/sci/ - Science & Math

>>9522070
>you should be able to square both sides of an equation twice and find its roots
no shit

>>9522077
What happens when you square both sides a second time, dumbass?

>>9522085
today: OP learns about "LEFT INVERSE ADDITION"

>>9522085
you wait for one second to move 107 to the other side of the equation before squaring a second time, double dumbass

such an obvious step shouldn't need to be spelled out for you

>>9522100
such a step is so obvious it didn't need to be mentioned
I guess you need instructions to wipe your own ass though

>>9522108

>>9522085
x^2 = 107 + (107 + x)^1/2
x^2 - 107 = (107+x)^1/2
(x^2 - 107)(x^2-107) = (107 + x)
x^4 - 214x^2 + (107)^2 = 107 + x
x^4 - 214x^2 + 11449 = 107 + x
x^4 - 214x^2 +11342 = x
x^4 - 214x^2 - x + 11342 = 0

Find roots
Plug in to original equation see which one follows restrictions on radical

>>9522109
not interested in your family photos pal

>>9522116
this guy right here
this guy knows how to multiply

>>9522070
x=1/2+sqrt(429)/2

ez

>>9522116
>finding the roots to a quartic equation

good luck, lol!

>>9522133
hmmm

>>9522133
just graph it

>>9522139
thanks for doing my homework

>>9522153
none of those variables are defined for you.
you're not a very good troll

>>9522133
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.

>>9522164
It = 4

>>9522130
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?

>>9522223
also for the other roots of the quartic x = (+/-5*sqrt(17)-1)/2, how do we know these are not solutions without testing?

>>9522250
See >>9522228 for two roots, factor, find other two roots

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

>>9522252
but how do you know the other positive root isn't a solution without testing?

>>9522263
?
[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

>>9522272
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.

>>9522294
I'll check, but this seems pretty suspicious
If x satisfies the first, by application of the definition, it satisfies the second

>>9522304
yeah that's what's perplexing to me, I don't see how squaring would add a false real root like that

>>9522317
[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

>>9522294
>= (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 >>9522348
107*4+1 = 429

>>9522348
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

>>9522373
The left one is not even a root for the original equation

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

>>9522116
>>9522133
This is very simple and does not need quartic equations.

>>9522424
yeah no shit, my question is why not? the quartic equation there is what you get when you square each side

>>9522613
disregard this I'm a brainlet

>>9522133
>>9522546
>"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?