Question 422034
{{{1 + 1/x^2 = 3/x}}} Start with the given equation.



{{{x^2 + 1 = 3x}}} Multiply EVERY term by the LCD {{{x^2}}} to clear out the fractions.



{{{x^2-3x+1=0}}} Subtract 3x from both sides.



Notice that the quadratic {{{x^2-3x+1}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-3}}}, and {{{C=1}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-3) +- sqrt( (-3)^2-4(1)(1) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-3}}}, and {{{C=1}}}



{{{x = (3 +- sqrt( (-3)^2-4(1)(1) ))/(2(1))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{x = (3 +- sqrt( 9-4(1)(1) ))/(2(1))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9-4 ))/(2(1))}}} Multiply {{{4(1)(1)}}} to get {{{4}}}



{{{x = (3 +- sqrt( 5 ))/(2(1))}}} Subtract {{{4}}} from {{{9}}} to get {{{5}}}



{{{x = (3 +- sqrt( 5 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (3+sqrt(5))/(2)}}} or {{{x = (3-sqrt(5))/(2)}}} Break up the expression.  



So the solutions are {{{x = (3+sqrt(5))/(2)}}} or {{{x = (3-sqrt(5))/(2)}}} 



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