3x^(1/2)+7x^(1/4)-20=0

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Solution for 3x^(1/2)+7x^(1/4)-20=0 equation: 


D( x )

x < 0

x < 0

x in <0:+oo)

3*x^(1/2)+7*x^(1/4)-20 = 0

t_1 = x^(1/4)

3*t_1^2+7*t_1^1-20 = 0

3*t_1^2+7*t_1-20 = 0

DELTA = 7^2-(-20*3*4)

DELTA = 289

DELTA > 0

t_1 = (289^(1/2)-7)/(2*3) or t_1 = (-289^(1/2)-7)/(2*3)

t_1 = 5/3 or t_1 = -4

t_1 = -4

x^(1/4)+4 = 0

1*x^(1/4) = -4 // : 1

x^(1/4) = -4

( -4 < 0 i 1/4 in (0:1) ) => x naleu017Cy do O

t_1 = 5/3

x^(1/4)-5/3 = 0

1*x^(1/4) = 5/3 // : 1

x^(1/4) = 5/3

x^(1/4) = 5/3 // ^ 4

x = 625/81

x = 625/81

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