Question 252147
I think you made a typo. The problem should be "Show that 2x^2-12x+23 is equal to 2(x-3)^2+5"



{{{2x^2-12x+23}}} Start with the given expression.



{{{2(x^2-6x+23/2)}}} 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 {{{-6}}} to get {{{-3}}}. In other words, {{{(1/2)(-6)=-3}}}.



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



{{{2(x^2-6x+highlight(9-9)+23/2)}}} Now add <font size=4><b>and</b></font> subtract {{{9}}} inside the parenthesis. Make sure to place this after the "x" term. Notice how {{{9-9=0}}}. So the expression is not changed.



{{{2((x^2-6x+9)-9+23/2)}}} Group the first three terms.



{{{2((x-3)^2-9+23/2)}}} Factor {{{x^2-6x+9}}} to get {{{(x-3)^2}}}.



{{{2((x-3)^2+5/2)}}} Combine like terms.



{{{2(x-3)^2+2(5/2)}}} Distribute.



{{{2(x-3)^2+5}}} Multiply.



So after completing the square, {{{2x^2-12x+23}}} transforms to {{{2(x-3)^2+5}}}. So {{{2x^2-12x+23=2(x-3)^2+5}}} for all real values of 'x'.