Question 210323
{{{(x+ 1/x)^2 -2(x+ 1/x)+1=0}}} Start with the given equation.



Take note that the term {{{x+1/x}}} is being repeated here. So one way to simplify this equation (for now) is to do a simple substitution



So let {{{z=x+1/x}}}. So this means that {{{z^2=(x+1/x)^2}}}



{{{z^2-2z+1=0}}} Replace each instance of {{{x+1/x}}} with 'z'



Now we have a much simpler equation to solve for.



Notice that the quadratic {{{z^2-2z+1}}} is in the form of {{{Az^2+Bz+C}}} where {{{A=1}}}, {{{B=-2}}}, and {{{C=1}}}



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



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



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



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



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



{{{z = (2 +- sqrt( 4-4 ))/(2(1))}}} Multiply {{{4(1)(1)}}} to get {{{4}}}



{{{z = (2 +- sqrt( 0 ))/(2(1))}}} Subtract {{{4}}} from {{{4}}} to get {{{0}}}



{{{z = (2 +- sqrt( 0 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{z = (2 +- 0)/(2)}}} Take the square root of {{{0}}} to get {{{0}}}. 



{{{z = (2 + 0)/(2)}}} or {{{z = (2 - 0)/(2)}}} Break up the expression. 



{{{z = (2)/(2)}}} or {{{z =  (2)/(2)}}} Combine like terms. 



{{{z = 1}}} or {{{z = 1}}} Simplify. 



So the only solution in terms of 'z' is {{{z = 1}}} 



However, we want to solve for 'x' (not 'z'). But do recall that we let {{{z=x+1/x}}}



{{{z=x+1/x}}} Start with the given equation.



{{{1=x+1/x}}} Plug in the solution (in terms of 'z') {{{z=1}}}



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



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



Notice that the quadratic {{{x^2-x+1}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-1}}}, 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 = (-(-1) +- sqrt( (-1)^2-4(1)(1) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-1}}}, and {{{C=1}}}



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



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



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



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



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



{{{x = (1 +- i*sqrt(3))/(2)}}} Simplify the square root  



{{{x = (1+i*sqrt(3))/(2)}}} or {{{x = (1-i*sqrt(3))/(2)}}} Break up the expression.  



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Answer:



So the solutions (in terms of x) are {{{x = (1+i*sqrt(3))/(2)}}} or {{{x = (1-i*sqrt(3))/(2)}}}



Note: since the degree of {{{(x+ 1/x)^2 -2(x+ 1/x)+1}}} turns out to be 4, this means that there are really 4 solutions. So what happened? The twist is that each solution is counted twice. You'll most likely hear that these solutions are of multiplicity 2.