General Addition

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hpineiro2
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General Addition

Postby hpineiro2 » Mon Aug 08, 2011 4:49 am UTC

Alright, my power went out as I nearly finished this last time.. Here's take two. I'm making it short and sweet.

So the other day, I proposed this question to myself: "How do you add radicals?" Well, specifically this question: "How do you add unlike radicals together and get an exact answer?" I did not want to settle for "x = √(y) + √(z)." So I played around with algebra for a night, trying to use logarithms and failing miserably. But eventually I came up with a General Addition Formula. Here's the math:

Again:
x = √(y) + √(z)

I wanted to solve the equation for x and get an exact answer. However, I decided to take a step back and examine this problem in terms of basic addition.

• Let x = the sum of y + z
• Let y & z = any two numbers (real, imaginary, etc. -- Works for everything that I know)

x = y + z

This is really simple when y and z are integers, and apparently even "simpler" when it comes to unlike radicals (because you just leave them the way they are), but I wanted to take it a step farther:

x² = (y + z)(y + z)
x² = y² + 2yz + z²
x = √(y² + 2yz + z²)

Of course, this could be simplified:

x = √((y + z)²)
x = y + z

But that's not going to help when it comes to adding two unlike radicals together. This General Addition Formula allows any two numbers to be added together.

Example:
Solve for x (exact answer).
x = √(2) + √(3)

By using the General Addition Formula:
x = √(5 + 2√(6))

Now, the main reason I bring this up here is to see if anybody knows where this extended addition would be useful. Is there a particular field in math (or a certain job) that this extended answer √(5 + 2√(6)) would be more practical than √(2) + √(3)? Or is this just some fun algebra that should be left as √(2) + √(3)? I just found it interesting how two unlike radicals could be added together resulting in an exact answer, as opposed to a calculator-approximated decimal answer. Thanks for reading!

Sagekilla
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Re: General Addition

Postby Sagekilla » Mon Aug 08, 2011 8:43 pm UTC

You have to be careful when saying something like [imath]x = \sqrt{x^2}[/imath] Because that's not right. It's [imath]\sqrt{x^2} = |x| \ne x[/imath]

I can also give a simple counterexample to this:

x = -1
y = -1

[imath]\sqrt{x} + \sqrt{y} = 2i[/imath]
[imath]\sqrt{x + y + 2 \sqrt{x y}} = \sqrt{-2 + 2} = 0[/imath]

No need to even introduce complex values of x or y, as this already fails for this simple case. If you restrict your domain to positive reals,
then this isn't an issue. Alternatively, you can define x to be any positive real number, and y to be any real (positive or negative).
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Dopefish
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Re: General Addition

Postby Dopefish » Mon Aug 08, 2011 9:24 pm UTC

For x=y=-1 their formula says [(-1)+(-1)] =-2] = [sqrt( (-1)^2 +2(-1)(-1) + (-1)^2 ) = 2], which is false because of the sqrt(x^2)= |x| not x as mentioned, but the magnitude should always be right (at least for real numbers anyway).

In your counter example you additionally sqrt'd everything under the sqrt for some reason which is why the magnitudes don't match.

I feel like I've seen this sort of thing done before for some specific reason, but I can't remember what. Perhaps some sort of program that required weird input or some such...

Yesila
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Re: General Addition

Postby Yesila » Tue Aug 09, 2011 12:53 am UTC

hpineiro2 wrote:Example:
Solve for x (exact answer).
x = √(2) + √(3)

By using the General Addition Formula:
x = √(5 + 2√(6))




Your 5 should be a 13, since [imath]2^2+3^2=13[/imath].

My intuition tells me that the only time your [imath]\sqrt{x}+\sqrt{y}=\sqrt{x^2+y^2+2\sqrt{xy}}[/imath] formula would be "useful" would be if the inner most radical was of a perfect square... which happens exactly when x=y. Or if the sum [imath]x^2+y^2 +2\sqrt{xy}[/imath] "happens to be" rational. In any other case you'll be dealing with the root of a root which seems to me to be much less helpful than adding together to roots -- if for no other reason than if you want to approximate the value you need keep track of fewer digits to reach the desired level of accuracy.

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NathanielJ
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Re: General Addition

Postby NathanielJ » Tue Aug 09, 2011 2:34 am UTC

Yesila wrote:My intuition tells me that the only time your [imath]\sqrt{x}+\sqrt{y}=\sqrt{x^2+y^2+2\sqrt{xy}}[/imath] formula would be "useful" would be if the inner most radical was of a perfect square... which happens exactly when x=y.


It happens plenty of other times too (e.g., x = 3, y = 12), though one of them has to be a divisor of the other, and the remainder after division has to be a perfect square so it simplifies right away anyway (e.g., [imath]\sqrt{3}+\sqrt{12}=\sqrt{3}(1 + \sqrt{4}) = 3\sqrt{3}[/imath]).
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Dopefish
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Re: General Addition

Postby Dopefish » Tue Aug 09, 2011 2:46 am UTC

The OP doesn't have any roots of roots in any of their equations...am I missing something? :?

hpineiro2
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Re: General Addition

Postby hpineiro2 » Tue Aug 09, 2011 3:05 am UTC

Thank you everybody for your input. I apologize for my mistake. Yes, I meant √(x²) = |x|. For some reason that slipped my mind; perhaps because I was thinking in terms of positive numbers. When it comes to negative numbers, there's a little extra thinking required but I'll explain what I mean below (with some examples).

First, I think there may be some misunderstanding for the example. I hope this clarifies it:

Example:
Solve for x (exact answer).

x = √(2) + √(3)

Here are the steps:

Given:
y = √(2) -- NOT 2
z = √(3) -- NOT 3

Problem:
x = √(2) + √(3)

Square both sides:
x² = ( √(2) + √(3) )²

Expand:
x² = ( √(2) + √(3) )*( √(2) + √(3) )

"FOIL":
x² = 2 + √(6) + √(6) + 3

Simplify:
x² = 5 + 2√(6)

Solution:
x = √(5 + 2√(6))

The General Addition Formula comes from this:

x = y + z

Not this formula:

x = √(y) + √(z)

The latter formula was just referring to "Radical Addition". In the General Addition Formula, y and z each comprise of the entire corresponding number, including the square root (if there is one).

Now for negative numbers (and subtraction): There are some extra negative signs throw into the final answer depending on the signs of each number. View spoilers for examples (with step-by-step explanations).

Example 1:

x = y - z , where y > z

Spoiler:
Solve for x (exact answer).

x = √(3) - √(2)

Steps:

Given:
y = √(3)
z = -√(2)

Problem:
x = √(3) - √(2)

Square both sides:
x² = ( √(3) - √(2) )²

Expand:
x² = ( √(3) - √(2) )*( √(3) - √(2) )

"FOIL":
x² = 3 - √(6) - √(6) + 2

Simplify:
x² = 5 - 2√(6)

Solution:
x = √(5 - 2√(6))


Example 2:

x = y - z , where z > y

Spoiler:
Solve for x (exact answer).

x = √(2) - √(3)

Steps:

Given:
y = √(2)
z = -√(3)

Problem:
x = √(2) - √(3)

Square both sides:
x² = ( √(2) - √(3) )²

Expand:
x² = ( √(2) - √(3) )*( √(2) - √(3) )

"FOIL":
x² = 2 - √(6) - √(6) + 3

Simplify:
x² = 5 - 2√(6)

Solution:
x = -√(5 - 2√(6))

Notice: This solution is negative opposed to the previous example's answer which was positive. If you need an explanation, read below, if not, skip it.

For the equation x = 3 - 2 the answer is a positive (+) 1. But in the equation x = 2 - 3 the answer is a negative (-) 1. To determine whether the final answer is positive or negative when adding radicals, check to see if y > z or if z > y. The latter condition results in a negative answer.

Also, two negative numbers added together will equal a negative number (Example 3a & 3b).


Example 3a:

x = y + z , where z & y < 0

Spoiler:
To show that your "counterexample" works.

Solve for x (exact answer).

x = -1 + (-1)

Steps:

Given:
y = -1
z = -1

Problem:
x = -1 + (-1)

Simplify:
x = -1 - 1

Square both sides:
x² = ( -1 - 1 )²

Expand:
x² = ( -1 - 1 )*( -1 - 1 )

"FOIL":
x² = 1 + 1 + 1 + 1

Simplify:
x² = 4

Take the Square Root of both sides:
x = -√(4)

Solution:
x = -2

To determine whether the answer is negative or positive, look at Example 2.


Example 3b:

x = y + z , where z & y < 0

Spoiler:
Solve for x (exact answer).

x = -√(2) + -√(3)

Steps:

Given:
y = -√(2)
z = -√(3)

Problem:
x = -√(2) + -√(3)

Simplify:
x = -√(2) - √(3) [You could factor out a negative, but I won't in this example]

Square both sides:
x² = ( -√(2) - √(3) )²

Expand:
x² = ( -√(2) - √(3) )*( -√(2) - √(3) )

"FOIL":
x² = 2 + √(6) + √(6) + 3

Simplify:
x² = 5 + 2√(6)

Solution:
x = -√(5 + 2√(6))

This answer is negative because two negative numbers added together equals a negative number.

Side note: Wow this answer is kinda close to -π.


Example 4:

x = y + z , where y or z is an imaginary number (i)

Spoiler:
Solve for x (exact answer).

x = i + 2

Steps:

Given:
y = i
z = 2

Problem:
x = i + 2

Square both sides:
x² = ( i + 2 )²

Expand:
x² = ( i + 2 )*( i + 2 )

"FOIL":
x² = -1 +2i + 2i + 4

Simplify:
x² = 3 + 4i

Solution:
x = √(3 +4i)

This example is just to show that imaginary numbers work. I think this answer isn't as sloppy as the answer in the other examples, however, it is definitely more complicated than i + 2.


I hope that this clears up what I was trying to get across. Thank you for reading and commenting!
Last edited by hpineiro2 on Tue Aug 09, 2011 3:50 am UTC, edited 4 times in total.

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Qaanol
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Re: General Addition

Postby Qaanol » Tue Aug 09, 2011 3:13 am UTC

Yes, you can certainly square ξ = a√x + b√y to get ξ2 = a2x + b2y + 2ab√(xy), and then take an appropriate square root.

The former is more computationally robust, whereas the latter makes clearer what the minimum polynomial of ξ is. Which to prefer depends, at a certain level, on whether you’re doing analysis or abstract algebra.
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