4((1/x^2)-(1/2^2))=16

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Solution for 4((1/x^2)-(1/2^2))=16 equation: 


D( x )

x^2 = 0

x^2 = 0

x^2 = 0

1*x^2 = 0 // : 1

x^2 = 0

x = 0

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

4*(1/(x^2)-(1/(2^2))) = 16 // - 16

4*(1/(x^2)-(1/(2^2)))-16 = 0

4*(1/(x^2)-2^-2)-16 = 0

4*(x^-2-1/4)-16 = 0

4*(x^-2-1/4)-16 = 0

4*x^-2-17 = 0

4*x^-2-17 = 0

4*x^-2 = 17 // : 4

x^-2 = 17/4

-2 < 0

1/(x^2) = 17/4 // * x^2

1 = 17/4*x^2 // : 17/4

4/17 = x^2

x^2 = 4/17 // ^ 1/2

abs(x) = (4/17)^(1/2)

x = (4/17)^(1/2) or x = -(4/17)^(1/2)

x in { (4/17)^(1/2), -(4/17)^(1/2) }

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