y=(5x-4x^2)^1/5

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Solution for y=(5x-4x^2)^1/5 equation:


x in (-oo:+oo)

y = ((5*x-(4*x^2))^1)/5 // - ((5*x-(4*x^2))^1)/5

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

y-((5*x-4*x^2)/5) = 0

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

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

5*y-1*(5*x-4*x^2) = 0

4*x^2-5*x+5*y = 0

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

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

4*x^2-5*x+5*y = 0

4*x^2-5*x+5*y = 0

DELTA = (-5)^2-(4*4*5*y)

DELTA = 25-80*y

25-80*y = 0

25-80*y = 0 // - 25

-80*y = -25 // : -80

y = -25/(-80)

y = 5/16

DELTA = 0 <=> t_3 = 5/16

x = 5/(2*4) i y = 5/16

x = 5/8 i y = 5/16

( x = ((25-80*y)^(1/2)+5)/(2*4) or x = (5-(25-80*y)^(1/2))/(2*4) ) i y > 5/16

( x = ((25-80*y)^(1/2)+5)/8 or x = (5-(25-80*y)^(1/2))/8 ) i y > 5/16

y-5/16 > 0

y-5/16 > 0 // + 5/16

y > 5/16

x in { 5/8, ((25-80*y)^(1/2)+5)/8, (5-(25-80*y)^(1/2))/8 }

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