Question 174884

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



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



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



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



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



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



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



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



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



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



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



{{{z = 2}}} or {{{z = 2}}} Simplify. 



So the solution is {{{z=2}}} (with a multiplicity of 2)