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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
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!
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) ,
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
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!
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) ,