Question 591352

{{{x^2-4x+4=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-4x+4}}} is 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 only solution is {{{x = 2}}}