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Add Radicals

We've seen how to simplify radicals, so now let's try adding and subtracting them. We can use our knowledge of what we learned before to help us add radicals together. For example, to add polynomials together, we just added the like terms of x squared and x squared. We have three of these things called x squared, and two of these things called x squared. So if we add them all together, we'll have 5 of these things that we call x squared. In a similar way, we added 3 7ths and 2 7ths to equal 5 7ths. And most recently, we learned how to add rational expressions. If our denominators were alike, then we could add the numerators to get 5 divided by x plus 1. So based on what you know before, what do you think would be the sum of these two expressions?

Add Radicals

The first one be five times the square root of x, and the second one be thirteen times the forth root of five. Nice thinking if you got those two correct. The key to understanding these problems is that we can only add the radicals if the radicands are the same and if the indexes are the same. Here we have square roots and both of the roots have a radicand of x. So we have three square roots of x, and two square roots of x. So all together, we'll have five square roots of x. We're simply combining like terms just as we would with polynomials or with fractions. Here we combined the x squareds, and here we added together sevenths. So in the second example, we're really adding six fourth roots of 5, and seven four roots of five. So all together, we'll have 13 fourth roots of 5. The type of radical doesn't change in each case. We still have the square root of x in the first one and the 4 through the 5 in the second one.

Subtract Radicals

So we know when we add radicals together, you really add the like terms or the same type of radical. So what do you think about this one? What do you think the answer would be?

Subtract Radicals

Now I know its really tempting to think that the answer is seven but this would be a mistake. We know that the square root of five is really just a number. So we have seven times one number minus one times another number. So if we have seven of one number and we subtract one of that number. Then we'll six of that number left. This is why this answer is really six times the square root of five. If we think about factoring the square root of 5 from both of these terms, we would be left with the quantity of 7 minus 1 times the square root of this reasoning is with an object. If you thought the answer was seven, then consider this, if we had seven cakes and subtracted a cake we wouldn't be left with seven cakes we'd really be left with six cakes. This is sort of an analogy for what's going on here we have seven cakes at root 5's minus one cake or 1 root 5. This leaves us left with 6 root 5.

Add and Subtract Radicals 1

Now that we've seen adding and subtracting radicals, I want you to try these four problems out. Keep in mind that we can only combine radicals that are alike. They need to have the same index and they need to have the same radical. When you think you know answers, put them on the boxes and if something can't be simplified, simply write in that expression as it is. Think carefully about each of these and good luck.

Add and Subtract Radicals 1

Here are the four answers. Great work if you've had at least two of those. For the second one you might have written it as negative root 6 or negative 1 times root 6. Both are correct. For this third one we can't add the radicals together since the radicants are different. We do have the same index since we have a square root. But this radicand is a 2 and this one's a 3. So we leave this as is. Here's a little bit more work to make it more clear. We have 4 root 3 minus I'm left with two of the root 3s. We see that if we factor out a root 3 from the first term and the second term. This would live us with the quantity 4 minus 2 time root 3. We simplify the number in front of our radical sign to get helps, you can definitely do it. Whenever I add and subtract radicals I just look at the numbers in front and then look at what I need to do with them. Should I add or subtract these? In this case I subtract, so I just have 4 minus want to make sure that we have a 1 in front of this, a cube root of 3. So we're really subtracting 9 minus 1, which equals 8. We want to make sure that we keep our root on the end so we have 8 times the cubed root of 3. I hope your getting the hang of this.

Add and Subtract Radicals 2

Let's try these three problems out. You want to add of subtract the like radicals, and then write your final answer in these boxes. Keep in mind, that in this first one, we can only combine the root 5s and the root 11s. Think carefully about these next two, and good luck.

Add and Subtract Radicals 2

These are our 3 answers. Fantastic work if you've figured those out. For our first problem we want to combine the routes of 5. So we'll have 2 square roots of 5 minus 5 square roots of 5. When we subtract 2 minus 5 we'll get negative 3 groups of 5. For the roots of 11 we'll have 4 root 11 minus 1 root 11. Keep in mind that there's a 1 in front of this square root of 11. So, when we subtract You want to pay attention to the index of the radicals and the radicands. Here the index is a 2. But here the index is a 3. This means that these 2 radicals are not like terms. We have a square root of 2 and a cubed root of 2. This means that we can't add the 2 terms together which means we just list them as a sum. It might have been easy to make the mistake of writing 10 root 2. Or 10 times the cubed root of 2. But again, the indexes are not the same. So these would not be correct. For our third one, we want to combine the 5th roots of x. The indexes are 5, all match. And the radicand are all x. So we have like terms. So I have 3 plus 5 plus 1 times the 5th root of x. This would make 9 times the 5th 3 of x and our answer. This positive 3 is not like terms with anything else so we just carry it on and add it to the sum in our final answer.

Add and Subtract Radicals 3

Let's try another round of practice. What do think the solutions to each of these would be? Take your time. And think about whether or not you can add or subtract the radicals. Also you'll want to keep and eye out for the index. Here it's a 4.

Add and Subtract Radicals 3

For this one, we'll have one square root of 5 plus one square root of 5. We simply add the coefficients in front of the radicals together, to get 2 square roots of 5 or 2 times the square root of 5. We do the second one in a similar way, to add 3 and 1 to get 4 times the square root of 2. Here, we're subtracting the same quantity from the original amount. So that would make 0. If you need a little bit more convincing on this one, keep in mind there's a 1 in front of each of the radicals, so we can rewrite this expression by factoring out a fourth root of x. We'll have 1 minus 1, which we know equals 0 and 0 times anything is 0. For the fourth one we subtract 1 from 3 to get 2 times the cube root of a times b. Here, are the indexes of 3 match and the radicands are same ab, this means we can actually subtract the two radicals. And finally for our last one, we have 1 root of 3 plus 1 root of 3 plus 1 root of 3. This gives us 3 roots of 3, or 3 times our square root of 3. I know these are pretty tricky, so great work if you got at east three of them correct.

Simplify for Like Radicals

In all the cases before, we've been adding like radicals together. And in a couple of the cases, we couldn't even combine the radicals, because the radicands were different. Sometimes radicals will appear to be different, but they're actually alike. We can simplify these radicals to determine whether or not we can add or subtract them together. For example, we can split up the square root of 20 into 4 times 5. Then, we can split this radical up into two separate radicals, using our product rule. We do the same thing for 80 and for square. And the same is true for 500. 100 is the largest perfect square that goes into 500. This allows me to simplify the radicals more quickly. Now that we have some perfect squares, we can simplify them. The square root of 4 equals the remaining factors in front of our radical to get 16 root 5, 12 root 5, and negative 20 root 5. Notice how we were able to simplify each of these radicals to radicals that all have a square root of 5. Because these radicals are identical, we can add and subtract them. We'll have 16 plus 12 minus 20, which equals 8 times the square root of 5. This is why we learn how to simplify radicals first, because even though these radicals don't appear to be like, we can simplify them and then combine them.

Simplify for Like Radicals 1

Let's see if you can add and subtract these radicals, by simplifying them first. What would be the solution for this one, and what would be the simplified form for this one?

Simplify for Like Radicals 1

The first one simplifies to negative 4 times the square root of 2, and the second one simplifies to 27 times the cube root of 3. Nice work if you got both of these correct. Simplifying roots isn't always the easiest, so let's see how we can do it. We can split up the root of 8 into the root of 4 times 2. We can also split up the root of 50 into the root of 25 times 2, and the root of 200 into the root of 100 times 2. This gives us perfect squares, which allows us to take square roots in our next line. Now that we're here, we can take the square root of 4, which equals 2, the square root of 25, which equals 5, and the square root of 100 which equals 10. We multiply the numbers in front of our radicals, to get 6 root 2, minus 30 root 2, plus 20 root 2. This leaves us with numbers together, we'll get negative 4 times the square root of 2, our final answer. For our second problem we want to break up these two numbers into perfect cubes. I know 81 is the same thing as 9 times 9 which is the same thing as 3 times 3, times 3, times 3. I rewrite 81 using 3 cubed times 3. We do this since we know if we have a factor that appears three times we can take it outside of the cubed root. So I used my product rule to split up this radical into two pieces, and then we take the cube root of 3 cubed, which just equals cube root of 3. Notice that I've rewritten 8 as 2 to the 3rd power. We do this so that we can take the cube root of 2 cubed, which gives us 2. We multiply 7 times 3, to get 21, and we multiply 3 times 2 to get 6. Now we have two radicals that have a matching index and a matching radicand. So this means we can add them together. We'll have 27 times the cube root of three.

Simplify for Like Radicals with Variables

We can use this same approach to simplify radicals that contain variables. We'll want to simplify these numbers first and pull anything out, or any factors out that we can. Remember since we have square roots here, we can only pull out factors that appear twice. Once we have like radicals, we can add them together. So what do you think for this one, what would be the answer?

Simplify for Like Radicals with Variables

one out, I know it was tough. Now there are a lot of ways to simplify this radical, I'm going to show you the quickest way. We start by splitting 108 into split it up into 25 times 3. Now we can apply the product rule and split up this radical into two separate radicals. The square root of 36 times the square root of 3y. We do the same thing for this radical. This allows us to take the square root of 36 which is 6. and the square root of 25, which equals 5. We know 5 times 6 equals 30, and 6 times 5 also equals 30. We have two like radicals, since they're both the square root of 3y, and we add 30 and 30 to get our final answer, 60 times the square root of 3y.

Simplify for Like Radicals 2

If you had trouble with the last one, then try these two out. And if you got the last one correct, then it's time to show off. What would be the answer to these two problems?

Simplify for Like Radicals 2

Here are the two solutions. Great work if you found both of those. First, we split up each of these numbers into two factors. We want to make sure that one of the factors is the largest square number that divides into it. For the first one, we know the square root of 16 equals 4, and then we're left with multiplying by the square root of 3y. We take the square root of 4 to you get 2 times the square root of 3y. And we take the square root of 9 to get 3 times the square root of y. Be sure you keep the signs the same as you work. Now we have like terms, since each one has a square root of 3y. We add 4 and 2, and subtract 3 to get 3 times the square root of 3y. For the second problem, we start by splitting up 8 into 4 times 2. And we split up 32 into 16 times 2. For our numbers, we take the square root of 4 which equals 2. And this 2 stays inside of our square root since it doesn't appear twice as a factor. For the variables, we want to remember power divided by root. We take the power of 2 and divide it by 2 to get 1. This means x to the 1 appears outside of our radical. For y, we'll do 3 divided by 2 because our index is a 2 here. We know and one y stays in. We saw this before when we're simplifying roots earlier. We used power divided by root, and this number indicates the power outside of the radical where the remainder indicates the power inside the radical. For our second radical, we take the square root of 16, which equals 4. The 2 stays inside of the square root and now we need to simplify y to the 3rd power. We use power divided by root, so 3 divided by 2 equals 1 remainder 1. So, we have one y on the outside and one y on the inside. We multiply the numbers together infront of each radical to get 6xy times square root of 2y minus 8xy times the square root of 2y. Now, we subtract these two radicals. We use 6xy minus 8xy to get negative 2xy times the square root of 2y, our final result. Keep in mind that there are two reasons why we could subtract here. We subtracted because the radicals were alike and we subtracted because these polynomials were like terms. We're really packing on the algebra now.

Simplify for Like Radicals 3

We can put all of our knowledge together about radicals, factors, variables and even exponents to simplify something like this. Try simplifying each of these cube roots and then see if you can subtract the radicals. Be sure that if you pull out any factor, you multiply it by what's out in front. This one's pretty tough so take your time on it and feel free to play around with the math. Good luck.

Simplify for Like Radicals 3

Our simplified answer is 5x cubed y cubed times the cube root of 2x squared y. Now that's pretty amazing if you got it right on your first try. If you stumbled a little bit, that's definitely okay, see if you can find your first mistake in the solution video. Then pause it to see if you can work toward this answer. We start by simplifying the cube root of 54 and the cube root of 16. We want to split this up into 27 times 2. This allows us to take the cube root of so we can take the cube root of 8. So for this first radical, we know the cube root of 27 equals 3. This 2 stays inside of the radical, since we don't have the 2 repeated 3 times. Now we can use power divided by root to simplify our variables. For x, we'll have 8 divided by 3, which equals 2 remainder 2. This means we'll have x squared on the outside of our radical, And x squared on the inside of a radical for the variable y we'll have 10 divided by 3 which equals y or just y on the inside of a radical for the second radical we want to take the cubed root of 8 which equals 2. This to just stays inside the radical. For the variable x we use power divided by root, this gives us one remainder two. So will have one x outside of a radical and x squared inside of a radical. For y will have two remainder one so y squared on the outside And Y to the 1 on the inside. We multiply these numbers together to get 9 x cubed, y cubed over here we multiply them as well. We'll have 4 x to the third and y cubed. Notice that this term and this term has X cubed Y cubed times the cube root of 2x squared y. This means that they are like. So really we're just going to do 9 minus 4 and then multiply it by the trailing variables and radical. 9 minus 4 equals 5, so this is our final result.

Complex Numbers

Let's wrap up this lesson by looking at adding and subtracting complex numbers. Any number that is the sum of a real number and an imaginary number is an example of a complex number. Complex numbers are in the form a plus bi or a minus bi. A and b are any real numbers and the second term is an imaginary number. It's imaginary because it has i or the square root of negative 1. Instead of writing the square root of negative 1 all the time, we usually just write i. So if we want to add or subtract any complex numbers together, we treat i just like a radical. We can combine any terms that both contain an i, since they both have a square root of negative 1. So what do you think these would be. If we added this complex number to this complex number, what would be the answer? And if we subtracted this complex number from this one, what would be the result? Keep in mind that you really need to just combine like terms. You're also need to remember what to do when subtracting a quantity. Good luck on this one.

Complex Numbers

Here the solution is 8 plus 7i, and here the solution is 3 plus 7i. Great work if you figured those two out. We're really adding these two complex numbers together, so we can drop the parentheses. Next, I want to combine my real parts and my imaginary parts. I know addition is commutative, so I can rewrite this line as 2 plus 6 plus 3i plus 4i, 2 plus 6 equals 8 And 3i plus 4i equals 7i. Remember I represents the square root of negative 1. So we have 3 square roots of negative 1 and 4 square roots of negative 1. That gives us a total of 7 square roots of negative 1 or just 7i. When we subtract a complex number, we want to distribute this negative 1 to each of the terms. So we'll have negative imaginary parts of 5i and 2i. This gives us a total of 3 plus 7i. Our final result.

Complex Numbers 2

What about these two problems? What would be the sum here and what would be the difference here.

Complex Numbers 2

For the first one we'll get 1 plus 11i, and for the second one we'll get 3 plus parentheses since we're not multiplying anything. Then we just add the real numbers together to get 4 minus 3 which equals 1. For the imaginary numbers we have 6i and 5i, and that sums to 11i. Don't make the mistake of adding these two together to get 12i. Remember this i represents the square root of negative number. When we subtract this complex number we want to distribute this negative 1 to each term this means positive 4 becomes negative 4 and negative our imaginary numbers together. We'll have 7 minus 4, which equals 3, and 3i plus 2i, which equals 5i. This gives us our final result of 3 plus 5i. Well done.