/sci/ - Science & Math

[Anonymous No.3062235]

Can /sci/ do Algebra?

Find X

[Anonymous No.3062321]

>>3062235
x=1.2055694

[Anonymous No.3062357]

>>3062321
Nope your wrong

[Anonymous No.3062364]

Who needs algebra when you can just graph that bitch?

[Anonymous No.3062380]

>>3062364
You aren't just a betaman in the real world. You are also a betanerd

[Anonymous No.3062383]

>>3062235
These posts should be banned if wolfram can answer
http://www.wolframalpha.com/input/?i=%28x%2Fx-1%29+-+1%2F%28%28x^2%29-x%29+%3D+1+%2F+%28%28x^3%29-3%
29

[Anonymous No.3062391]

>>3062321

This is correct.

I was pissed when I tried to simplify all of that and found out the answer wasn't rational.

[Anonymous No.3062393]

Op here damn it I fucked up the problem

here is the updated Problem

[Anonymous No.3062408]

>>3062393
X=2
Obvious just by looking at it. No algebra needed.

[Anonymous No.3062413]

>>3062393

but yeah, fucking use the internets (wolfram alpha) for such things

[Anonymous No.3062417]

>>3062380
>implying I actually enjoy math and would want to waste my time working something out on paper when i could get it done in 10 seconds on a calculator.

[Anonymous No.3062435]

[Anonymous No.3062464]

>>3062235
lol op, there is a bunch of Xs in there, how you didn't found X?

[Anonymous No.3062466]

>>3062393

Shift the term on the right side of the equation to the left, the second term on the left to the right.

You can easily get a common denominator between (x/(x-1)) and 1/((x^3)-x), and it conveniently works out so that (x^2)-1 is in both the numerator and the denominator, leaving you with 1/x = 1/((x^2)-x, which you can then get to x^2 = 2x

x = 2

[Anonymous No.3062476]

>>3062464
damn great job

[Anonymous of Germany !WvzhzxDiIQ No.3062494]

<span class="math">(x^2 - x)(x^3-x) - (x^2-1)(x^3-x) = (x^2-1)(x^2-x)[/spoiler]
<span class="math"> x^5-x^4 -x^3 +x^2 - x^5 + 2 x^3 - x = x^4 - x^3 -x^2 +x [/spoiler]
<span class="math"> 2 \cdot x^4 -2 \cdot x^3 - 2 \cdot x^2 + x = 0[/spoiler]
<span class="math"> 2 x^3 - 2x^2 -2x + 1 = 0[/spoiler]
<span class="math"> \Rightarrow x = - 0.855 ; x = 0,403 ; x = 1,452 [/spoiler]

[Anonymous No.3062497]

Hold on while I TeX...

[Anonymous No.3062516]

>>3062494

lol, no. your solution is neither correct nor elegant.

[t-shirt !pants1bKj2 No.3062536]

i got x=2 and x=-1

>mfw f(x=-1)

[Anonymous No.3062572]

>>3062497
<span class="math"> \displaystyle{ \frac{x}{x-1} - \frac{1}{x^2-x} = \frac{1}{x^3-x} } [/spoiler]

<span class="math"> \displaystyle{ \frac{x}{x-1} - \frac{1}{x^2-x} - \frac{1}{x^3-x} = 0 } [/spoiler]

<span class="math"> \displaystyle{ x(x^2-x)(x^3-x)-(x-1)(x^2-x)(x^3-x)-(x-1)(x^2-x)(x^3-x)}{(x-1)(x^2-x)(x^3-x)} } [/spoiler]

<span class="math"> \displaystyle{ x^3(x-1)(x^2-1)-2x^2(x-1)^2(x^2-1) = 0 } [/spoiler]

<span class="math"> \displaystyle{ (x^4-x^3)(x^2-1)-(2x^4-2x^2)(x^2-2x+1) = 0 } [/spoiler]

<span class="math"> \displaystyle{ x^6-x^5-x^4+x^3-2x^6+4x^5-2x^4+2x^4-4x^3+2x^2 = 0 } [/spoiler]

<span class="math"> \displaystyle{ -x^6+3x^5-x^4-3x^3+2x^2 = 0 } [/spoiler]

<span class="math"> \displaystyle{ -x^2(x-2)(x-1)^2(x+1) } [/spoiler]

<span class="math"> \displaystyle{ x=-1,0,1,2 } [/spoiler]

[Anonymous No.3062593]

<span class="math"> \displaystyle{ \frac{x}{x-1} - \frac{1}{x^2-x} = \frac{1}{x^3-x} } [/spoiler]

<span class="math"> \displaystyle{ \frac{x}{x-1} - \frac{1}{x^2-x} - \frac{1}{x^3-x} = 0 } [/spoiler]

<span class="math"> \displaystyle{ x(x^2-x)(x^3-x)-(x-1)(x^2-x)(x^3-x)-(x-1)(x^2-x)(x^3-x)}{(x-1)(x^2-x)(x^3-x) } [/spoiler]

<span class="math"> \displaystyle{ x^3(x-1)(x^2-1)-2x^2(x-1)^2(x^2-1) = 0 } [/spoiler]

<span class="math"> \displaystyle{ (x^4-x^3)(x^2-1)-(2x^4-2x^2)(x^2-2x+1) = 0 } [/spoiler]

<span class="math"> \displaystyle{ x^6-x^5-x^4+x^3-2x^6+4x^5-2x^4+2x^4-4x^3+2x^2 = 0 } [/spoiler]

<span class="math"> \displaystyle{ -x^6+3x^5-x^4-3x^3+2x^2 = 0 } [/spoiler]

<span class="math"> \displaystyle{ -x^2(x-2)(x-1)^2(x+1) } [/spoiler]

<span class="math"> \displaystyle{ x=-1,0,1,2 } [/spoiler]

[Anonymous No.3062628]

>>3062593

Pretty sure I expanded wrong somewhere :/

[Anonymous No.3062643]

>>3062391
1.2055694 is rational

[Anonymous No.3062656]

>>3062516

Notmathfag here. What do mathfags mean by 'elegant', I don't get it. I've heard them say stuff like "This proof is elegant"

[Anonymous No.3062761]

<span class="math"> \displaystyle{ \frac{x}{x-1} - \frac{1}{x^2-x} = \frac{1}{x^3-x} } [/spoiler]

<span class="math"> \displaystyle{ \frac{x}{x-1} - \frac{1}{x(x-1)} = \frac{1}{x(x-1)(x+1)} } [/spoiler]

<span class="math"> \displaystyle{ \left[ \frac{x}{x-1} - \frac{1}{x(x-1)} = \frac{1}{x(x-1)(x+1)} \right] [x(x-1)(x+1)] } [/spoiler]

<span class="math"> \displaystyle{ \frac{x^2(x-1)(x+1)}{x-1} - \frac{x(x-1)(x+1)}{x(x-1)} = \frac{x(x-1)(x+1)}{x(x-1)(x+1)} } [/spoiler]

<span class="math"> \displaystyle{ x^2(x+1)-(x+1)=1 } [/spoiler]

<span class="math"> \displaystyle{ x^3+x^2-x-2=0 } [/spoiler]

<span class="math"> \displaystyle{ (x-2)(x^2+x+1)=0 } [/spoiler]

<span class="math"> \displaystyle{ x=2, -\frac{1 \pm \sqrt{3}}{2} } [/spoiler]

[Anonymous No.3062891]

>>3062761
Close, but (x - 2) is not a factor.

2 |  1   1   -1   -2
   |____2__6__10_
      1    3    5    8

[Anonymous No.3062934]

ok so obviously nobody double checks their algebra by plugging in their answers, very fucking typical, and everybody goes rushing to the quadratic formula like a hoard of trained zombies.

x=0, I know this is undefined once plugged back in but I stand by it because I used nothing but algebraic manipulations.

[Anonymous of Germany !WvzhzxDiIQ No.3063021]

>>3062516

Guess I fucked up somewhere. Whatever.

[Anonymous No.3063075]

why the fuck are people saying 2? plugging in 2 would make it say 1 and 1/2 = 1/6

[Anonymous No.3063097]

>implying this is actual algebra and not high school arithmetic
>implying you know what algebra is

[Anonymous No.3063117]

>>3062891
My calculator said it was :/

[Anonymous No.3063118]

This!

[Anonymous No.3063139]

>>3062891
>>3063117
I guess I plugged it in wrong, x=1.296 and some imaginary number

[Anonymous No.3063147]

>>3063139
1.206*

[Anon No.3063599]

x*x/(x-1)*x - 1/(x^2 - x) = 1/(x^3-x)

(x^2 - 1)/(x^2 - x) = 1/(x^3 -x)

(x^2 - 1)/(x^2 - x) = 1/(x^2 -1)x

(x^2 - x)/(x^2 -1) = x(x^2 -1)
(x-1)/(x^2 -1) = (x^2 - 1)
x -1 = (x^2 -1) ^2
x - 1 = x^4 -x^2 - x^2 + 1
-2 = x^4 - 2x^2

now lets use substitution that y=x^2,

-2 = y^2 - 2y
0 = y^2 - 2y + 2

a=1, b=-2, c=2

[2 +/- (4 - 8)^.5]/2

both roots are imaginary

[Anon No.3063608]

>>3063599
I didn't bother taking square root after making substitution. I'm sure you gentlemen get the idea though.