Infinity free: square roots/cube roots

in reality, all of this has been a total load of old bollocks

Correct...?

Yes
1
25%
No
3
75%
 
Total votes: 4

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C
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Infinity free: square roots/cube roots

Postby C » 05 Jun 2018, 18:37

A square root has two solutions.

It follows that a cube root has three solutions.

Is this correct?




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Last edited by C on 05 Jun 2018, 19:16, edited 1 time in total.
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harvey k-tel
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Re: Infinity free: square roots/cube roots

Postby harvey k-tel » 05 Jun 2018, 18:41

*fires elastic band at C's head when his back is turned*
Tempora mutatur et nos mutamur in illis

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Charlie O.
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Re: Infinity free: square roots/cube roots

Postby Charlie O. » 05 Jun 2018, 18:43

If you mean in the sense that a square root has a negative as well as a positive solution, then no.

negative x negative x negative = negative
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C
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Re: Infinity free: square roots/cube roots

Postby C » 05 Jun 2018, 19:12

Harvey K-Tel wrote:*fires elastic band at C's head when his back is turned*


:lol:




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Re: Infinity free: square roots/cube roots

Postby sloopjohnc » 05 Jun 2018, 21:13

Harvey K-Tel wrote:*fires elastic band at C's head when his back is turned*


:lol:

You hit him low, I'll hit him high.
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C
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Re: Infinity free: square roots/cube roots

Postby C » 05 Jun 2018, 21:45

The cube root of negative 8 is negative 2

ie
-2 x -2 x -2 = -8

One root. So incorrect?





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Lord Rother wrote:And there was me thinking you'd say "Fair enough, you have a point Bob".

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C
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Re: Infinity free: square roots/cube roots

Postby C » 07 Jun 2018, 16:45

Pansy Puff wrote:Cube roots of -8 are -2, 1 + root(3)i and 1 - root(3)i


Where i is the square root of -1

So the square root of -4 is equal to square root of 4 multiplied by square root of -1 (can we do this? check with something you know - is the square root of 36 equal to square root of 4 multiplied by square root of 9...? Yes it is!)

= square root of 4 x i

so the two roots are 2i and -2i [called conjugate roots]

The process of square roots of negative numbers takes us away from real numbers into imaginary numbers or complex numbers

So, applying this allows us to solve cube roots.

All real numbers (except zero) have exactly one real cube root and a pair of complex conjugate cube roots







.
Last edited by C on 07 Jun 2018, 16:55, edited 1 time in total.
Lord Rother wrote:And there was me thinking you'd say "Fair enough, you have a point Bob".

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Re: Infinity free: square roots/cube roots

Postby C » 07 Jun 2018, 17:02

Good lad

True – and I agree. I was about to edit out ‘except zero’ when you replied.

Those that would include are only considering the end product 0 x i =0 and not the big 'open' picture.

It takes me back to my earlier point of 1=3/3 etc



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Lord Rother wrote:And there was me thinking you'd say "Fair enough, you have a point Bob".

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harvey k-tel
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Re: Infinity free: square roots/cube roots

Postby harvey k-tel » 07 Jun 2018, 17:59

*looks around for more elastic bands*
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Re: Infinity free: square roots/cube roots

Postby harvey k-tel » 07 Jun 2018, 18:31

Noooooo!!
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Belle Lettre
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Re: Infinity free: square roots/cube roots

Postby Belle Lettre » 07 Jun 2018, 19:42

C wrote:
Pansy Puff wrote:Cube roots of -8 are -2, 1 + root(3)i and 1 - root(3)i


Where i is the square root of -1

So the square root of -4 is equal to square root of 4 multiplied by square root of -1 (can we do this? check with something you know - is the square root of 36 equal to square root of 4 multiplied by square root of 9...? Yes it is!)

= square root of 4 x i

so the two roots are 2i and -2i [called conjugate roots]

The process of square roots of negative numbers takes us away from real numbers into imaginary numbers or complex numbers

So, applying this allows us to solve cube roots.

All real numbers (except zero) have exactly one real cube root and a pair of complex conjugate cube roots







.

For some reason I find this rather pleasing.
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C
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Re: Infinity free: square roots/cube roots

Postby C » 07 Jun 2018, 21:44

C wrote:A square root has two solutions.

It follows that a cube root has three solutions.

Is this correct?




.


Correct




.
Lord Rother wrote:And there was me thinking you'd say "Fair enough, you have a point Bob".


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