Solving A Very Special Quartic | x^4+1=0

⭐ Join this channel to get access to perks:
#shorts #shortsvideo #shortsyoutube
via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

[MichaelRothwell]

Ferranti's method works well here, though this was an unexpected approach. The fourth roots of -1 are usually found in polar form using de Moivre's theorem as cis(kπ/4), k=1, 3, 5, 7, then converted to Cartesian form to give 1/√2(±1±i). But now I notice that Ferranti's method is actually quicker if you want the answers in Cartesian form.

[mcwulf]

I like your method. I started by putting (x^2)^2 = -1
x^2 = i
x = +/- √i = the answers you have.

[verdienthusiast]

Isn't this also a Sophie Germain identity? x^4 + 4((1/√2)^4)