An Interesting Polynomial Equation

[@NadiehFan]

Fifth method: (1 + x²)² = 4x(1 − x²) (x² + 1)² + 4x(x² − 1) = 0 (x² − 1)² + 4x(x² − 1) + 4x² = 0 (x² − 1 + 2x)² = 0

[@SyberMath]

Nice!

[@vacuumcarexpo]

The logo has changed before I knew it.

[@gregevgeni1864]

Great video! Thanks for sharing!

[@black_eagle]

The expanded equation is p(x) = x^4 + 4x^3 + 2x^2 - 4x + 1 = 0. Notice that (x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1, which has 3 terms in common. So p(x) = (x+1)^4 - 4x^2 - 8x = (x+1)^4 - 4x(x+2). Let u = x + 1: p(u) = u^4 - 4(u-1)(u+1) = u^4 - 4(u^2-1) = u^4 - 4u^2 + 4 = (u^2 - 2)^2 p(u) = 0 => u^2 = 2 => x+1 = +/- sqrt(2) => x = +/- sqrt(2) - 1.

[@stephenshefsky5201]

I solved it the same way. Should we call it "completing the quartic"?

[@SyberMath]

Very good!

[@BarchinoyG]

This channel is the best ngl

[@SyberMath]

Thank you! Plz spread the word! 😍

[@farhansadik5423]

The logo changing was the biggest plot twist of the year 😭 i loved the old one though

[@SyberMath]

That guy looks old! 😆😂

[@farhansadik5423]

@SyberMath yes but he looked cool, just as you would!

[@rachid6342]

How can i check if cos téta is differente than 0 ?

[@giuseppemalaguti435]

Se pongo x=tgθ, risulta facilmente l'equazione finale 1=sin4θ...4θ=π/2..θ=π/8..x=tg22,5=√2-1

[@scottleung9587]

Nice - I got the same answers!

[@SyberMath]

Awesome! 👍

[@scottleung9587]

@ yea I used a method similar to your first, except I used the logic that @black_eagle used

[@yoav613]

The man and the methods strikes again!😊💥👌💪💯

[@SyberMath]

🙌

[@dan-florinchereches4892]

Did you change the channel icon? I think letting u=x^2+1 will transform equation into u^2=4x(2-u) u^2+4xu-8x=0 Does not factor nicely so back to brute force if x=tan u 1/(cos^2u)^2=4sinu/cosu(cos^2u-sin^2u)/cos^2u 1=4sin u*cos u * (cos^2 u-sin^2 u) 1= 2sin (2u) * cos(2u) 1= sin(4u) => 4u €{ π/2+2kπ) u€{ π/8+kπ/2 | k€ Z} x=tan u € { tan(π/8), tan(5π/8) } Because the angle between the 2 tangents is 90 degrees when considering them slopes of lines then tan(5π/8)=-1/tan(π/8) tan(π/4)=2tan(π/8)/(1-tan^2(π/8)) If tanπ/8=t then 1=2t/(1-t^2) -t^2-2t+1=0 t1,2=(-2+-√(4+4))/2=-1+-√2 Since π/8 is in first quadrant we can accept tanπ/8=t=-1+√2 as solution So x €{-1+√2; -1-√2}

[@SyberMath]

Yess! Nice work!

[@MrGeorge1896]

x⁴ + 4x^3 + ... this is screaming really loud for (x + 1)⁴ isn't it? If we try it we get (x + 1)⁴ -4 (x + 1)² + 4 = 0 ( (x + 1)² - 2)² = 0 x + 1 = ± √2