Question 260559

{{{2x^2-6x+5}}} Start with the given expression.



{{{2(x^2-3x+5/2)}}} Factor out the {{{x^2}}} coefficient {{{2}}}. This step is very important: the {{{x^2}}} coefficient <font size=4><b>must</b></font> be equal to 1.



Take half of the {{{x}}} coefficient {{{-3}}} to get {{{-3/2}}}. In other words, {{{(1/2)(-3)=-3/2}}}.



Now square {{{-3/2}}} to get {{{9/4}}}. In other words, {{{(-3/2)^2=(-3/2)(-3/2)=9/4}}}



{{{2(x^2-3x+highlight(9/4-9/4)+5/2)}}} Now add <font size=4><b>and</b></font> subtract {{{9/4}}} inside the parenthesis. Make sure to place this after the "x" term. Notice how {{{9/4-9/4=0}}}. So the expression is not changed.



{{{2((x^2-3x+9/4)-9/4+5/2)}}} Group the first three terms.



{{{2((x-3/2)^2-9/4+5/2)}}} Factor {{{x^2-3x+9/4}}} to get {{{(x-3/2)^2}}}.



{{{2((x-3/2)^2+1/4)}}} Combine like terms.



{{{2(x-3/2)^2+2(1/4)}}} Distribute.



{{{2(x-3/2)^2+1/2}}} Multiply.



So after completing the square, {{{2x^2-6x+5}}} transforms to {{{2(x-3/2)^2+1/2}}}. So {{{2x^2-6x+5=2(x-3/2)^2+1/2}}}.



So {{{2x^2-6x+5=0}}} is equivalent to {{{2(x-3/2)^2+1/2=0}}}.



Now let's solve {{{2(x-3/2)^2+1/2=0}}} 



{{{2(x-3/2)^2+1/2=0}}} Start with the given equation.



{{{2(x-3/2)^2=0-1/2}}} Subtract {{{1/2}}} from both sides.



{{{2(x-3/2)^2=-1/2}}} Combine like terms.



{{{(x-3/2)^2=(-1/2)/(2)}}} Divide both sides by {{{2}}}.



{{{(x-3/2)^2=-1/4}}} Reduce.



{{{x-3/2=""+-sqrt(-1/4)}}} Take the square root of both sides.



{{{x-3/2=sqrt(-1/4)}}} or {{{x-3/2=-sqrt(-1/4)}}} Break up the "plus/minus" to form two equations.



{{{x-3/2=(1/2)i}}} or {{{x-3/2=-(1/2)i}}}  Take the square root of {{{-1/4}}} to get {{{(1/2)i}}}.



Note: {{{i=sqrt(-1)}}}



{{{x=3/2+(1/2)i}}} or {{{x=3/2-(1/2)i}}} Add {{{3/2}}} to both sides.



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Answer:



So the solutions are {{{x=3/2+(1/2)i}}} or {{{x=3/2-(1/2)i}}}.