times as the index of the root, then that means we can pull one of it out. That was why we could pull out one x, one y and one y. We multiply all those factors together and leave any single or lonely factors inside of the radical. This will give us our final answer.

So let's see if you have the hang of it. Try simplifying these two radicals and put your answers here. These are a little bit more difficult, so don't worry if you stumble. Just try your best as always.

Here are our two answers. Great work if you found even one of those. For the first one, we want to list all the factors of our radicand. We know 2 times 2 times 3 equals 12. Then we have 3 x's multiplied together. And 2 y's multiplied together. We're taking the square root and the index is at 2 here. So we want to find factors that repeat twice. We have two 2s, two x's and two y's, so we can pull one 2 out, one x out and one y out. This 3 and this x are factors that need to stay underneath the square root symbol, and it's because they are not paired. This leaves us with our final answer, 2xy times the square root of 3x. For the second one, we want to find factors that repeat three times, since our index is a 3. We have three 3's and three c's, so we can pull out one 3 and one c from our radical. These factors don't appear in a triple, so we have to leave them inside of the cube root. We'll add 2a squared times c and our cube root. This gives us our final result. Keep in mind that this 3 is the index on the root, it's not the exponent of the c. So be careful when you're writing your expressions.

There is a quicker way to think about our process though. When simplifying radicals that involve variables, we want to remember power divided by root. For example, if we wanted to simplify the square root of x to the fourth, we would do the power of the 4 divided by the root 2, so we would get x squared. We're dividing this power by our index of 2, since it's the square root. This should also make sense because x to the 4th can be rewritten as x squared times x squared. We have a repeated factor here so we can pull out x squared. We can use this same approach of power divided by root for higher order roots like cube roots, fourth roots and fifth roots. So how do you think we could simplify this radical? Write your answer here.

The cube root of x to the 12th would equal x to the 4th. Nice work if you figured that one out. We use power divided by root again to get x to the 12 divided by 3. 12 is our power and 3 is our root, or the index. It tells us what type of root we are taking. This should also make sense as our answer since we can expand x to the 12th as x to the 4th times x to the 4th times x to the 4th. x to the 4th appears 3 times underneath our cube root, so that means we can pull one of its factors out. So we're left with just x to the 4th.

## Remainders with Power Divided by Root

In general it's easy to use power divided by root rather than to expand the factors of a variable. But not all powers are evenly divisible by a root. We know 3 doesn't divide evenly into 14. We could rewrite x to the 14th as x to the 12th times x squared. We use our product tool to split this up into two separate cube roots. And since we know that cube root of x to the 12 is equal to x to the 4th, this would be our final answer. An easier way to get to this more quickly is to use power divided by root and to look at the remainder. We know this radical can be written as x to the 14 divided by 3. So when I divide the power by the root, I get 14 divide 3, which equals 4 remainder 2. Notice that the 4 appears as the exponent of our factor outside of the radical, whereas the remainder 2, is the power inside of the radical. So, using this ideal of the remainder, I want you to see if you can simplify this radical. Use the idea of power divided by root to figure out what would be on the outside of the radical and what would stay on the inside. When you think you have it put your answer here.

## Remainders with Power Divided by Root

To solve this problem, we want to use power divided by root for each of our factors. So for x we'll have 7 divided by 3. For y we'll have 5 divided by 3. And for z we'll have 10 divided by 3. Since 7 divided by 3 is 2 remainder 1, we'll have x squared on the outside and x to the 1 on the inside of our cubed root. For y, we'll have 1y on the outside and we'll have y squared left on the inside. And finally for z, we'll have z cubed on the outside, and 1z on the inside of our cube root. This is our final result. We could have also completely expanded this radicand. Once we have all the factors listed out, we want to find the factors that appear three times. That means we can pull it out. Here's a group of three xs. And here's a group of three xs. So we can pull out an x and an x. We can also pull out one y, and we can pull out three zs. These other factors don't appear in a triple, so we leave them inside of our cube root. X, y squared, z. And finally, we just rewrite the front part of our radical to get x squared y z cubed times the cubed root of xy squared z.

## Simplify Roots Practice

Let's wrap up this lesson with some practice. I want you to simplify each of these roots and write your answer in the corresponding box. You'll want to use your knowledge of simplifying numbers and variables. Also, pay particular attention to the type of root you're taking. In this last one, you're taking a fifth root.

## Simplify Roots Practice

Here are our final answers. Great work if you got those correct. This cube root and this 5th root might have been tricky, so if you struggled with them, that's okay. For the first one, we split 32 into 2 times 16. The square root of 16 equals 4, and the square root of x squared is x, and the square root of y squared is y. This leaves us with a 2 in the radical, which gives us our final result. For the second one, we want to split up 75 into 3 times 25. The square root of 25 equals 5, and then we leave the 3 inside of the square root since it's not a perfect square. Next, we use our idea of power divided by root. We have a square root, so we want to take this power and divide it by 2, 3 divided by 2 equals 1 with a remainder of 1. So, one a will be on the outside, and one a will be on the inside. For b, we'll have 8 divided by 2 which is 4, so e to the 4. And finally, for c, we'll have c squared since c to the 4th divided by 2 equals c squared. This is our final result. For the next problem, we rewrite a as 2 cubed. We know that cube root of 2 cubed is just 2 since that this factor appears three times. Now we us power divided by route to simplify each of these variables. 5 divided by 3 is 1 remainder 2. For c, 4 divided by 3 equals 1 remainder 1. And for d, we have 6 divided by 3 which equals d squared, and no d's inside the radicals since the remainder is 0. And finally, for our last one, we can't take the 5th root of 36, since we don't have a factor that appears 5 times. Nothing simplifies. But for our variable factors, we can simplify 10 divided by 5 equals 2. So, we'll have w squared, with no w's inside our radical, since the remainder is 0. For x, we'll have 6 divided by 5 which equals 1 remained 1. And for y, we'll have 12 divided 5 which equals 2 remainder 2. Here's our final answer.