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Why not simply multiply the equation by x^2(x-1)^2 to obtain x^4 + 4 x^2 = 1? From which one finds x^2 = -2 ± sqrt(5) directly.
Because then you have a sub five minute video, not 27. Imagine setting 10 of these to be done in under an hour. This guy would be up til 3am!
Got the same four solutions but did it by using the difference of two squares which led to the equation x^4+ 4(x^2)- 1= o etc.
Nice
Yo reduje y llegué a una ecuación bicuadrada x^4 +4x^2-1=0 y obtuve 2 soluciones reales y 2 soluciones complejas 😊😊
I said a =(x+1)/x and b = (x+1)/(x-1) and that a^2-b^2 =1 and that a/b + a =2. Seems easier to solve is anyone with me?
Had to say for the solution of b: (b^2+1/b^2)+2(b+1/b)-2=0 make b+1/b = z and then solve for z^2-2z-4=0. Gets tough afterwards.
It requires patience and experience 😍💕🙏
(x+1)² [1/x² -1/(x-1)²] =1 difference of 2 squares(x+1)² [(1/x -1/(x-1)] [1/x + 1/(x-1)]=1 unite bases(x+1)² [(x-1-x) /(x(x-1))] [(x-1+x/(x(x-1))] =1 multiply both sides by (x(x-1))²(x²+2x+1) (-1) (2x-1) = x²(x² -2x+1) disterbutex⁴ +4x² -1= 0 x² =-2±√5x= ±√(2±√5)
{x^2+2}/x^2 ➖{ x^2+2}/(x^2 ➖ 2)=2x^2/x^2 ➖ 2x^2/{x^0+x^0➖ }={2x^2/x^2 ➖ 2x^2/x^1}={0+0 ➖ }x^{0+x^0 ➖}/x^1=1x^1/x^1=1x^1 (x ➖ 1x+1).
let x^2=u , u^2+4u-1 , u=(-4+/-V(16+4))/2 , u=-2+/-V5 , x^2=-2+/-V5 , x= +V(-2+V5) , -V(-2+V5) , i*V(2+V5) , -i*V(2+V5) ,
Why not simply multiply the equation by x^2(x-1)^2 to obtain x^4 + 4 x^2 = 1? From which one finds x^2 = -2 ± sqrt(5) directly.
Because then you have a sub five minute video, not 27. Imagine setting 10 of these to be done in under an hour. This guy would be up til 3am!
Got the same four solutions but did it by using the difference of two squares which led to the equation x^4+ 4(x^2)- 1= o etc.
Nice
Yo reduje y llegué a una ecuación bicuadrada x^4 +4x^2-1=0 y obtuve 2 soluciones reales y 2 soluciones complejas 😊😊
I said a =(x+1)/x and b = (x+1)/(x-1) and that a^2-b^2 =1 and that a/b + a =2. Seems easier to solve is anyone with me?
Had to say for the solution of b: (b^2+1/b^2)+2(b+1/b)-2=0 make b+1/b = z and then solve for z^2-2z-4=0. Gets tough afterwards.
It requires patience and experience 😍💕🙏
(x+1)² [1/x² -1/(x-1)²] =1 difference of 2 squares
(x+1)² [(1/x -1/(x-1)] [1/x + 1/(x-1)]=1 unite bases
(x+1)² [(x-1-x) /(x(x-1))] [(x-1+x/(x(x-1))] =1 multiply both sides by (x(x-1))²
(x²+2x+1) (-1) (2x-1) = x²(x² -2x+1) disterbute
x⁴ +4x² -1= 0
x² =-2±√5
x= ±√(2±√5)
{x^2+2}/x^2 ➖{ x^2+2}/(x^2 ➖ 2)=2x^2/x^2 ➖ 2x^2/{x^0+x^0➖ }={2x^2/x^2 ➖ 2x^2/x^1}={0+0 ➖ }x^{0+x^0 ➖}/x^1=1x^1/x^1=1x^1 (x ➖ 1x+1).
let x^2=u , u^2+4u-1 , u=(-4+/-V(16+4))/2 , u=-2+/-V5 , x^2=-2+/-V5 , x= +V(-2+V5) , -V(-2+V5) , i*V(2+V5) , -i*V(2+V5) ,