/sci/ - Science & Math

OP here, one direction I tried was: (a + b)(a - b) = a2 - b2

but it didn't work

http://www.wolframalpha.com/

>>5682086
worlframalpha is a POS right now, you need registration or facebook to use that shit

(a+b)^2=a^2+b^2+2ab

>>5682102
doesn't work, already tried

start with <span class="math">(a+b)^2[/spoiler]

>>5682105
It does work, I just did it. You must persist.

Squaring the equations for a and b (keeping in mind that both are nonnegative to preserve equivalence) yields:
a^2=4-sqrt(10+2sqrt(5)) and b^2=4+sqrt(10+2sqrt(5)).
Adding the two gives a^2+b^2=8, which can be reformed to (a+b)^2=8+2ab.
Multiplying the equations of a and b gives ab=sqrt(6-2sqrt(5)) by the third binomial theorem (Is it called that way in English? "(x+y)(x-y)=x^2-y^2), which is the square root of the sum of a rational and an irrational, which is the square root of an irrational, which is irrational since if the squareroot of an irrational could be rational, the square of that root would have to be rational too. Since ab is irrational, 8+2ab and therefore (a+b)^2 must be irrational too. This eliminates the answers a) through c).
Now there are only two answers left, which can be quite trivially tested for truth by inserting them into the equation for (a+b)^2.

you and your 'phd' friends are fucking idiots.

i hope none of them work in mathematics or physics, because they don't deserve to.

>>5682095
Someone should register on there and set up a proxy for us to use it.

How is 1/a^3 the same as a^-3?

I used to solve things like this in high school. Seriously.

>>5682125
we're all VERY proud of you

>>5682124
Or at least what rule is it called?

>>5682124
lets sat A=2.
1/2^3 = 1/8
2^-3 = 1/8
i accidentally mathed. forget what this is called

>>5682129
i don't think it has an official name, it's just the definition of a negative exponent

>>5682131
his point isn't that he's smart, it's that you're a fucking moron destined for walmart.

>>5682137
There are circumstances that delay people's progress in math other than being a moron.

<span class="math"> (a+b)^2=a^2+b^2+2ab=4-\sqrt{10+2\sqrt{5}}+4+\sqrt{10+2\sqrt{5}}+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{(\sqrt{5}-1)^2}=6+2\sqrt{5}=(1+\sqrt{5})^2 [/spoiler]

OP here, so far only:
>>5682102
>>5682109
>>5682113
discussed the question in matter.
keep focused, /sci/

>>5682133
>>5682133
That shows that 1/2^3 = 2^-3 but I'm asking why 1 divided by any number to a power is the same as that number to the negative of that power.

>>5682137
It's like when people take a picture of something, but make sure to include their cleavage or their abs in the frame. They sorta want you to look at the thing, but really they want you to be impressed with them. Academia is full of people like this. I just want him to know he is loved for being so smart, so he can get on with his life.

<span class="math"> (a+b)^2=a^2+b^2+2ab=4- \sqrt{10+2 \sqrt{5}}+4+ \sqrt{10+2 \sqrt{5}}+2 \sqrt{6-2 \sqrt{5}}=8+2 \sqrt{( \sqrt{5}-1)^2}=6+2 \sqrt{5}=(1+ \sqrt{5})^2 [/spoiler]

>>5682151
This

>>5682148
Looks like the answer is that it can't be solved with plain algebra. They know you can do algebra, so they are trying to see if you can think creatively to find your answers, like with the techniques >>5682113 mentioned.

<div class="math"> \sqrt{a + \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2-b}}{2}} + \sqrt{\frac{a - \sqrt{a^2-b}}{2}} </div>

That'll get real big real quick...hopefully you can cancel something.

>>5682152

<div class="math">\left(\sqrt{4 - \sqrt{10+2 \sqrt{5}}} + \sqrt{4 + \sqrt{10+2 \sqrt{5}}}\right)^2 =
\left(4 - \sqrt{10+2 \sqrt{5}} \right) + 2 \cdot \sqrt{4 - \sqrt{10+2 \sqrt{5}}} \cdot \sqrt{4 + \sqrt{10+2 \sqrt{5}}} + \left(4 + \sqrt{10+2 \sqrt{5}}\right) = </div><div class="math">
8 + 2 \cdot \sqrt{\left(4 - \sqrt{10+2 \sqrt{5}}\right)\left(4 + \sqrt{10+2 \sqrt{5}}\right)} = 8 + 2 \cdot \sqrt{16 - (10+2 \sqrt{5}}) = 8 + 2\sqrt{6 - 2\sqrt{5}}
= 6 + 2 \sqrt{5}= 5 + 2 \sqrt{5} + 1 = (\sqrt{5} + 1)^2
</div>

a+b=opisafag

>>5682167
this is wrong

>>5682191
be more specific or you just a troll

>>5682199
5th step onward

>>5682206
extra slow for you:
<div class="math">8 + 2\sqrt{6 - 2\sqrt{5}} = 8 + 2\sqrt{\left(\sqrt{5} - 1\right)^2} = 8 - 2 + 2\sqrt{5} = 6+2\sqrt{5} = 5 + 2 \sqrt{5} + 1 = (\sqrt{5} + 1)^2</div>
so
<span class="math">(a+b)^2 = (\sqrt{5} + 1)^2[/spoiler]

>>5682270
oh fuck me

Is it possible just to raise everything to the 1/2 power and chug from there, like it would be a fuck ton easier than squaring it.

>>5682270

nice, anon, I gave it a shot, but couldn't recognize 'em high school formulas fast enough, been a while.

>>5682095
No you don't. I use it all the time.

/thread

>>5682586
yeah, um, during the exam they don't allow calculators, so, how would you go about doing those decimal approx's?

>>5682590
http://www.mathwords.com/r/radical_rules.htm

>>5682590
en dot wikipedia dot org/wiki/Methods_of_computing_square_roots#Digit-by-digit_calculation

>>5682113
You don't happen to be french, do you?

>>5682600
Didn't they teach you that ?