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}
The logo has changed before I knew it.
Great video!
Thanks for sharing!
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.
I solved it the same way. Should we call it "completing the quartic"?
This channel is the best ngl
The logo changing was the biggest plot twist of the year 😭 i loved the old one though
The man and the methods strikes again!😊💥👌💪💯
Nice - I got the same answers!
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}
Se pongo x=tgθ, risulta facilmente l'equazione finale 1=sin4θ...4θ=π/2..θ=π/8..x=tg22,5=√2-1
problem
(1+x²)² = 4x(1-x²)
Let
y = 1+x
x = y-1
x² = (y-1)²
The equation becomes
[ 1+ (y-1)² ]² = 4 (y-1) [ 1- (y-1)² ]
= 4 (y-1) [ 1- (y-1)² ]
= 4 (y-1) [1- (y-1)] [1+ (y-1)]
= 4 y (y-1) (2 - y)
= -4y (y - 1) (y - 2)
[ y² -2y+2 ]² + 4y³ -12y² +8y = 0
y⁴-4y³+8y²-8y+4+4y³ -12y²+8y = 0
y⁴+8y²+4-12y² = 0
y⁴- 4 y² + 4 = 0
(y²-2) ² = 0
y² = 2
y = ± √2
x = y - 1
= ±√2 - 1
answer
x ∈ { -√2 -1, √2 -1 }