You should be able to solve this

x=1/2+sqrt(429)/2

ez

>finding the roots to a quartic equation

good luck, lol!

hmmm

just graph it

thanks for doing my homework

none of those variables are defined for you.
you're not a very good troll

Plug and chug.

x = (-3*b+s*(sqrt(3*(3*b^2-8*a*c+2*a*cbrt(4*(2*c^3-9*b*c*d+27*a*d^2+27*b^2*e-72*a*c*e+sqrt((2*c^3-9*b*c*d+27*a*d^2+27*b^2*e-72*a*c*e)^2-4*(c^2-3*b*d+12*a*e)^3)))+2*a*cbrt(4*(2*c^3-9*b*c*d+27*a*d^2+27*b^2*e-72*a*c*e-sqrt((2*c^3-9*b*c*d+27*a*d^2+27*b^2*e-72*a*c*e)^2-4*(c^2-3*b*d+12*a*e)^3)))))+t*sqrt(3*(3*b^2-8*a*c+2*a*(-1+sqrt(-3))/2*cbrt(4*(2*c^3-9*b*c*d+27*a*d^2+27*b^2*e-72*a*c*e+sqrt((2*c^3-9*b*c*d+27*a*d^2+27*b^2*e-72*a*c*e)^2-4*(c^2-3*b*d+12*a*e)^3)))+2*a*(-1-sqrt(-3))/2*cbrt(4*(2*c^3-9*b*c*d+27*a*d^2+27*b^2*e-72*a*c*e-sqrt((2*c^3-9*b*c*d+27*a*d^2+27*b^2*e-72*a*c*e)^2-4*(c^2-3*b*d+12*a*e)^3))))))+t*sgn((sgn(-b^3+4*a*b*c-8*a^2*d)-1/2)*(sgn(max((2*c^3-9*b*c*d+27*a*d^2+27*b^2*e-72*a*c*e)^2-4*(c^2-3*b*d+12*a*e)^3,min(3*b^2-8*a*c,3*b^4+16*a^2*c^2+16*a^2*b*d-16*a*b^2*c-64*a^3*e)))-1/2))*sqrt(3*(3*b^2-8*a*c+2*a*(-1-sqrt(-3))/2*cbrt(4*(2*c^3-9*b*c*d+27*a*d^2+27*b^2*e-72*a*c*e+sqrt((2*c^3-9*b*c*d+27*a*d^2+27*b^2*e-72*a*c*e)^2-4*(c^2-3*b*d+12*a*e)^3)))+2*a*(-1+sqrt(-3))/2*cbrt(4*(2*c^3-9*b*c*d+27*a*d^2+27*b^2*e-72*a*c*e-sqrt((2*c^3-9*b*c*d+27*a*d^2+27*b^2*e-72*a*c*e)^2-4*(c^2-3*b*d+12*a*e)^3))))))/(12*a)

Directions: Take s=-1,1 and t=-1,1. Use real cube roots if possible, and principal roots otherwise.

It = 4

yeah it's this
is there a way we know there aren't any other roots? is there a way we know 1/2(1-sqrt(429)) isn't a solution other than testing?

...

also for the other roots of the quartic x = (+/-5*sqrt(17)-1)/2, how do we know these are not solutions without testing?

See for two roots, factor, find other two roots

[math]x=\frac{1}{2}+\frac{\sqrt{429}}{2}[/math]

but how do you know the other positive root isn't a solution without testing?

?
[eqn] \sqrt{107 + x} = x [/eqn] has two roots in the complex plain. Obviously these two roots are shared with [eqn] \sqrt{107+\sqrt{107+x}} = x [/eqn]
Factor this second equation into a binomial and find the other two roots in the complex plain

Not sure about this positive root negative root stuff. I'm just a simple minded bumpkin

the second real root doesn't solve the equation. If you substitute x = (5*sqrt(17)-1)/2 into sqrt(107+sqrt(107+x)) - x, you get 1.

I'll check, but this seems pretty suspicious
If x satisfies the first, by application of the definition, it satisfies the second

yeah that's what's perplexing to me, I don't see how squaring would add a false real root like that

[eqn] \sqrt{a + x} = x \implies x = \frac{1}{2}\Big\{1\pm\sqrt{1+4a}\Big\}[/eqn]
Note squaring this gives
[eqn] x^2 = \frac{1}{2}\Big\{2a+1\pm\sqrt{1+4a}\Big\} [/eqn]
Substituting [math] x [/math]
[eqn] \sqrt{a+\sqrt{a+x}} = \sqrt{a+\sqrt{\frac{1}{2}\Big\{2a+1\pm\sqrt{1+4a}\Big\}}} \\
= \sqrt{a+\sqrt{\Big\{\frac{1}{2}\pm\frac{1}{2}\sqrt{1+4a}\Big\}^2}} [/eqn]

Wait a minute, why are you using [math] x = (5*\sqrt(17)-1)/2 [/math]
It's suppose to be the negative of that

>= (5*sqrt(17)-1)/2 into sqrt(107+sqrt(107+x)) - x, you get
I don't even think the negative of that is a root 107*4+1 = 429

if you just use the original equation and evaluate the quartic, you get the expression
(107+x)=(x^2-107)^2
x^4-2*107*x^2-x+107*106=0
(x^2+x-106)(x^2-x-107)=0
x=(+/-5sqrt(17)-1)/2,(+/-sqrt(429)+1)/2
but only one of the positive roots is consistent with the original equation

The left one is not even a root for the original equation

I can’t believe you guys waste your time doing these

This is very simple and does not need quartic equations.

yeah no shit, my question is why not? the quartic equation there is what you get when you square each side

disregard this I'm a brainlet

>"The notes left by Évariste Galois prior to dying in a duel in 1832 later led to an elegant complete theory of the roots of polynomials, of which this theorem was one result."

You dare to look down on this level of badassery?