Question 180885
{{{(2x - 1 )^2 -4(2x -1)+ 2 =0}}} Start with the given equation.


Let {{{z=2x-1}}} (notice there are two terms that contain {{{2x-1}}})


{{{z^2 -4z+ 2 =0}}} Plug in {{{z=2x-1}}}



Notice we have a quadratic equation in the form of {{{az^2+bz+c}}} where {{{a=1}}}, {{{b=-4}}}, and {{{c=2}}}



Let's use the quadratic formula to solve for z



{{{z = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{z = (-(-4) +- sqrt( (-4)^2-4(1)(2) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-4}}}, and {{{c=2}}}



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



{{{z = (4 +- sqrt( 16-4(1)(2) ))/(2(1))}}} Square {{{-4}}} to get {{{16}}}. 



{{{z = (4 +- sqrt( 16-8 ))/(2(1))}}} Multiply {{{4(1)(2)}}} to get {{{8}}}



{{{z = (4 +- sqrt( 8 ))/(2(1))}}} Subtract {{{8}}} from {{{16}}} to get {{{8}}}



{{{z = (4 +- sqrt( 8 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{z = (4 +- 2*sqrt(2))/(2)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{z = (4)/(2) +- (2*sqrt(2))/(2)}}} Break up the fraction.  



{{{z = 2 +- 1*sqrt(2)}}} Reduce.  



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



So the answers (in terms of z) are {{{z = 2+sqrt(2)}}} or {{{z = 2-sqrt(2)}}} 



However, we need the solutions in terms of "x". Remember, we let {{{z=2x-1}}}, so...



{{{z = 2+sqrt(2)}}} or {{{z = 2-sqrt(2)}}} Go back to the solutions (in terms of "z")



{{{2x-1 = 2+sqrt(2)}}} or {{{2x-1 = 2-sqrt(2)}}} Plug in {{{z=2x-1}}}



{{{2x = 2+sqrt(2)+1}}} or {{{2x = 2-sqrt(2)+1}}} Add 1 to both sides (for each equation).



{{{2x = 3+sqrt(2)}}} or {{{2x = 3-sqrt(2)}}} Combine like terms.



{{{x = (3+sqrt(2))/2}}} or {{{x = (3-sqrt(2))/2}}} Divide both sides by 2 (for each equation) to isolate "x".



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Answer:



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