Fast way to find roots?

[Dillinger4]

Is there a fast way to find roots? I have to take a college entrance exam and on the sample test i was given one of the questions is Sqr(18) + Sqr(8). Besides using a calculator how do you find what the root is without having to plug in number after number. :)

[MathImagics]

The values can be reduced to 3 * sqr(2) + 2 * sqr(2) which gives you 5 * sqr(2) which is sqr(50) or just a tad over 7 (approx 7.07) ...

If you're not allowed to use a calculator, I think the point of the question is to identify that the formula can be simplified so that it is easy to approximate..

Dr Memory

[Dillinger4]

Right 5 * sqr(2). But how is Sqr(18) + Sqr(8) broken down to 3 * sqr(2) + 2 * sqr(2). I think i am missing somthing. :(

[MathImagics]

18 is 9 * 2 is 3^2 * 2, so sqr(18) = sqr(3^2 * 2) = 3 *sqr(2)

ditto for 8 = 2^2 * 2

If x = y^2 * z, then sqr(x) = y * sqr(z), OK?

[bugzpodder]

yes if you want to simiply sqrt(x), find the biggest square that divides into x, lets say y^2. so sqrt(x)=y*sqrt(z), where z=x/(y^2)

[MathImagics]

..... and Bob's your uncle!!!!

[Dillinger4]

So let me see if i can get this. Sqr(21) + Sqr(33)

42 * 1.3125 where 1.3125 = 21 / 42
52 * 1.32 where 1.32 = 33 / 52

So we should have 9 * Sqr(2)?

[bugzpodder]

lol, forgot to mention something, x,y,z has to be non-negative integers! so no. in case of root(33), x=33,y=1,z=33. so we have root(33), which indicates that it cannot be simiplified. lets say root(18), x=18,y=3,z=2. so it becomes 3*root(2)

[NotLKH]

Originally posted by Dilenger4
So let me see if i can get this. Sqr(21) + Sqr(33)

42 * 1.3125 where 1.3125 = 21 / 42
52 * 1.32 where 1.32 = 33 / 52

So we should have 9 * Sqr(2)?
That can't be "simplified", depending on your definition of simplification.

Baqsically, all these tests are doing is trying to see how observant you are in combination with standards rules and regs.

So, with something like sqr(18) + sqr(8) you must first recognize that 18 and 8 share 2 as a common factor. so sqr(18) + sqr(8) = sqr(9*2) + sqr(4*2).

so, you can see this is a SIMPLE relation. sqr(9)*sqr(2) + sqr(4)*sqr(2).
You should know how to do it. But evidently, By the number of Q's you have about exponential identities, you need some tuition.

Lets start with the basics.

1) What is xy times xz
2) what is {(a(bx))1/x} times {(a(cx)1/x}

[NotLKH]

Originally posted by Dilenger4
So let me see if i can get this. Sqr(21) + Sqr(33)

42 * 1.3125 where 1.3125 = 21 / 42
52 * 1.32 where 1.32 = 33 / 52

So we should have 9 * Sqr(2)?
OK, lets see.
21 = 3*7
33 = 3*11
So:
Sqr(21) + Sqr(33) ==>
Sqr(3*7) + sqr(3*11)==>
{Sqr(7)+Sqr(11)}*Sqr(3)

Can't go any furthur!

;)

[Dillinger4]

Math is not my forte as you can tell but i still find it interesting. :)
So Sqr(18) can be broken down as Sqr(2 * 32) but how do i find the actual value without having to plug in number after number? I am assuming Sqr(18) works out to somthing like 4.242. I was originally refering to finding the square root of a number that is not a prefect square but i guess the only way to go about it is to estimate.

[bugzpodder]

lol without a calcuator don't expect to find the a close approximation of the roots. how close do you want to estimate? first you better memorize the roots of all the prime numbers <15 to that estimate (so if you want an approximation to 100 decimal place, you better memorize the roots of 2,3,5,7,11,13 to 100 decimal place) and then do what we did before and multiply them together by hand (requires patience and good multiplying skills)

so if you want to find root(6), you'd do root(2)*root(3)

root(18) would give you 3*root(2).

however you can't find values like root(34) unless you memorize further. there is the newton's square root method if you don't memorize

if you want to find the sqrt of 36, you'd take 1 and divide by 36
then you get the answer (which is 36) and average it with 1, and now you get 18.5. you repeat the process until both numbers are very close. (so you would take 36/18.5 and average with 18.5). it takes only a few tries.

[Starman]

(I have posted this before somewhere, do a search on 'square root' there are several methods)

Here is a manual method example, it’s similar to long division :
Personally I find it easier than the guess/average method as I don't have to continually multiply the answer to check the result.
You can continue as long as you like, and each iteration provides the next digit (not an approximation).


Pick a number, any number 18.6624

Draw a line above the number and a vertical line down the left hand side.

Starting at the decimal point, draw vertical lines grouping the numbers into pairs.

|---|-|---|------------
|18|.|66|24

Take the approx square root(lower than) of the leftmost pair of digits 18, which is 4.

Write this number above the leftmost pair (add the decimal point if it’s next) this is the partial answer.

4 .
|---------------------
|18|.|66|24


Write the square of this partial answer under the first pair and subtract.

4 .
|---------------------
|18|.|66|24
|16
| --
| 2

Bring down the next pair of digits

4 .
|---------------------
|18|.|66|24
|16
| --
| 266


Multiply the partial answer by 2 and write it next to the new line, with an underline to the right of it.

4 .
|---------------------
|18|.|66|24
|16
| --
8_ | 266

Find a digit (x) that will replace the underline so that (10 times double the partial answer plus x) multiplied by x is less than the present remainder
or in this case so that x times eighty-x is less that 266.
4 . 3
|---------------------
|18|.|66|24
|16
| --
83 | 266
| 249
add this new digit to the partial answer on the top line and multiply the 83 by the new digit.

Subtract this new value from the line above it and bring down the next pair of digits

4 . 3
|---------------------
|18|.|66|24
|16
| --
83 | 266
| 249
|------
| 1724


Again, multiply the partial answer by 2 and leave an underline to the right

4 . 3
|---------------------
|18|.|66|24
|16
| --
83 | 266
| 249
|------
86_ | 1724

Find a replacement for the underline similar to above, add this to the partial answer, and multiply

(the 862 by 2)


4 . 3 2
|---------------------
|18|.|66|24
|16
| --
83 | 266
| 249
|------
862 | 1724
| 1724
----------
0

Subtract and continue until you have enough digits in the partial answer for your purposes.


In this case ( as I cheated) the result of the subtraction is zero so the answer is a proper square root.

4.32 = sqrt(18.6624)

(If the font knackers my format, just line up the vertical pipes and hopefully it will become clearer)

[Starman]

Sorry, it did. I know there's a way to post without having all the whitespace removed, but I hope you can work out what's there.

[Dreamlax]

hold up... I was just taught that square roots and products of square roots can be treated like normal numbers (kinda thing). ie, SQRT ( 12 ). Make it twelve... then find two factors, one of which being a perfect square, 3, and 4. So, 12 = 3 * 4, and SQRT ( 12 ) = SQRT ( 3 ) * SQRT ( 4 ). From there, you can find SQRT ( 4 ) easily enough and multiply it by SQRT ( 3 ) to find a 'simpler' way of writing SQRT ( 12 ).

It's basically what's been said here, only told in a different way.

[Evil_Giraffe]

I think the method you'd be after is to represent it as prime roots (I forget the proper name for them)
for example: 18
18 as a product of primes is 2*3*3
=> sqr(18) = sqr(2*3*3) = sqr(2) * sqr(3) * sqr(3) = sqr(2) * 3
another example: 23
23 as p.o.p. = 23 (ummmm, k, so this isn't a great example)
=> sqr(23) = sqr(23)
k lets say 21
21 as p.o.p. = 3 * 7
sqr(21) = sqr(3*7) = sqr(3) * sqr(7)

Get the idea? It's the most exact way to represent square roots (if not the most useful). Where you have a pair of numbers, you can multiply them together so the sqr disappears, and make it a coefficient.
27 = 3*3*3
sqr(27) = sqr(3) * sqr(3) * sqr(3) = 3 * sqr(3)

Isn't maths great after popping down to the pub for lunch?