Annoying math problem

[Sar]

Ok, I'm still trying to study math and now I have a goal in it! I'm going to start taking lessons to get a silly little degree... eventually, someday, maybe, perhaps (as if I'd ever)

Anyway I'm a bit baffled by this problem.

First example

sqrt(225) the answer is given as sqrt(225) = 15

That's pretty straight forward and after feeling accomplished for the fact that I managed to push the correct buttons in a calculator I get to do this one

2nd example

sqrt(75) but this time the correct answer was 3*sqrt(5)

Why is the answer shown this way in this case? I think it has something to do with the fact that the answer is an infinite decimal, but how am I supposed to calculate that sort of answers?

Kinda annoying that they give you the problem and the solution, but not the way you're supposed to get them.

 

[Artlav]

No, that's 5*sqrt(3).

sqrt(75)=sqrt(3*25)=sqrt(25)*sqrt(3)=5*sqrt(3)

Pretty simple.

And yes, sqrt(3) is an irrational (infinite ratio) number.

 

[Fabri91]

It can be annoying, but with the time you'll get the hang of it...you just have to look everytime if the term you have to calculate the square root of is a multiple of another whose square root is "simpler": in this case, as you've mentioned, 75 is 3*25, and the square root of 25 is 5, so you can write 5*sqrt(3).

 

[Sar]

But how do you come up with the numbers 3 and 25 in the first place? It can't be so that I'd need to remember each number with a simple square root.

(and yes I'm probably not supposed to think this as hard as I do at the moment, but I've found that if I get to the bottom of the things, learning the next things is easier)

 

[Quick_Nick]

Chances are, you ARE supposed to remember perfect squares.

 

[Johnman]

And even if you don't, you will eventually recognize them. After all, you know 5*5 = 25, 10*10 = 100 or 3*3 = 9 and I'm sure you didn't need to make a conscious effort to remember those. It just comes with practice. Don't give that too much though.

 

[fireballs619]

From what I gathered, you are talking about simplifying radicals. Here's how I do it.

Let's say you are to simplify [math]\sqrt{112}[/math].

You are going to want to find the prime factors of 112. For simplicity sake, I always divide by two if it is a even number, which 112 is. So we now have:
[math]\sqrt{56*2}[/math]
58 is not prime, but 2 is, so let's keep factoring 56. 56 is also even, so again let's divide by two.

[math]\sqrt{28*2*2}[/math]
Again, 28 is not prime, so onward we go.

[math]\sqrt{14*2*2*2}[/math]
Once more now!

[math]\sqrt{7*2*2*2*2}[/math]
Okay, so now we're down to prime numbers. This part can be a a little bit confusing. You want to look for pairs of numbers, and you move those to the outside of the radical. When you move them out side, however, only one ends up outside. Just imagine one is lost in transit. If you'll look at our factorization, you'll see we have two pairs of twos (four twos). There are two pairs, therefore, two two's end up outside the radical. This results in:

[math]2*2\sqrt{7}[/math]
That can also be written as:

[math]4\sqrt{7}[/math]
So, we have simplified:
[math]\sqrt{112}[/math]to
[math]4\sqrt{7}[/math]
If you check their decimal approximations, you'll find that both are about equal to 10.58301.

Now for practice, try to simplify this:
[math]\sqrt{120}[/math]
Of course, if you weren't asking about simplifying radicals, disregard this post.

 

[n72.75]

You can approximate roots with Newton's method.

 

[NOMAD]

I think you should remember some important conclusions.
For example :
sqrt(4)=2
sqrt(9)=3
sqrt(25)=5
sqrt(36)=6
sqrt(225)=15
sqrt(625)=25
etc;
This may help you work out the problem.
For example :

sqrt(12)=sqrt(4)*sqrt(3),sqrt(4)=2,so sqrt(12)=4*sqrt(3).

sqrt(15625)=?
First you should remember that 5 is an aliquot part of a number whose last number is 5 OR 0. So 15625=5*3125=5*5*625.
Now,we can know the answer:
sqrt(15625)=sqrt(5)*sqrt(5)*sqrt(625),
sqrt(5)*sqrt(5)=5,sqrt(625)=25
therefor,sqrt(15625)=5*25=125.

If a number whose last number is 2,4,6,8,you can try to divide the number by 2,4,6 or 8.
For example:
sqrt(384)=?
384=4*96=4*(4*24)=4*(4*(4*6))
so,
sqrt(384)=sqrt(4)*sqrt(4)*sqrt(4)*sqrt(6)=4*2*sqrt(6)=8*sqrt(6).

I think these will help work out your problems.Thanks.

 

[Sar]

So

[math]\sqrt{120} = \sqrt{60*2} = \sqrt{30*2*2} = 2*\sqrt{30}[/math]
I stop at the thirty since 15*2*2*2 would lead to something like

[math]\sqrt{15*2*2*2} = 3*\sqrt{15}[/math]
and that doesnt come out right.

Anyway, the lost in translation I take it that it works like this:

[math]n+n+n+n/n[/math]
Thank you guys, this has been really helpfull!

is there any source for the math code and how to use it? seems like a nice way of making expressions.

n72.75 : Yeah, I bet I can, but I'm studying the first course solo and approximation comes in another course atleast I hope so since that looks like fun. Now I only have equations, exponents, roots, and stuff like that.

 

[Artlav]

Wrong unless typo -
[math] \sqrt{15*2*2*2} = 2*\sqrt{2*15} [/math]It's multiplication -
[math] \sqrt{a*b} = \sqrt{a}*\sqrt{b} [/math]
I wonder, here we learn multiplication tables early at school - products of all permutations of numbers 1 through 9.

Do you?

 

[Notebook]

Showing my age here...we learned our "times tables" up to 12x12 by rote in school. Of course that was in LSD days(Pounds, Shillings, Pence).

N.

 

[Sar]

Wrong unless typo -
[math] \sqrt{15*2*2*2} = 2*\sqrt{2*15} [/math]It's multiplication -
[math] \sqrt{a*b} = \sqrt{a}*\sqrt{b} [/math]
I wonder, here we learn multiplication tables early at school - products of all permutations of numbers 1 through 9.

Do you?

Yes we do, I was using the logic I explained

[math] n+n+n+n/n [/math]
Which is flawed it seems.

Taking that [math] 2 * \sqrt{2*15} = 2*\sqrt{30}[/math]
[math]\sqrt{280} = \sqrt{140*2} = \sqrt{70*2*2} = \sqrt{35*2*2*2} = \sqrt{7.5 *2*2*2*2}[/math]
But that's already a decimal so I take it upwards to a nondecimal so

[math]\sqrt{35*2*2*2} = 2*\sqrt{35*2} = 2*\sqrt{70}[/math]
This logic seems legit, at least wolfram agrees with the result.

 

[NOMAD]

I stop at the thirty since 15*2*2*2 would lead to something like

\sqrt{15*2*2*2} = 3*\sqrt{15}

and that doesnt come out right.

Yes, you're right.
sqrt(120)=sqrt(4)*sqrt(30)=2*sqrt(30)
and
sqrt(30)=sqrt(2)*sqrt(15)=sqrt(2)*( sqrt(3)*sqrt(5) )
but
sqrt(2) ,sqrt(3) and sqrt(5) are irrational numbers,so we don't need to simplify it.
The final result is 2*sqrt(30)

I also curious about that why don't you try 4,because 120=4*30,it is convenient .
Like this :
sqrt(256)=sqrt(4)*sqrt(64)=2*8=16
sqrt(280)=sqrt(4)*sqrt(70)=2*sqrt(70)
You don't have to always use number 2,try other number like 4,6,8,it will help you simplify more quickly and accurately.

 

[Sar]

I have experimented with the numbers up to a point

[math]\sqrt{2048} = 32\sqrt{2} = \sqrt{(32^2)*2}[/math]
I think I've managed to grasp the idea of square root quite nicely (as nicely as one can in one day with limited tutoring)

I assume that the same system works on nth root aswell?

 

[RGClark]

I have experimented with the numbers up to a point

[math]\sqrt{2048} = 32\sqrt{2} = \sqrt{(32^2)*2}[/math]
I think I've managed to grasp the idea of square root quite nicely (as nicely as one can in one day with limited tutoring)

I assume that the same system works on nth root aswell?

Quite so.

Bob Clark

 

[Urwumpe]

Of course it does.

Also: The Newton method is for most simple squares like shooting with 46 cm cannons on pigeons. There is an even older method, which you will often find in computer programming tutorials as example of iterative solving: Herons method. It only uses the four basic operators and is pretty fast (unless you have a FPU, like all modern microprocessors have and which can do that MUCH faster)

https://secure.wikimedia.org/wikipedia/en/wiki/Methods_of_computing_square_roots#Babylonian_method

If you just need to calculate accurate for 3-4 digits, this should be much better than doing the more secure way with Newtons general solver for non-linear functions (because this method doesn't care which function you throw at it, it can solve almost everything, given enough time and good enough initial values)

 

[n72.75]

No...

Newton's method would be much faster.

e.g. take a look at this function: http://www.wolframalpha.com/input/?i=x^(5/4)-10

To find the nth root of m, where n and m are in Re ---> x^n-m solve for the positive zero using Newton's method to find the principal root.

 

[Stoat]

I think there could be the possibility of you becoming swamped by too many willing helpers. If you go to this website, I think you'll find that this guy is wonderful at explaining maths.

Go to downloads and download algebra. Then have a check through his extras. There's one pdf on commonly made mistakes which is wonderful.

http://tutorial.math.lamar.edu/

 

[Urwumpe]

No...

Newton's method would be much faster.

Actually, if you apply Newtons method on

[math]\sqrt{x} = a \rightarrow 0 = a^2 - x[/math]
you would get the same iteration step as Herons method. :lol:

Newtons method is the general iterative solution that works best in most cases - but having already prepared special solutions ready is better, because the paperwork before you can iterate with Newton, is annoying.