Algebra 2 Operations with Radical Expressions

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Today, we will be covering operations with radical expressions.0002

And in earlier lessons in Algebra I, we talked about square roots and properties when working with square roots.0006

And now, we are going to go on to talk about other roots.0012

I will be starting out with reviewing properties of radicals and applying these properties0017

to roots other than square roots: cube roots, fourth roots, sixth roots, and on.0021

It is the same properties, such as the quotient property.0025

And the quotient property says that, if you have the root (it could be a square root;0029

it could be a cube root) of a/b, this is equivalent to the root of a divided by the root of b,0034

with the restriction that b cannot be equal to 0, because if b did equal 0,0045

the square root of that would be 0 (or cube root, or whichever root).0050

And then, you would end up with 0 in the denominator, which would result in an undefined expression.0053

OK, so first looking at this quotient property with an example: the fifth root of 17x¹⁰, all of that divided by 12y⁴.0060

This is the fifth root of 17x¹⁰, divided by 12y⁴.0081

This can be split up into this: the fifth root of 17x¹⁰ over the fifth root of 12y⁴.0087

And the quotient property, along with the product property, allows us to simplify radical expressions,0099

which we are going to talk about in a few minutes.0105

Here, I have the product property: this would be something along the lines of0108

the fourth root of 4x⁶ equals the fourth root of 4, times the fourth root of x⁶.0114

And again, this helps us to simplify--the same idea as working with square roots, only now the index is a different number.0131

There is also the restriction that, if n is even, a and b are greater than or equal to 0.0139

And that would be to avoid this situation: look at something like the square root of 10.0143

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."0150

So, I have an even index, and I end up with this--not a problem.0160

However, I could also say, "Well, the square root of 10...10 also factors out to -5 times -2."0164

So, I follow the product property, and I end up with the square root of -5, times the square root of -2.0175

And this is not what we are looking for when we talk about a radical; we are staying with real numbers.0183

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.0188

Again, reviewing some concepts that were learned in Algebra I and applied to square roots,0205

but applying them to other roots this time: simplifying: a radical expression is simplified if the index is as small as possible,0210

the radicand contains no n^th powers and no fractions, and no radicals are in the denominator.0221

Let's go ahead and look at this part: the radicand contains no n^th powers.0227

Starting out with square roots, just to illustrate this: if you have something like the square root of 18,0240

this can be rewritten, using the product property, as 9 (which is 3²) times 2.0248

Well, since the root here, the index, is 2, here I have n = 2.0257

I look in the radicand, and I have an n^th power; I have 2.0268

So, this could be further simplified to say "the square root of 3², times the square root of 2, equals 3√2."0273

We did this earlier on; but we just restricted this discussion to square roots.0282

I could look at a more complicated example, using an index of 3, looking for the cube root.0287

If I had something like the cube root of 27, times x⁷: this is not in simplest form,0294

because I have an index of 3, and I could rewrite this as 3³...well, x⁷ is equal to x⁶ times x.0303

And x⁶ is equal to (x²)³; so if I rewrite this as x⁶(x),0318

I could then go on and write this as the cube root of 3³,0332

times...since x⁶ is equal to...(x²)³, times x.0341

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.0349

So now, what I can do is say, "OK, this 3 essentially cancels out that 3; and I end up with just a 3."0362

The cube root of 3³, of 27, is 3; the cube root of x², cubed, is x².0369

And that just leaves an x behind, like this.0376

OK, so simplest form means that the radicand contains no n^th powers.0380

Look at the index and make sure that you can't factor out something that would be raised to that same power.0385

Also, the radicand cannot contain fractions--no fractions if it is in simplest form.0393

Something such as the fourth root of x, over 2z, is not in simplest form,0404

because there is a fraction under this fourth root sign.0414

And we can use the quotient property to simplify this; and we will talk in a few minutes0422

about how you go about getting rid of fractions that are under the radical sign.0429

OK, the other thing is: no radicals in the denominator.0436

So, if I have something such as 3y divided by the cube root of 2y, this is also not in simplest form.0447

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.0455

So, the index is as small as possible; that is what we are going to be working with (this is usually not an issue;0464

usually the index is as small as it can be); the radicand contains no n^th powers (so if there is an0469

n^th power inside, as part of that radicand, we need to take the root of that and remove it from under the radical;0474

the same here with the cube root; in addition, the radicand contains no fractions (so something like this0482

is not in simplest form); and there are no radicals in the denominator (so this is also not in simplest form).0490

You can go through this checklist in your mind, when you think you are done simplifying,0498

and make sure that the expression you are working with meets these conditions.0502

OK, so I mentioned that you cannot have radicals in the denominator if something is going to be in simplest form.0508

So, getting rid of radicals in the denominator is known as rationalizing the denominator.0516

So, now we are going to talk about rationalizing denominators.0522

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.0525

So, to eliminate radicals in the denominator, multiply the numerator and the denominator by a quantity, so that the radicand is an n^th power.0532

What does this mean? 2 divided by √5xy: what I want to do is make this radicand (5xy)²,0543

because here, n equals 2; so I want to get 5xy, squared, as the radicand.0558

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.0579

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.0589

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 1.0601

So, the numerator is 2 times √(5xy), divided by...looking at the product property:0612

the product property tells me that, if I multiply these two,0622

I am going to end up with √(5xy times 5xy), which equals √(5xy²).0626

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.0647

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.0658

So, this is going to equal 2√5xy, divided by 5xy.0667

Now, this is in simplest form, because I no longer have a radical in the denominator.0675

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).0679

And I could take the square root of that to just get 5xy.0691

But you need to make sure that you multiply both the numerator and the denominator by the same term,0694

so that you are actually just really multiplying this entire thing by 1.0701

OK, if the expression in the denominator is a radical, but it is a binomial, you need to use a different technique.0708

We just talked about rationalizing a denominator when the radical was the monomial.0716

But if you have a binomial, then what you need to do is work with a conjugate.0721

So, first let's just review what a conjugate is.0727

Conjugates are the sum and difference of two terms--not even worrying about the radical sign right now.0731

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 - 1.0742

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.0757

OK, so these radical expressions, a√b and c√d and a√b - c√d--the only difference here is in the sign.0769

These are called conjugates; so they can be used to rationalize denominators that are binomials.0782

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 + 3.0792

So, these two are conjugates; this is a conjugate pair.0810

So, what I need to do is multiply 5, divided by the square root of 7, minus 3, times √7 + 3, over √7 + 3.0815

Now, since the numerator and the denominator are the same, I am really just multiplying by 1; so again, this is allowed.0832

All right, so in the numerator, this gives me...using the distributive property...5√7, plus 5 times 3, which is 15.0842

The denominator: recall that here, if you look at what I am doing, it is multiplying a sum and a difference.0861

So, if I had a + b times a - b, this is going to end up giving me a²...the outer term is -ab;0866

the inner term is positive ab; so that is going to cancel out, and then I am going to get -b².0877

b times b is b², but I have a negative sign in front of it.0886

So, I am going to end up with a² - b²; and in this case, multiplying √7 - 3 and √7 + 3,0890

a equals √7 in this situation, and b equals 3; so this is going to give me0899

(√7)² - (I am using this format) 3², which equals 5√7 + 15.0906

OK, I have (√7)², √7 times √7; it is just going to give me 7.0920

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.0932

So, this is going to give me 5√7 + 15, divided by 2.0944

And now, the radical is gone from the denominator.0949

So, I have rationalized the denominator when I had a radical, and I had a situation where it was part of a binomial.0952

I had a denominator that was a binomial.0961

And I did that by multiplying both the numerator and the denominator by the conjugate of the denominator.0963

Another review of a concept you may have learned earlier on, which is adding and subtracting radicals:0973

radicals are like radicals if they have the same index and the same radicand.0980

For example, 4 times the cube root of 5x, plus 2 times the cube root of 5x:0985

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.0997

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."1006

And that leaves behind 4 + 2; so this becomes the cube root of 5x times 6, or 6 times the cube root of 5x.1017

All I did is add 4 and 2; and you can really just look at this as a variable, almost, with the radical.1034

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.1043

You could just look at it this way: that this whole thing is like a variable.1055

And you can add these two together, but the variables are the same,1058

because it is just saying 4 y's and 2 y's equal 6 y's, and that is the same idea here.1063

OK, subtraction: the same thing--you just have to be careful (as always, when you are working with subtraction) with the signs.1070

So, if I am subtracting something like 5 times the fourth root of 7yz, minus 3 times the fourth root of 7yz,1081

I check and see that I have the same index (which is 4) and the same radicand (7yz).1097

So, this becomes pulling out the same factor, which is the fourth root of 7yz, leaving behind 5 - 3.1103

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.1119

So, adding and subtracting like radicals is pretty straightforward.1131

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.1135

Multiplying: with multiplying radicals, we are going to use the product property.1144

And if we are going to multiply sums or differences of radicals, we will be using the distributive property.1150

So, let's just start out with multiplying two monomials that involve radicals,1155

the fifth root of 2x³ times the fifth root of x².1160

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.1171

So, what I am doing down here, instead of going from left to right--I am going from right to left.1180

I already have these two split up, but the product property tells me I can combine them.1187

So, this would actually be the fifth root of 2x³, times x².1192

Recall that, if you are multiplying exponents with a like base, then here, I can just add these exponents.1202

So, this is going to give me 2 times x⁵.1215

Now, I see that what I have (using the product property again) is the fifth root of 2, times the fifth root of x⁵,1224

which equals...well, the fifth root of x⁵ is just x, times the fifth root of 2.1237

So, you can see how multiplication and using the product property allowed me to actually simplify this.1243

First, I used the product property to multiply these two together.1251

Then, I used my property of exponents that says I add the exponents, since there are like bases here.1255

That gave me 2x⁵; and I saw that this is not in simplest form,1262

because I have a radicand that contains the n^th power, the fifth power.1267

So then, I further simplified.1274

OK, so that is if you have monomials; now, for multiplying sums or differences of radicals, we need to use the distributive property.1276

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,1283

minus 2, times the square root of 5, we are going to use FOIL.1294

The first terms (multiplying the first terms, because I am multiplying two binomials--FOIL--First terms):1300

that is 2 times the square root of 3x, times the square root of x.1305

Plus the outer terms--that is 2√3x, times -2√5.1313

Inner terms are √2 times √x.1324

And finally, the last terms are √2 times -2√5.1332

OK, using the product property right here tells me that √a times √b is √ab.1339

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;1347

so I am going to rewrite this as -4; √3x times 5...3 times 5x, plus the square root of 2x.1358

And then here, I have a -2 out in front; and then, that is the square root of 2 times 5.1373

I am doing some simplification: this equals 2 times 3x², minus 4 times √15x, plus √2x, minus 2√10.1380

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.1396

So, I can pull this out, and I need to remember to use absolute value bars, because this was an even index,1404

and when I took that root of this, I ended up with an odd power, 1, so I need to use absolute values.1416

And I can't simplify any further, because I can't combine these, since they are not like radicals.1427

They are to the same powers, but none of them have the same radicands, so I can't add or subtract them.1433

So again, multiplication with sums or differences of radicals--you just use the distributive property,1439

like we have earlier on, when working with numbers or variables.1445

OK, in this first example, we are going to simplify this expression; and it is the fifth root of x⁸y⁹z⁵, divided by 243.1452

So, I am going to use the quotient property, because I know that this equals, according to the quotient property,1465

the fifth root of x⁸y⁹z⁵, all divided by the fifth root of 243.1473

So, recall that, in order to be in simplest form, a radical expression needs to have an index that is as small as possible;1484

no n^th powers; no fractions under the radical; and no radicals in the denominator.1492

So, you should be familiar with these rules.1509

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.1520

I then took care of that by using the quotient property.1529

I no longer have a big radical sign where I have this fraction under it; I split it up.1534

The only problem is that I now have a radical in the denominator.1539

So, this is still not in simplest form.1544

In addition, I also have some n^th powers under here; when I work on this,1548

I will see that there are some terms here that are to the fifth power.1554

So, I can rewrite this as x⁵ times x³, because these have like bases, so I add the exponents.1561

That would give me x⁸ back, so I can see here that I have n = 5, and the radicand contains some fifth powers.1572

y⁹ would be y⁵ times y⁴, because I would add these to get 9 back; and then leave z as it is, z⁵.1582

243 is not totally obvious, but it turns out, if you work this out, that 3 to the fifth power is 243.1593

So, I also have a fifth power as part of the radicand in the denominator.1604

OK, I am going to use the product property to rewrite this with my fifth powers all together here:1610

x⁵y⁵z⁵ times what is left over (that is the fifth root of x³y⁵).1616

OK, what this gives me is the fifth root of these; well, these are all to the fifth power.1641

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.1649

Since this is odd, I don't have to worry about absolute value bars.1658

We only worry about that when the index is even.1663

And this is times the fifth root of x³; this should actually be y⁴ right here,1667

and y⁵ here and y⁴ here, divided by...well, this is the fifth root, and this is raised to the fifth power.1675

So, that just becomes 3; or I could rewrite this as xyz divided by 3, all that times the fifth root of xy⁴.1686

So, I double-check: is this in simplest form?1702

There are no n^th powers in the radicand; what I have left is xy⁴; there are no fifth powers here.1705

There are no fractions under this radical sign; there are no radicals in the denominator.1713

So, this is in simplest form.1720

Here, we are asked to add and subtract some square roots; recall, though, that you can only add or subtract radicals1725

if they have the same index (which these all do) and the same radicand (which they don't).1734

However, they are not in simplest form yet; it is important to always simplify first.1739

The problem is that I have some perfect squares left here as part of the radicand that I could pull out.1744

So, I am going to rewrite this with the perfect squares factored out.1752

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;1756

54 is 9 times 6, so that is another perfect square; and then, 75 is 25 times 3.1769

According to the product property, the square root of ab equals the square root of a times the square root of b.1777

So, I can rewrite this as √4 times √6, minus √16 times √3, plus √9 times √6, minus √25 times √3.1786

And as you get better at this, you might not need to write out every step.1802

But for now, it is a good idea, just to keep track.1806

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, √3.1809

Now, I am looking, and I actually do have some like radicals that I can add, because they all have the same index;1825

and now I see that these two (let's rewrite it like this): 2√3 + 3√6, have the same radicand.1832

And then, I have -4√3 - 5√3.1847

In this situation, what I am going to do is add the 2 and the 3, and this is going to give me 5√6.1855

You are looking at this the same way that you would a variable: if I had 2y and 3y, it would become 5y.1865

Plus...this is going to be -4 and -5; that is going to be -9, and then √3, which is 5√6 - 9√3.1873

So, this is now in simplest form; and when I looked at this, it first looked like I could not combine these.1889

But when I went about my checklist of how to simplify, I didn't have to worry about radicals in the denominator1897

or fractions under the radical sign, but I did have to get rid of the perfect squares that were part of the radicand.1902

And I used the product property to achieve that.1909

Once I did that, I saw that I actually could combine some of these radicals to get this simplest form.1912

OK, simplify: this time, I am asked to multiply two binomials.1922

And I am going to use FOIL, just like I normally would.1927

I am rewriting this here; use FOIL just as though you are multiplying any other two binomials.1932

So, this is going to give me 2√3 times the other first term, which is 6√3,1942

plus 2√3 (the outer terms) times 2√5.1950

Now the inner terms are -4√5 times 6√3; and then the last terms are -4√5 times 2√5.1958

OK, I am making sure that I have everything correct...inner, and then last.1975

OK, we can use the product property; and the product property tells me that √ab = √a times √b.1979

And I am actually moving from right to left here, because these are separated, and I want to put them together.1992

So, this is going to give me 2 times 6, √3 times 3, plus 2 times 2, and then this is √3 times 5,1998

plus -4 times 6, and this is √5 times 3, plus -4 times 2, times √5 times 5.2015

This gives me 12; and I could rewrite this as 3² + 4; this is 15.2029

-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 5².2038

OK, I am not done simplifying yet, because recall that I look at the index; it is 2;2049

and I see that I do have a term here and here that are to the power of 2.2055

I can take 3²; that square root is just going to be 3; so this gives me 12 times 3.2062

This does not have any perfect squares within it as factors, so I leave it alone.2069

The same with this term; here, I do have 5², so the square root of 5 squared is just 5.2076

Continuing to simplify: 12 times 3 is 36; -8 times 5 is -40.2085

I can combine these two, 36 - 40; that is going to give me -4.2096

I also can combine these two, because these are the same index, 2, and they are the same radicand.2106

So, this would be the same as 4 - 24 times √15,2116

which is going to give me -4; and then 4 - 24 is just going to give me -20√15.2128

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.2140

There are no radicals in the denominator; and I don't have any n^th powers here; I don't have any perfect squares under here.2151

So, I first multiplied these two binomials out, using the distributive property.2160

I got down to here; then I took my perfect squares; I took the square root of this number, 9, which is 3.2167

I took this square root of 5², which is 5.2176

I did some more simplifying, and then combined these two radicals that had like radicands and had the same index.2181

OK, Example 4: Thinking about my rules, is this is simplest form?2194

No; it doesn't have any fractions under the radical; however, there is a radical in the denominator.2200

So, a radical expression is not in simplest form if there is a radical in the denominator.2208

Recall that, to simplify a radical binomial expression, you multiply both the numerator and the denominator by the conjugate of the denominator.2214

So here, I have 2 + √3; the conjugate of that is going to be 2 - √3.2223

So, these are conjugates; this is a conjugate pair.2232

I am going to take 4 - √3, divided by 2 + √3; and I am going to multiply that times this conjugate, 2 - √3.2238

This is just the same as multiplying this by 1.2255

I am going to have to use the distributive property, because I am multiplying these binomials.2259

So here, I am just going to have to use FOIL, as usual.2266

The first two terms are going to give me 4 times 2; the outer is going to give me 4 times -√3.2269

The inner two terms--that is -√3 times 2; and then, the last two terms are -√3 times -√3.2279

The denominator is a little bit easier, because this denominator is in the form (a + b) (a - b).2294

It is the product of a sum and a difference, which gives me a² - b².2303

Here, a = 2, and b = √3; so that is going to give me a², which is 2², minus (√3)².2307

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√3.2325

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.2346

OK, all divided by...2² is 4; minus...well, the square root of 3 squared is just 3.2356

OK, so this gives me 8 - 4√3 - 2√3.2368

Well, this √3 squared is also 3, so I am going to change that to a 3...divided by 4 - 3, which is 1...2383

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√3.2391

Since these have the same index and the same radicand, I can combine these two to get -6√3.2406

So, we started out with something not in simplest form, because it had a radical in the denominator.2414

And since that was a binomial, I multiplied both the numerator and the denominator by the conjugate of the denominator.2419

I got this whole thing; I then just continued to simplify.2427

And when I got to here, I saw that I had two radicals that could be combined to give me -6√3.2434

And then, I combined my constants.2444

And this is now in simplest form, because there are no radicals in the denominator;2446

there are no n^th powers in this radicand (no perfect squares, in this case);2450

and no fractions under the radical sign; and the index is the smallest power that it can be.2458

That concludes this session of Educator.com; thanks for visiting!2466