Question 388773


{{{x^2-4x+6}}} Start with the given expression.



Take half of the {{{x}}} coefficient {{{-4}}} to get {{{-2}}}. In other words, {{{(1/2)(-4)=-2}}}.



Now square {{{-2}}} to get {{{4}}}. In other words, {{{(-2)^2=(-2)(-2)=4}}}



{{{x^2-4x+highlight(4-4)+6}}} Now add <font size=4><b>and</b></font> subtract {{{4}}}. Make sure to place this after the "x" term. Notice how {{{4-4=0}}}. So the expression is not changed.



{{{(x^2-4x+4)-4+6}}} Group the first three terms.



{{{(x-2)^2-4+6}}} Factor {{{x^2-4x+4}}} to get {{{(x-2)^2}}}.



{{{(x-2)^2+2}}} Combine like terms.



So after completing the square, {{{x^2-4x+6}}} transforms to {{{(x-2)^2+2}}}. So {{{x^2-4x+6=(x-2)^2+2}}}.



So {{{x^2-4x+6=0}}} is equivalent to {{{(x-2)^2+2=0}}}.



Now let's solve {{{(x-2)^2+2=0}}}



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



{{{(x-2)^2=0-2}}}Subtract {{{2}}} from both sides.



{{{(x-2)^2=-2}}} Combine like terms.



{{{x-2=""+-sqrt(-2)}}} Take the square root of both sides.



{{{x-2=sqrt(-2)}}} or {{{x-2=-sqrt(-2)}}} Break up the "plus/minus" to form two equations.



{{{x-2=i*sqrt(2)}}} or {{{x-2=-i*sqrt(2)}}}  Simplify the square root.



{{{x=2+i*sqrt(2)}}} or {{{x=2-i*sqrt(2)}}} Add {{{2}}} to both sides.



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Answer:



So the solutions are {{{x=2+i*sqrt(2)}}} or {{{x=2-i*sqrt(2)}}}.



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