Question 626499


{{{x^2-16x+64=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-16x+64}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-16}}}, and {{{C=64}}}



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



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



{{{x = (-(-16) +- sqrt( (-16)^2-4(1)(64) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-16}}}, and {{{C=64}}}



{{{x = (16 +- sqrt( (-16)^2-4(1)(64) ))/(2(1))}}} Negate {{{-16}}} to get {{{16}}}. 



{{{x = (16 +- sqrt( 256-4(1)(64) ))/(2(1))}}} Square {{{-16}}} to get {{{256}}}. 



{{{x = (16 +- sqrt( 256-256 ))/(2(1))}}} Multiply {{{4(1)(64)}}} to get {{{256}}}



{{{x = (16 +- sqrt( 0 ))/(2(1))}}} Subtract {{{256}}} from {{{256}}} to get {{{0}}}



{{{x = (16 +- sqrt( 0 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (16 +- 0)/(2)}}} Take the square root of {{{0}}} to get {{{0}}}. 



{{{x = (16 + 0)/(2)}}} or {{{x = (16 - 0)/(2)}}} Break up the expression. 



{{{x = (16)/(2)}}} or {{{x =  (16)/(2)}}} Combine like terms. 



{{{x = 8}}} or {{{x = 8}}} Simplify. 



So the only solution is {{{x = 8}}}



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