Ako riešite x ^ (2/3) - 3x ^ (1/3) - 4 = 0?

Ako riešite x ^ (2/3) - 3x ^ (1/3) - 4 = 0?
Anonim

odpoveď:

nastaviť Z = x ^ (1/3) Keď nájdete Z korene, nájsť X = z ^ 3

Korene sú 729/8 a -1/8

vysvetlenie:

nastaviť X ^ (1/3) = z

X ^ (2/3) = x ^ (1/3 * 2) = (x ^ (1/3)) ^ 2 = z ^ 2

Tak sa rovnica stáva:

Z ^ 2-3z-4 = 0

Δ = b ^ 2-4ac

Δ=(-3)^2-4*1*(-4)

Δ=25

z_ (1,2) = (- b + -sqrt (Δ)) / (2a)

z_ (1,2) = (- (- 4) + - sqrt (25)) / (2 * 1)

z_ (1,2) = (4 + -5) / 2

Z_1 = 9/2

Z_2 = -1/2

Riešiť X:

X ^ (1/3) = z

(X ^ (1/3)) ^ 3 = z ^ 3

X = z ^ 3

X 1 = (9/2) ^ 3

X 1 = 729/8

X_2 = (- 1/2) ^ 3

X_2 = -1/8

odpoveď:

x = 64 alebo x = -1

vysvetlenie:

poznač si to (x ^ (1/3)) ^ 2 = x ^ (2/3)

Factorising x ^ (2/3) - 3x ^ (1/3) - 4 = 0 dáva;

(x ^ (1/3) - 4) (x ^ (http: // 3) + 1) = 0

rArr (x ^ (1/3) - 4) = 0 alebo (x ^ (1/3) + 1) = 0

rArr x ^ (1/3) = 4 alebo x ^ (1/3) = - 1

„kockovanie“ oboch strán dvojice rovníc:

(x ^ (1/3)) ^ 3 = 4 ^ 3 a (x ^ (1/3)) ^ 3 = (- 1) ^ 3

rArr x = 64 alebo x = - 1