√ doesn't make sense, where is Einstein when you need him?

Question

Mutiply and Simplify
√20√18=2√5*3√2=6√10

1)How do you do √20√18 this please explain. I don't just give me the anwser, I want to know how to do it please.

2) How do you do √3³*5³

It doesn't make sense cause it would have to be √3*√5*√3*√5*√3*√5 and this would make no sense. Cause a √x² is like √x*√x

3) √432=√2^4*3³=4*3*√3=12√3
This doesn't make sense can someone explain this to me PLEEASE/Show steps.

Answer 1

Let's start with the first. But let me first say you have nothing incorrect in any of your question material.

1) You can do this one the way stated under "Multiply and Simplify" as it is of course perfectly valid to resolve each √ and then multiply the results and resolve that for the answer. It's not what I would do though, as my preference is to avoid extra resolving steps unless the numbers are big (like √264 * √1322 = √349008 !) and obnoxious looking. Then I DO like to resolve, multiply, resolve. So this one, I'd multiply the values under the two radicals: √20 * √18 = √360 and resolve that by seeing it equals √36 * √10 = 6 * √10 = 6√10.

2) No easy path on this one. You do have the right idea for expanding the powers under the radical and just need to look at them for a second to realize you can group them a little to ease our way: √(3*3) * √(5*5) * √(3*5) and since the √ of 3^2 = 3 and of 5^2 = 5: 3 * 5 * √15 = 15√15. If you just multiply them out first and resolve second, like I did in 1), you get the fairly obnoxious √3375 to resolve. Besides, taking valid shortcuts is always correct in mathematics as it leaves less detail work for you to do and fewer of the errors possible in detail work as well as taking less time leaving you more time for actually hard problems.

3) This is a very handy way to resolve this one. A first step you might show to better understand it would be showing the factoring of 432 into prime factors before the grouping that gave us the "√2^4*3^3" step: √432 = √(2*2*2*2*3*3*3). Because, as you extract each factor you get: 432...216...108...54...27...9...3...1 giving the factors 2...2...2...2...3...3...3. AFTER writing out this step, I think you'll agree the next step, the grouping, makes a great deal more sense. And then the next step is also intermediate between that one and the "4*3*√3" step: show the detailed result of resolving the factor grouping step: 2*2*3*√3 and THEN multiply it all or one pair, one pair, one pair, etc. until done. That gives an explanation for the "4" that appeared as if from nowhere. Adding these two steps will, I think, make the answer shown seem clearer to you. And that's the point, eh?

If you have a knack for it, remembering various operation results and other figures can give you a fantastic head start in handling most class problems. Less so in the job world, but get through class first! Think of it like the addition and multiplication tables you learned in 1st and 2nd grade. The squares of the first 10-25 integers is one set. For instance, knowing 12^2 is 144 just from memory not only helps you quickly resolve a square root involving 144, but it does it without having to fully factor a big number. √14400 is now easy as it is √144 * √100 = 12 * 10 = 120. You could factor it out and combine and resolve, but look at how quick it was just knowing a couple very common squares. A second useful idea is knowing multiples of some common numbers. 144-288-432-576-720 would let you swiftly resolve √432 without any worry about your result. 144 is, by the way, a dozen dozens, or a gross, and common in school math problems as well as (not so usual) real life and jobs. I know how all this sounds, but it's really only 25-50 things to learn all told and it can be VERY helpful as time you do NOT spend on problems made simple by knowing these things is time you DO have for hard ones. And you get to rack the points for these with little effort making letter grade jumps easy to make. (Don't tell mom and dad how easy it was, just let them think you are still slaving away.)

But even if you don't, remember to always seize shortcuts when you can (when you understand them!) and to check a problem early to see if there are any. Then multiply this kind of thing out and look at the result to see which you'd rather do: resolve the multiplied out result or resolve the starting material, multiply and resolve. Pick the EASIEST or CLEAREST for YOU. There is no real rule of thumb for it. Take the moments, and look for yourself and go the route that looks best. What works for your friends or the teacher may not work for you so why slavishly follow their choices?

Answer 2

First, you must understand the rule that the square root of a product is the same as the product of the square roots. Said algebraically:

√(a·b) = √a · √b
For example: √36 = √(4·9) = √4 ·√9 = 2 · 3 = 6

So:
#1
√20·√18
√(4·5)·√(9·2) <== factor out largest perfect squares
√4·√5·√9·√2 <== separate into roots of all factors
2·√5·3·√2 <== simplify roots of perfect squares
2·3·√5·√2 <== rearrange so rational numbers (the 2 and 3) are together
6·√5·√2 <== combine integers
6·√(5·2) <== combine radicals under same sign
6·√10 <== simplify under radical

#2
You're on the right track.
√(3³*5³) = √(3²·3·5²·5)
= √3²·√3·√5²·√5
= 3·√3·5·√5
= 3·5·√3·√5
= 15·√15

#3
Once again, you need to factor out the largest square. If it isn't self-evident, try factoring out smaller squares. Factoring down to the prime factors is one way
So:
√432 = √(2^4·3³)
= √(2²·2²·3²·3)
=√2²·√2²·√3²·√3
=2·2·3·√3
=12√3

Answer 3

The function "square root" has relatively simple rules when using real numbers:

You cannot take the square root of a negative number.

A multiplication of roots is the same as the root of the multiplication.

A division of roots is the same as the root of the division

A root of a number is "the same" as raising that number to the power 1/2

You can factor the number inside a root.

-----

√20*√18 = √(20*18) =
√(4*5*9*2) =
√4*√5*√9*√2 = 2*√5 * 3*√2 =
2*3 * √(5*2) = 6*√10

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I assume that all the powers are inside the root:
√(3*3*3*5*5*5) =
√9*√3*√25*√5 =
3*√3 * 5*√5 = 15*√15


x = √(x^2) is exactly like √x * √x = √(x*x)

-----

Find the factors of 432:

2*2*2*2*3*3*3

Therefore
√432 = √(2*2*2*2*3*3*3) =
√(2*2) * √(2*2) * √(3*3) * √3

As we just saw, √(x*x) = x
We are left with:
2*2*3*√3 = 12*√3 (often written without the *, as 12√3)

Answer 4

You could avoid your expression of "this doesn't make sense". The fact is, IT DOES make sense and you simply just have to understand it.

We'll show that

√20√18 = (√4*5)√(9*2).

Now √4 = 2 and √9 = 3.

So (√4*5)√(9*2) = (2√5)*3√2.

√3³*5³ does makes sense!

You're right it is equivalent to √3*√5*√3*√5*√3*√5 - which reduces to 3*5*√3*√5 or 15√15.

Finally, √432 = √2*2*2*2*3*3*3 - we simply factored 432!

This gives us √432 = √2^4*3^3 or √432 = 2^2*3√3 = 12√3.

I suggest we listen to our mentors and understand.