# Simplifying radicals

Page 1 of 1 [ 13 posts ]

mr_bigmouth_502
Veteran

Joined: 12 Dec 2013
Age: 27
Gender: Male
Posts: 7,002
Location: Alberta, Canada

10 Mar 2015, 9:59 pm

This is driving me nuts. I cannot for the life of me understand how to fully simplify radicals. Like, say I'm trying to simplify the cube root of 1280; the root itself is 10.85767047, but I have no idea how I'm supposed to express this as a radical, other than just saying cube root of 1280. I know that's not the answer they want, because they want it in lowest terms.

I've always struggled with putting things in lowest terms in the first place, and I've always found it to be rather pointless as well. "Simplest terms" my ass, they want to make things more complicated than they have to be.

Anyway, if anyone here knows what I'm talking about, I'd like to have a detailed, step-by-step explanation of what I'm supposed to do. I'm going batty over this!

naturalplastic
Veteran

Joined: 26 Aug 2010
Age: 66
Gender: Male
Posts: 26,408
Location: temperate zone

10 Mar 2015, 10:42 pm

Are you asking about math? Or are you asking about stenography?

Do you mean "how do I write 'the cube root of 1280' on a normal computer keyboard that doesn't have a radical symbol?"

Is the fact that a person can't type a radical symbol the whole issue? Or is this some deeper mathematical question?

Since 1280^2 is taken to mean "1280 squared" couldn't you just type 1280^1/3 to mean "1280 raised to the power of one third" which is the same thing as "the cube root of 1280"?

mr_bigmouth_502
Veteran

Joined: 12 Dec 2013
Age: 27
Gender: Male
Posts: 7,002
Location: Alberta, Canada

11 Mar 2015, 12:54 am

naturalplastic wrote:
Are you asking about math? Or are you asking about stenography?

Do you mean "how do I write 'the cube root of 1280' on a normal computer keyboard that doesn't have a radical symbol?"

Is the fact that a person can't type a radical symbol the whole issue? Or is this some deeper mathematical question?

Since 1280^2 is taken to mean "1280 squared" couldn't you just type 1280^1/3 to mean "1280 raised to the power of one third" which is the same thing as "the cube root of 1280"?

I'm talking about math. I know there's a way to type a root symbol, but I forgot the alt code for it.

EDIT: It's alt+251 for the root symbol, and alt+0179 for an exponent of 3.

Anyway, I need to find out how to turn ³√1280, which equals 10.85767047, into a simplified radical, like 2√5.

Orangez
Deinonychus

Joined: 15 Nov 2014
Age: 28
Gender: Male
Posts: 320
Location: British Columbia

11 Mar 2015, 1:20 am

It is quite simply you just have to remember your perfect cubes. It is the same logic as simplifying square roots.
1280= 64*20
and
4^3=64
Therefore,
³√1280= ³√(64*20)
= ³√64 * ³√20
= 4*³√20

mr_bigmouth_502
Veteran

Joined: 12 Dec 2013
Age: 27
Gender: Male
Posts: 7,002
Location: Alberta, Canada

11 Mar 2015, 3:00 am

Orangez wrote:
It is quite simply you just have to remember your perfect cubes. It is the same logic as simplifying square roots.
1280= 64*20
and
4^3=64
Therefore,
³√1280= ³√(64*20)
= ³√64 * ³√20
= 4*³√20

Thank you! Finding a list of perfect cubes online is easy enough, what I forgot were actually the different steps involved.

If it weren't for calculators or the Internet, I'd be so lost right now.

heavenlyabyss
Veteran

Joined: 9 Sep 2011
Gender: Male
Posts: 1,393

12 Mar 2015, 5:26 pm

Sometimes it's easiest to start with the lowest possible cube.

I'd start with 8 first (2 cubed) You can make a tree.

1280 = 8*160.

Then see if 160 is divisible by 8 again. Sure enough it is. 160 = 8*20.

So cubed root of 1280 = cubed root of (8*8*20) = cubed root of 8 * cubed root of 8 * cubed root of 20 (since you can separate the cubed root this way). Then simplify this gives 2*2*cubed root 20 equals 4 times the cubed root of 20.

It's really a pain to do. You have to use guess and check. But it won't take so long because you only have to check 8, 27, 64, 125, and so on. The numbers get big fast.

Edit: If 8 didn't work, I'd try 27 next and then 64. Of course you can jump straight to 64 but I find it's easiest to start low and then work up.

Jaywalker
Emu Egg

Joined: 19 Feb 2015
Posts: 3

12 Mar 2015, 11:29 pm

heavenlyabyss wrote:
Sometimes it's easiest to start with the lowest possible cube.

I'd start with 8 first (2 cubed) You can make a tree.

1280 = 8*160.

Then see if 160 is divisible by 8 again. Sure enough it is. 160 = 8*20.

So cubed root of 1280 = cubed root of (8*8*20) = cubed root of 8 * cubed root of 8 * cubed root of 20 (since you can separate the cubed root this way). Then simplify this gives 2*2*cubed root 20 equals 4 times the cubed root of 20.

It's really a pain to do. You have to use guess and check. But it won't take so long because you only have to check 8, 27, 64, 125, and so on. The numbers get big fast.

Edit: If 8 didn't work, I'd try 27 next and then 64. Of course you can jump straight to 64 but I find it's easiest to start low and then work up.

We just need to check for divisivibility by the cubes of the primes: 2^3, 3^3, 5^3, etc. Consider the cube of non-prime 4^3=64:
4^3 = (2*2)^3 = (2^3)*(2*3)
= 8*8, which we've already factored out.

Likewise, 216 = 6^3 = (2*3)^3
= (2^3)*(3^3) = 8*27, which will already have been factored out.

When I teach the process, I generally start with a finding the prime factorization of the radicand, which is the number under the radical. That makes the cubes easy to find without needing to have a list of cubes handy.

Consider 1296^(1/3):
1296^(1/3)
= (2^4 * 3^4)^(1/3)
obtaining prime factorization of 1296
= (2^3 * 2 * 3^3 * 3)^(1/3) pulling out the perfect cubes
= (2^3 * 3^3 * 2 * 3)^(1/3) reordering
= [(2*3)^3 * (2*3)]^(1/3) a^c * b^c = (a*b)^c
= (6^3 * 6)^(1/3) simplifying
= (6^3)^(1/3) * 6^(1/3) (a*b)^c = a^c * b^c
= 6 * 6^(1/3) simplifying

It's not really as tedious as it looks. Once you understand the process, you can combine steps.

I hope this helps.

heavenlyabyss
Veteran

Joined: 9 Sep 2011
Gender: Male
Posts: 1,393

13 Mar 2015, 12:16 am

The above post is correct.

The answer I gave was not simplified all the way.... been too many years since I've done this stuff. Sorry for the misinformation.

Jaywalker
Emu Egg

Joined: 19 Feb 2015
Posts: 3

14 Mar 2015, 11:33 am

heavenlyabyss wrote:
Sorry for the misinformation.

There was no misinformation in your post. The method you described will always work. Further, if I was teaching to factor out the cubes of primes, I would first teach to factor out all of the cubes, as you did, so that my students might understand why we need to concern ourselves only with the cubes of the primes.

ruveyn
Veteran

Joined: 21 Sep 2008
Age: 84
Gender: Male
Posts: 31,502
Location: New Jersey

14 Mar 2015, 12:51 pm

mr_bigmouth_502 wrote:
This is driving me nuts. I cannot for the life of me understand how to fully simplify radicals. Like, say I'm trying to simplify the cube root of 1280; the root itself is 10.85767047, but I have no idea how I'm supposed to express this as a radical, other than just saying cube root of 1280. I know that's not the answer they want, because they want it in lowest terms.

I've always struggled with putting things in lowest terms in the first place, and I've always found it to be rather pointless as well. "Simplest terms" my ass, they want to make things more complicated than they have to be.

Anyway, if anyone here knows what I'm talking about, I'd like to have a detailed, step-by-step explanation of what I'm supposed to do. I'm going batty over this!

1280 is not a perfect cube, so its cube root is not a rational number. The best you can do is write it as cuberoot(1280) of 1280^1/3.

You will have the same problem with the square root of 2.

ruveyn

heavenlyabyss
Veteran

Joined: 9 Sep 2011
Gender: Male
Posts: 1,393

15 Mar 2015, 12:00 am

Jaywalker wrote:
heavenlyabyss wrote:
Sorry for the misinformation.

There was no misinformation in your post. The method you described will always work. Further, if I was teaching to factor out the cubes of primes, I would first teach to factor out all of the cubes, as you did, so that my students might understand why we need to concern ourselves only with the cubes of the primes.

Well, I appreciate the kind words but I gave an answer of 4 * the cubed root of 20, which isn't as simplified as 6 * the cubed root of 6 ( we want the number under the radical as small as possible, right?) In order for my method and the other poster's method to work, you have to make sure you pull out the largest possible cube right from the get go, correct? So instead of pulling out either 8 or 64, we should have pulled out 216. Which I actually think is easiest for numbers that are not very large. However, I think the method you describe would work better for very large numbers.

The problem here is let's say you start with 4* the cubed root of 20.... how can we simplify this without first converting it back into the original problem (that is cubed root of 1296?) If we try to simplify the cubed root of 20, we have no where to go since it is already as simple as possible.

So, again to the OP, if it still matters, don't start with the lowest numbers. This doesn't work. You have to pull out the largest possible cube or use jaywalker's method.

Jaywalker
Emu Egg

Joined: 19 Feb 2015
Posts: 3

15 Mar 2015, 3:28 am

heavenlyabyss wrote:
Jaywalker wrote:
heavenlyabyss wrote:
Well, I appreciate the kind words but I gave an answer of 4 * the cubed root of 20, which isn't as simplified as 6 * the cubed root of 6 ( we want the number under the radical as small as possible, right?) In order for my method and the other poster's method to work, you have to make sure you pull out the largest possible cube right from the get go, correct?

You simplified 1280^(1/3). I simplified 1296^(1/3). I intentionally chose a radicand (i.e., the number under the radical) that was a product of the fourth powers of two primes. 1296 is the smallest such number.

Because every integer has a unique prime factorization, your method would reach an identical result as mine if you started with the same radicand.

It doesn't matter what order you factor out the cubes; you'll ultimately get the same result (assuming no mistakes). If you start with larger cubes first, there is no need to check any cubes greater than half the radicand.

The reason I start with a prime factorization is that the procedure generalizes for any nth root, where n is a positive integer.

heavenlyabyss
Veteran

Joined: 9 Sep 2011
Gender: Male
Posts: 1,393

15 Mar 2015, 7:19 pm

Oops, lol. I didn't realize that.

I did two different problems and of course came up with 2 different answers.

Never mind. I was confusing myself.