x^(4/3)-10x^(2/3)+25=0

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Solution for x^(4/3)-10x^(2/3)+25=0 equation: 


x in (-oo:+oo)

x^(4/3)-(10*x^(2/3))+25 = 0

x^(4/3)-10*x^(2/3)+25 = 0

t_1 = x^(2/3)

1*t_1^2-10*t_1^1+25 = 0

t_1^2-10*t_1+25 = 0

DELTA = (-10)^2-(1*4*25)

DELTA = 0

t_1 = 10/(1*2)

t_1 = 5 or t_1 = 5

t_1 = 5

x^(2/3)-5 = 0

1*x^(2/3) = 5 // : 1

x^(2/3) = 5

x^(2/3) = 5 // ^ 3

x^2 = 125 // ^ 1/2

abs(x) = 5*5^(1/2)

x = 5*5^(1/2) or x = -(5*5^(1/2))

t_1 = 5

x^(2/3)-5 = 0

1*x^(2/3) = 5 // : 1

x^(2/3) = 5

x^(2/3) = 5 // ^ 3

x^2 = 125 // ^ 1/2

abs(x) = 5*5^(1/2)

x = 5*5^(1/2) or x = -(5*5^(1/2))

x in { 5*5^(1/2), -(5*5^(1/2)), 5*5^(1/2), -(5*5^(1/2)) }

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