The real solution is pretty easy to get because, when you factor t cubed minus t, it breaks down to (t-1)(t)(t+1), and 4, 5, and 6 are the highest set of three consecutive factors of 120.
You can express x³ + x² + x - 16 = 0 as a depressed cubic equation and you get (1 + x/3)³ + (2/3)(1 + x/3) - 439/27 = 0. You do this by letting u = 1 + x/3 in this case as the coefficient of x² is 1. If it were 2, then you'd let u = x + 2/3. So, u³ + (2/3)u - 439/27 = 0, which is the depressed cubic equation. Now you can use Cardano's cube root formula to find a real root of u. The other two solutions can be found by either multiplying the real root by the three roots of unity ( i.e. 1, ω, ω²), or by using synthetic division to find the depressed quadratic polynomial, but it's best to use the cube root of unity approach. Once you have found all three roots in terms of u, then it is easy to find them in terms of x.
The real solution is pretty easy to get because, when you factor t cubed minus t, it breaks down to (t-1)(t)(t+1), and 4, 5, and 6 are the highest set of three consecutive factors of 120.
sqrt(x^3) - sqrt(x) = 120
x^3 - 2sqrt(x^3)sqrt(x) + x = 14400
x^3 - 2sqrt(x^3 * x) + x = 14400
x^3 - 2sqrt(x^4) + x = 14400
x^3 - 2x^2 + x = 14400
(1)x^3 + (-2)x^2 + (1)x + (-14400) = 0
I cheated and put this through a cubic equation calculator and got the following answers:
x1 = -11,5 + 21,06537i
x2 = -11,5 - 21,06537i
x3 = 25
Hocam x³+ x² + x = 16
Bu denklemin çözümü gelsin. Çok uğraşmama rağmen ben çözemedim.
It has one irrational root and 2 complex, you cannot find it by default methods
You can express x³ + x² + x - 16 = 0 as a depressed cubic equation and you get
(1 + x/3)³ + (2/3)(1 + x/3) - 439/27 = 0.
You do this by letting u = 1 + x/3 in this case as the coefficient of x² is 1. If it were 2, then you'd let u = x + 2/3.
So, u³ + (2/3)u - 439/27 = 0, which is the depressed cubic equation.
Now you can use Cardano's cube root formula to find a real root of u.
The other two solutions can be found by either multiplying the real root by the three roots of unity ( i.e. 1, ω, ω²), or by using synthetic division to find the depressed quadratic polynomial, but it's best to use the cube root of unity approach.
Once you have found all three roots in terms of u, then it is easy to find them in terms of x.
Hocam bu kadar sık video atmayın. Hangi birini izleyeceğimi şaşırıyorum.
All that time, and you didn’t simplify your fractions?!
Aslında soru hiç de zor değil.Sadece uzun işlemler var.
x does NOT equal 25.
Why not?
25³ = 5⁶ = 125² = 15625, so
√15625 = 125,
√25 = 5,
125 - 5 = 120.
let u=x^(1/2) , -> u^3 |+/-| n*u^2-u-125=0 , by faktoring , / trial , 120=5*24 , n=5 / , u^3-5u^2 +5u^2-25u +24u-120=0 ,
u^2(u-5)+5u(u-5)+24(u-5)-120=0 , (u-5)(u^2+5u+24)=0 , u-5=0 , u=5 , u=x(1/2) x=u^2 , u^2=25 , x=25 ,
/// for complex , u^2+5u+24=0 /// , test , 25^(3/2)-25^(1/2)=125-5 , 125-5=120 , OK ,
(-23+5*√71*i)/2