(x^2+2x+1)/(1-x)=1.05

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Solution for (x^2+2x+1)/(1-x)=1.05 equation: 


D( x )

1-x = 0

1-x = 0

1-x = 0

1-x = 0 // - 1

-x = -1 // * -1

x = 1

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

(x^2+2*x+1)/(1-x) = 1.05 // - 1.05

(x^2+2*x+1)/(1-x)-1.05 = 0

(x^2+2*x+1)/(1-x)+(-1.05*(1-x))/(1-x) = 0

x^2-1.05*(1-x)+2*x+1 = 0

x^2+3.05*x-0.05 = 0

x^2+3.05*x-0.05 = 0

x^2+3.05*x-0.05 = 0

DELTA = 3.05^2-(-0.05*1*4)

DELTA = 3.05^2+0.2

DELTA = 9.5025

DELTA > 0

x = ((3.05^2+0.2)^(1/2)-3.05)/(1*2) or x = (-(3.05^2+0.2)^(1/2)-3.05)/(1*2)

x = ((3.05^2+0.2)^(1/2)-3.05)/2 or x = (-((3.05^2+0.2)^(1/2)+3.05))/2

(x+((3.05^2+0.2)^(1/2)+3.05)/2)*(x-(((3.05^2+0.2)^(1/2)-3.05)/2)) = 0

((x+((3.05^2+0.2)^(1/2)+3.05)/2)*(x-(((3.05^2+0.2)^(1/2)-3.05)/2)))/(1-x) = 0

((x+((3.05^2+0.2)^(1/2)+3.05)/2)*(x-(((3.05^2+0.2)^(1/2)-3.05)/2)))/(1-x) = 0 // * 1-x

(x+((3.05^2+0.2)^(1/2)+3.05)/2)*(x-(((3.05^2+0.2)^(1/2)-3.05)/2)) = 0

( x+((3.05^2+0.2)^(1/2)+3.05)/2 )

x+((3.05^2+0.2)^(1/2)+3.05)/2 = 0 // - ((3.05^2+0.2)^(1/2)+3.05)/2

x = -(((3.05^2+0.2)^(1/2)+3.05)/2)

( x-(((3.05^2+0.2)^(1/2)-3.05)/2) )

x-(((3.05^2+0.2)^(1/2)-3.05)/2) = 0 // + ((3.05^2+0.2)^(1/2)-3.05)/2

x = ((3.05^2+0.2)^(1/2)-3.05)/2

x in { -(((3.05^2+0.2)^(1/2)+3.05)/2), ((3.05^2+0.2)^(1/2)-3.05)/2 }

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