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.
@okkescinar14323 күн бұрын
Hocam bu kadar sık video atmayın. Hangi birini izleyeceğimi şaşırıyorum.
@martingibbs89723 күн бұрын
All that time, and you didn’t simplify your fractions?!
@despa77263 күн бұрын
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
@jeffw1267Күн бұрын
x does NOT equal 25.
@okkescinar14323 күн бұрын
Aslında soru hiç de zor değil.Sadece uzun işlemler var.
@okkescinar14323 күн бұрын
Hocam x³+ x² + x = 16 Bu denklemin çözümü gelsin. Çok uğraşmama rağmen ben çözemedim.
@itzrealzun3 күн бұрын
It has one irrational root and 2 complex, you cannot find it by default methods
@davidbrisbane72063 күн бұрын
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.