(2x)^2/(1-x)^2=40

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Solution for (2x)^2/(1-x)^2=40 equation: 


D( x )

(1-x)^2 = 0

(1-x)^2 = 0

(1-x)^2 = 0

1-x = 0 // - 1

-x = -1 // * -1

x = 1

x in (-oo:1) U (1:+oo)

((2*x)^2)/((1-x)^2) = 40 // - 40

((2*x)^2)/((1-x)^2)-40 = 0

(4*x^2)/((1-x)^2)-40 = 0

(4*x^2)/((1-x)^2)+(-40*(1-x)^2)/((1-x)^2) = 0

4*x^2-40*(1-x)^2 = 0

80*x-36*x^2-40 = 0

80*x-36*x^2-40 = 0

4*(20*x-9*x^2-10) = 0

20*x-9*x^2-10 = 0

DELTA = 20^2-(-10*(-9)*4)

DELTA = 40

DELTA > 0

x = (40^(1/2)-20)/(-9*2) or x = (-40^(1/2)-20)/(-9*2)

x = (2*10^(1/2)-20)/(-18) or x = (-2*10^(1/2)-20)/(-18)

4*(x-((2*10^(1/2)-20)/(-18)))*(x-((-2*10^(1/2)-20)/(-18))) = 0

(4*(x-((2*10^(1/2)-20)/(-18)))*(x-((-2*10^(1/2)-20)/(-18))))/((1-x)^2) = 0

(4*(x-((2*10^(1/2)-20)/(-18)))*(x-((-2*10^(1/2)-20)/(-18))))/((1-x)^2) = 0 // * (1-x)^2

4*(x-((2*10^(1/2)-20)/(-18)))*(x-((-2*10^(1/2)-20)/(-18))) = 0

( 4 )

4 = 0

x belongs to the empty set

( x-((-2*10^(1/2)-20)/(-18)) )

x-((-2*10^(1/2)-20)/(-18)) = 0 // + (-2*10^(1/2)-20)/(-18)

x = (-2*10^(1/2)-20)/(-18)

( x-((2*10^(1/2)-20)/(-18)) )

x-((2*10^(1/2)-20)/(-18)) = 0 // + (2*10^(1/2)-20)/(-18)

x = (2*10^(1/2)-20)/(-18)

x in { (-2*10^(1/2)-20)/(-18), (2*10^(1/2)-20)/(-18) }

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