Question:
Simplify the radical √128?
Kira
2017-01-16 12:02:14 UTC
I know how to simplify radicals like √27 that result in = 3√3

But I don't know what to do when both of the factors are composite. For example, the factors of 128 are 16 and 8. Of course, √16 is just 4. So should the answer be 4√8? Or can you further simplify √8 to √4•2 etc?
Nine answers:
nbsale (Freond)
2017-01-16 18:37:52 UTC
You are better off not looking at just pairs of factors like 16 and 8, but in the prime factors. 128 = 2^7.



The largest even power in that is 2^6, which is (2^3)^2, so 2^3 = 8 is the largest square in 128.



Therefore √128 = √(8^2)2 = 8√2.



4√8 was close, but you missed the fact that 8 contains a perfect square 4. When you take that out, you get 4√8 = 8√2.
Zac Z
2017-01-16 14:13:34 UTC
Hi Kira!



All your suggestions are correct, i.e. √128 = 4√8 = 4√4•2

But you can simplify further.

Let me explain the process as I have the impression that some things are not clear for you.



It's really quite simple if you know just one identity:

√(a•b) = √a • √b

In other words, the square root of a product is equal to the product of the respective square roots*.



What this means is that you can "extract" any factor that is a square number.

Example: √12 = √(4•3) = √4 • √3 = 2√3



The last piece in the puzzle is the exponents of the factors.

Applying the square root to any (non-negative) factor is equivalent to dividing their exponents by 2.**

Example: √5⁴ = 5⁽⁴´²⁾ = 5²



We can apply these things in the following way.

1. To simplify square roots of integers, first factorize them.

2. Extract the square root of factors with an even exponent by dividing the exponent in half.

3. Extract the square root of factors with an odd exponent by factoring out one factor to obtain an even exponent. Apply step 2 for the even exponents; the single factor remains under the root.

Example:

√15,006,250 = √(2•5⁵•7⁴) = √(2•5•5⁴•7⁴) = 5²•7²•√(2•5) = 1,225•√10



Finally, let's tackle your problem! ;-)

If you realize that 128=2⁷ then you're done very quickly:

√128 = √2⁷ = √(2•2⁶) = 2³√2 = 8√2

Alternatively, you can also simplify in steps:

√128 = √(16•8) = 4√8 = 4√(4•2) = 4•2√2 = 8√2









* This only works for non-negative numbers but I don't think you're worried about this at the moment.

** This can be easily shown using the identity (a^m)^n = a^(m•n)

For n=2 we get: (a^m)² = a^(2m)

Now we extract the square root:

√[a^(2m)] = √[(a^m)²] = √[(a^m)•(a^m)] = a^m

As a last step, one could substitute 2m with u to get the identity:

√(a^u) = [...] = a^(u/2)
Como
2017-01-16 19:21:23 UTC
√ 128 = √ [ 16 x 8 ]

√ 128 = 4 x 2√2 = 8√2
cidyah
2017-01-16 13:08:22 UTC
√128 = √64√2 = 8√2
anonymous
2017-01-17 12:25:48 UTC
128 = 2 * 64 = 2^2 * 32 = 2^3 * 16 = 2^4 * 8 = 2^5 * 4 = 2^6 * 2 = 2^7...



so, 128 = 2^7...



---> sqrt (128) = sqrt (2^7) = sqrt (2^6 * 2^1) = (2^6 * 2)^(1/2) = 2^(6/2) * 2^(1/2) = 2^3 * sqrt (2) = 8 * sqrt (2)...



ans...: 8 * √2...
busterwasmycat
2017-01-16 12:34:03 UTC
the big thing in your particular example is that you failed to recognize that 4*16=8*8, which would eliminate your problem. 8*sqrt(2) is the simplified form.



To answer your exact question, you must further simplify if it is possible. if you continued your simplification by seeing that 8=2*4, you would have ended up with the correct answer: 4*2*sqrt(2)=8*sqrt(2)
anonymous
2017-01-18 22:14:05 UTC
8 root 2
anonymous
2017-01-23 15:25:02 UTC
√ 128 = √ [ 16 x 8 ]

√ 128 = 4 x 2√2 = 8√2
Colin
2017-01-16 12:09:14 UTC
128 = 2*2*2*2*2*2*2

Can we split that into identical groups?

Not quite...

= 2*(2*2*2)*(2*2*2)

Thus the square root is

Sqrt(2)*(2*2*2)

= 8*sqrt(2)


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