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#2 you are wrong.....(x^2+2)(x^2-1) is not x^4+x^2=1.....what FACTOR??? totally wrong...
you have x^4+x^2=1
1. you can solve the ecuation like this
x^2(x^2+1)=1 in wich we have the following possible results for x: x^2=1 meaning x=-1 or x=+1 and X^2+1=1 meaning x^2=0 so x=0.
2. you can replace x^2 with y and the ecuation becomes y^4+y^2=1 => y^4+y^2+1=0 and you solve it like a 2gr ecuation....
there is no real solution, cuz at some point you will either half to find the root of negative one, which is i, or the answer you get will not work when plugged into the equation
#2 rofl total failed, +/-1?? ;D 1^4 + 1^2 = 1? Math, it'f for the lose for you, and im not even talking about sqrt(2) cause its already bigger that 1, total failed
Comments 12
t1=0.618033988...; t2=-1.61803398 => x1=+0.7861513... x2=-0.7861513... x3=-1.272019....*i x4=+1.272019..*i
t^2 + t = 1
t^2 +t - 1 = 0
D = 1 + 4 = 5
t1,2 = -1 +- sqrt(5) / 2
x^2 = -1 +- sqrt(5) / 2
x = +- sqrt (-1 +- sqrt(5) / 2)
you have x^4+x^2=1
1. you can solve the ecuation like this
x^2(x^2+1)=1 in wich we have the following possible results for x: x^2=1 meaning x=-1 or x=+1 and X^2+1=1 meaning x^2=0 so x=0.
2. you can replace x^2 with y and the ecuation becomes y^4+y^2=1 => y^4+y^2+1=0 and you solve it like a 2gr ecuation....
Subtract 1 from each side =
x^4+x^2-1=0
FACTOR
(x^2+2)(x^2-1)=0
Solve each separately;
x^2+2=0
x^2=-2
x=+/-(i)sqrt(2)
x^2-1=0
x=+/-1
Present as single solution:
x=+/-(i)sqrt(2), +/-1
Math. It's for the win.
substitute y for x² -> quadratic equation
y = -1/2-sqrt(1/4-(-1))
x = sqrt(y) = sqrt((sqrt(5)-1)/2) = 0.786151378
easy. boring poster.