Question 169989


{{{a^2-4a+5}}} Start with the left side of the equation.



Take half of the {{{a}}} 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}}}



{{{a^2-4a+highlight(4-4)+5}}} Now add <font size=4><b>and</b></font> subtract {{{4}}}. Make sure to place this after the "a" term. Notice how {{{4-4=0}}}. So the expression is not changed.



{{{(a^2-4a+4)-4+5}}} Group the first three terms.



{{{(a-2)^2-4+5}}} Factor {{{a^2-4a+4}}} to get {{{(a-2)^2}}}.



{{{(a-2)^2+1}}} Combine like terms.



So after completing the square, {{{a^2-4a+5}}} transforms to {{{(a-2)^2+1}}}. So {{{a^2-4a+5=(a-2)^2+1}}}.



So {{{a^2-4a+5=0}}} is equivalent to {{{(a-2)^2+1=0}}}.



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{{{(a-2)^2+1=0}}} Start with the given equation.



{{{(a-2)^2=0-1}}}Subtract {{{1}}} from both sides.



{{{(a-2)^2=-1}}} Combine like terms.



{{{x-2=0+-sqrt(-1)}}} Take the square root of both sides.



{{{a-2=sqrt(-1)}}} or {{{a-2=-sqrt(-1)}}} Break up the "plus/minus" to form two equations.



{{{a-2=i}}} or {{{a-2=-i}}} Replace {{{sqrt(-1)}}} with "i". Remember, {{{i=sqrt(-1)}}}



{{{a=2+i}}} or {{{a=2-i}}} Add {{{2}}} to both sides.



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Answer:



So the solutions are {{{a=2+i}}} or {{{a=2-i}}}.