(1-3x)^2=9x^2-7/x

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Solution for (1-3x)^2=9x^2-7/x equation:


D( x )

x = 0

x = 0

x = 0

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

(1-(3*x))^2 = 9*x^2-(7/x) // - 9*x^2-(7/x)

(1-(3*x))^2-(9*x^2)+7/x = 0

(1-3*x)^2-9*x^2+7/x = 0

(x*(1-3*x)^2)/x+(-9*x*x^2)/x+7/x = 0

x*(1-3*x)^2-9*x*x^2+7 = 0

x-6*x^2+7 = 0

x-6*x^2+7 = 0

x-6*x^2+7 = 0

DELTA = 1^2-(-6*4*7)

DELTA = 169

DELTA > 0

x = (169^(1/2)-1)/(-6*2) or x = (-169^(1/2)-1)/(-6*2)

x = -1 or x = 7/6

(x+1)*(x-7/6) = 0

((x+1)*(x-7/6))/x = 0

((x+1)*(x-7/6))/x = 0 // * x

(x+1)*(x-7/6) = 0

( x+1 )

x+1 = 0 // - 1

x = -1

( x-7/6 )

x-7/6 = 0 // + 7/6

x = 7/6

x in { -1, 7/6 }

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