3 = 0

[No Name No.3959049]

x² + x + 1 = 0 so x ≠ 0

subtract x and 1

x² = -x - 1

divide by x (we can do this because x ≠ 0)

x = -1 - 1/x

plug x back in

x² + (-1 - 1/x) + 1 = 0

simply -1 and +1

x² - 1/x = 0

add 1/x

x² = 1/x

multiply by x

x³ = 1

cube root both sides

x = 1

plug x back in

(1)² + (1) + 1 = 0

simply

3 = 0

Walla, proof that 3 = 0

[Anonymous No.3959070]

anyone?

[Anonymous No.3959091]

>
>plug x back in
>x² + (-1 - 1/x) + 1 = 0

No. Adding 1 + 1/x to both sides yields x^2 +1 + 1/x + 1 = 0.

[No Name No.3959100]

>>3959091
There was no addition there. Only simplification. I distributed the implicit one in front of the parenthesis and I canceled out the minus one and plus one.

[Anonymous No.3959099]

x³ = 1
doesn't imply
x=1
it implies
x=1 or x=cos(2pi/3)+i sin(2pi/3) or x=cos(4pi/3)+i sin(4pi/3)

[No Name No.3959105]

>>3959099
True, there are three answers. Two imaginary, one real. But they are still answers.

[Anonymous No.3959150]

>plug back in
>into the x only and not into the x squared

full retard.

[Anonymous No.3959158]

You could make this much shorter

x² + x + 1 = 0

multiply with (x-1)

(x² + x + 1) (x-1)= 0

simplify

x^3 =1

[Anonymous No.3959170]

>>3959150
Since I proved x = -1 - 1/x, you can swap them anywhere in the equation and it "should" hold true.

[Anonymous No.3959176]

>>3959158
Thank you.

[Anonymous No.3959194]

Fundemental theorem of algebra, all polynomials can be reduced to linear and quadratic terms over R[x]. To find more roots using radicals is impossible. However over C[x] all polynomials can be factored to linear terms. -Gauss 1799

[Anonymous No.3959210]

In other words, you fucked up because you aren't using complex numbers. You can brute force solutions in R via the quadratic formula, but until you define sqrt/-1 you don't have anymore solutions to the equation $ x^2+x+1$.

[No Name No.3959248]

>>3959210
sqrt(-1) is defined. It's defined as i :)

[Anonymous No.3959264]

>>3959248

Nope. 'i' is defined in the following manner:

i^2 = -1

[No Name No.3959282]

Well isn't that a boner kill.

[Anonymous No.3959287]

>>3959248
isn't (-i)^2 = -1 too, eh?

[Anonymous No.3959296]

>>3959287
Correct, that is why when you take the square root of something you must add plus or minus. But not for taking the cube root.

[Anonymous No.3959608]

>x² = -x - 1

>divide by x (we can do this because x ≠ 0)

>x = -1 - 1/x

3/10 would rage again

[Anonymous No.3959623]

>>3959049
x² + x + 1 = 0
Multiply by x
x^3 + x^2 + x = 0
Now x = 0 is a solution! Super magic!

[Ha No.3959642]

OP i am proud.

[Anonymous No.3959666]

>>3959623
>Doesn't know about negatives

[Anonymous No.3959696]

>>3959049
>x² + x + 1 = 0 so x ≠ 0 and x ≠ 1 as 3 ≠ 0
Fixed.

Anyway.

x³ = 1

=> x = 1, -1/2 + √(3)i/2, -1/2 - √(3)i/2

The last two are solutions of x² + x + 1 = 0, but you get the third one (1) because you multiplied by x, giving you a cubic, which has three roots by definition. The discriminant of the quadratic is negative so it will have two complex roots, which also satisfy the equation x³ = 1.

YES, I SEE WHAT YOU DID THERE.

[Anonymous No.3959715]

>>3959696
it isn't the multiplying by x

x = 1 becomes a solution when the substitution back in is made at x² + (-1 - 1/x) + 1 = 0, which is effectively multiplying both sides by (x-1)