How do I solve this?

TI-89

>>8650405
You can't. Galois theory proves that this is unsolvable.

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>>8650405

this is like algebra 2 shit come on man

>>8650405
[eqn] \sqrt{\left(\sqrt{2}+\sqrt{8}\right)^2} = \sqrt{2}+\sqrt{8} = \sqrt{2} + 2\sqrt{2} = 3\sqrt{2} [/eqn]

>>8650415
read the sticky

>>8650419
Its part of a distance formula equation

my answer did not equal the books,, why does the (power of 2) not cancel out the root things on 2 and 8?

chewing for you:

[math]
\sqrt{x^{2}} = x^{2 \cdot \frac{1}{2}} = x^{1} = x
\\
\sqrt{8} = \sqrt{2 \cdot 2 \cdot 2} = \sqrt{2^{2} \cdot 2} = 2\sqrt{2}

[/math]

>>8650427
>falling for the bait
The square can be used to cancel the outer root. Using the root on the other two would be like multiplying (√2 + √8) × (√2 + √8) which ends up being (√2√2 + √2√8 + √8√2 + √8√8) which is equal to (2 + 2√2√8 + 8)
If you need more help just ask

>>8650436
How do I write in math?

>>8650443
read the sticky

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>>8650439
I needed to find the Distance between these points

the books answer is
>square root of 83

I did not get this answer. How would this be calculated using the distance formula?

>>8650411
i once reported a thread in /sp/ with a picture of a naked black kid urinating on a woman and got banned for "Submitting false or misclassified reports".

i will never report anything again. fuck jannies, mods, and the gook moot

>>8650405
Finally someone post something challenging.
It took me about 30 minutes to solve, it was tougher than I originally thought. Here's my solution:
[eqn]\sqrt{(\sqrt{2}+\sqrt{8})^2} = \sqrt{(2+2\sqrt{2}\sqrt{8} +8)=}[/eqn]
[eqn]\sqrt{10+2\sqrt{2}\sqrt{8}}= \sqrt{10+(2^1*2^{1/2}*8^{1/2})=}[/eqn]
[eqn]\sqrt{10+(2^1*2^{1/2}*(2^3)^{1/2})} = \sqrt{10+(2^1*2^{1/2}*2^{3/2})} = [/eqn]
[eqn]\sqrt{10+(2^{2/2}*2^{1/2}*2^{3/2})}=\sqrt{10+(2^{6/2})}[/eqn]
[eqn]\sqrt{10+(2^{3}})= \sqrt{10+(8})= \sqrt{18}=\sqrt{3*3*2}[/eqn]
[eqn]\sqrt{3^2*2}= (3^2*2)^{1/2}= (3^2)^{1/2}*2^{1/2}=3^{2/2}*2^{1/2}=
3^1*2^{1/2}=3\sqrt{2} [/eqn]
Thanks anon for posting real problems!!!!!

>>8650471
find the distance between x1 and x2, find the distance between y1 and y2, now you got both catheti, use pythagoras

>>>/sci/sqt

>>8650443
[math]With[/math] [math]science[/math]

>>8650511
ok I got the correct answer now

I had forgot to do something

>>8650495
>Julio Profe

Kek

>>8650405
The answer is 3x root2

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>>8650415
nope
/r9k/ may work for you though

sqrt(2)+sqrt(8)

Or you can perform the square and get sqrt(2+2(4)+8) = sqrt(18) = 3sqrt(2)

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>all these brainlets saying sqrt(x^2)=x and not |x|

[math]\sqrt(\sqrt2+\sqrt8)^{2}[/math]
[math]\sqrt(2+\sqrt16+8)[/math]
[math]\sqrt14[/math]
I felt ridiculous typing this out.

>>8652274
>[math]\displaystyle\sqrt{\left(x^2\right)}=-|x| : x\in\mathbb{Z^+}[/math]
Yeah okay.

>>8652293
It's [math]\displaystyle2\sqrt{16}[/math]

>>8652293
even more ridiculous when you got it wrong senpai

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>>8650415
kek

The sqrt cancels the square.

You're left with sqrt(2)+sqrt(8)

Which is equal to sqrt(2)+2sqrt(2)

combine terms, 3sqrt(2)

>>8650495
You're an asshole. I like u tho.

>>8650443
Admin privileges.

>>8650443
You buy a 4chan pass and the math panel will display in your post/ reply box

Isn't this just the square root of 10? What am I missing?

>>8652653
Dude I think the square root and the square cancel out right??
so it should be Root 2+ Root 8...
Some dude got that in this thread except he showed his work which was unnecessary.

>>8650423
wrong
[math] \displaystyle
\sqrt{\left(\sqrt{2}+\sqrt{8}\right)^2} = \left | \sqrt{2}+\sqrt{8} \right | = \sqrt{2} + 2\sqrt{2} = 3\sqrt{2}
[/math]

>>8652653
>What am I missing
a brain

>>8652885
> assuming non-positive roots of positive numbers
shiggy

>>8652894
>convention is assumption
donatello

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>>8652894
https://en.wikipedia.org/wiki/Square_root

>>8653004
What?
I'm assuming that the square roots are gonna give positive numbers because I see no complex numbers.
[math] \sqrt{x^2}: \mathbb{R}^+ \to \mathbb{R}^+ [/math] is the identity function
>>8653042
But x > 0 you tard

>>8653110
the point is to assume nothing.

ABS is in the definition, use it always

>>8653115
> ABS is in the definition, use it always
it's not. but w/e

>>8650521

>>8650405
just type it into your calculator or something

>>8650405

[eqn]\sqrt{(\sqrt{2}+\sqrt{8})^{2} } [/eqn]
[eqn]= \sqrt{(\sqrt{2}+\sqrt{8})(\sqrt{2}+\sqrt{8}) } [/eqn]
[eqn]= \sqrt{(2 + \sqrt{16} + \sqrt{16} + 8)} [/eqn]
[eqn]= \sqrt{(10+4+4)} [/eqn]
[eqn]= \sqrt{18} [/eqn]
[eqn]= \sqrt{9\cdot 2} [/eqn]
[eqn]= 3\sqrt{2} [/eqn]

>>8650427
>cancel out
L0Lno fgt pls

>>8655564
retard

>>8650405
[math]
\sqrt{\left (\sqrt 2 + \sqrt 8\right ) ^2} =
\sqrt{\left (\sqrt 2 + 2 \sqrt 2\right ) ^2}
[/math]
[math]
= \sqrt {\left (3 \sqrt 2
\right ) ^2 } = \sqrt {\left ( \sqrt 18\right )^2}
[/math]
[math] = \sqrt {18} = \sqrt {2 \cdot 9} = 3 \sqrt 2
[/math]

>>8650415
How do you get banned from /pol/ I thought being a shitty board has become the core, frame and shell of /pol/ so thoroughly that the mods gave up?

>>8652235
>Weishaupt's plateau
I still don't get it