(32y^6-20y^4)/(4y^3)

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Solution for (32y^6-20y^4)/(4y^3) equation:


D( y )

4*y^3 = 0

4*y^3 = 0

4*y^3 = 0

4*y^3 = 0 // : 4

y^3 = 0

y = 0

y in (-oo:0) U (0:+oo)

(32*y^6-(20*y^4))/(4*y^3) = 0

(32*y^6-20*y^4)/(4*y^3) = 0

32*y^6-20*y^4 = 0

4*y^4*(8*y^2-5) = 0

8*y^2 = 5 // : 8

y^2 = 5/8

y^2 = 5/8 // ^ 1/2

abs(y) = (5/8)^(1/2)

y = (5/8)^(1/2) or y = -(5/8)^(1/2)

4*y^4*(y-(5/8)^(1/2))*(y+(5/8)^(1/2)) = 0

(4*y^4*(y-(5/8)^(1/2))*(y+(5/8)^(1/2)))/(4*y^3) = 0

( 4*y^4 )

4*y^4 = 0 // : 4

y^4 = 0

y = 0

( y+(5/8)^(1/2) )

y+(5/8)^(1/2) = 0 // - (5/8)^(1/2)

y = -(5/8)^(1/2)

( y-(5/8)^(1/2) )

y-(5/8)^(1/2) = 0 // + (5/8)^(1/2)

y = (5/8)^(1/2)

y in { 0}

y in { -(5/8)^(1/2), (5/8)^(1/2) }

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