Question 187842
{{{y^4-4y^2+4=0}}} Start with the given equation.



Let {{{x=y^2}}}. This means that {{{x^2=y^4}}}



{{{x^2-4x+4=0}}} Make the proper substitutions 



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



Let's use the quadratic formula to solve for x



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



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



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



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



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



{{{x = (4 +- sqrt( 0 ))/(2(1))}}} Subtract {{{16}}} from {{{16}}} to get {{{0}}}



{{{x = (4 +- sqrt( 0 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (4 +- 0)/(2)}}} Take the square root of {{{0}}} to get {{{0}}}. 



{{{x = (4 + 0)/(2)}}} or {{{x = (4 - 0)/(2)}}} Break up the expression. 



{{{x = (4)/(2)}}} or {{{x =  (4)/(2)}}} Combine like terms. 



{{{x = 2}}} or {{{x = 2}}} Simplify. 



So the solution is {{{x=2}}}



However, we want to solve for "y"



{{{x=y^2}}} Go back to the second equation.

  
  
{{{2=y^2}}} Plug in {{{x=2}}}



{{{y^2=2}}} Rearrange the equation



{{{y=sqrt(2)}}} or {{{y=-sqrt(2)}}} Take the square root of both sides (don't forget the "plus/minus")



So the solutions are {{{y=sqrt(2)}}} or {{{y=-sqrt(2)}}}