5y^(1/2)+y^(1/4)-22=0

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Solution for 5y^(1/2)+y^(1/4)-22=0 equation: 


D( y )

y < 0

y < 0

y in <0:+oo)

5*y^(1/2)+y^(1/4)-22 = 0

t_1 = y^(1/4)

5*t_1^2+1*t_1^1-22 = 0

5*t_1^2+t_1-22 = 0

DELTA = 1^2-(-22*4*5)

DELTA = 441

DELTA > 0

t_1 = (441^(1/2)-1)/(2*5) or t_1 = (-441^(1/2)-1)/(2*5)

t_1 = 2 or t_1 = -11/5

t_1 = -11/5

y^(1/4)+11/5 = 0

1*y^(1/4) = -11/5 // : 1

y^(1/4) = -11/5

( -11/5 < 0 i 1/4 in (0:1) ) => y naleu017Cy do O

t_1 = 2

y^(1/4)-2 = 0

1*y^(1/4) = 2 // : 1

y^(1/4) = 2

y^(1/4) = 2 // ^ 4

y = 16

y = 16

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