I found out how to solve this quartic equation

[@_P_a_o_l_o_]

Really cool technique!

[@JPiMaths]

@@_P_a_o_l_o_ thank you!

[@alphazero339]

Interesting problem thanks for the video. Another way to think about it is iterated function: if 𝙛(𝙭) = 𝙭² - 5 then the equation becomes 𝙛(𝙛(𝙭)) = x so in other words we are looking for when 𝙛(𝙭) is equal to 𝙭, which yields the equation you used 𝙭² - 5 = 𝙭. You encountered the form 𝙛(𝙭) = 𝙖𝙭² + 𝙘, I tried to generalize it to 𝙛(𝙭) = 𝙖𝙭² + 𝙗𝙭 + 𝙘 and after some expanding what I got is that the equation (𝙖³)𝙭⁴ + (2𝙖²𝙗)𝙭³ + (2𝙖²𝙘 + 𝙖𝙗² + 𝙖𝙗)𝙭² + (2𝙖𝙗𝙘 + 𝙗² - 1)𝙭 + 𝙖𝙘² + 𝙗𝙘 + 𝙘 = 0 will have (if there are any) solutions of 𝙖𝙭² + (𝙗-1)𝙭 + 𝙘 = 0. (Note the -1 appearing in coefficient of 𝙭¹ in both equations since the original form was 𝙛(𝙛(𝙭)) = 𝙭 and we moved the 𝙭 from RHS to LHS). If I hadn't made any silly mistake in expanding it should be correct, but I didn't test it

[@JPiMaths]

Interesting approach. It should be noted that f(f(x))=x doesn't imply that f(x)=x (e.g. consider the function f(x)=-x)

[@alphazero339]

@JPiMaths thanks that's true, I probably used incorrect notation. If I switch RHS and LHS it makes more sense: 𝙭 = 𝙛(𝙛(𝙭)), 𝙭∈ℝ implies 𝙭 = 𝙛(𝙭). (The f(x)=x isn't supposed to mean definition but that it's true at least for x)