Question 161351


{{{2x^2+5x-1}}} Start with the given expression.



{{{2(x^2+(5/2)x-1/2)}}} 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 {{{5/2}}} to get {{{5/4}}}. In other words, {{{(1/2)(5/2)=5/4}}}.



Now square {{{5/4}}} to get {{{25/16}}}. In other words, {{{(5/4)^2=(5/4)(5/4)=25/16}}}



{{{2(x^2+(5/2)x+highlight(25/16-25/16)-1/2)}}} Now add <font size=4><b>and</b></font> subtract {{{25/16}}} inside the parenthesis. Make sure to place this after the "x" term. Notice how {{{25/16-25/16=0}}}. So the expression is not changed.



{{{2((x^2+(5/2)x+25/16)-25/16-1/2)}}} Group the first three terms.



{{{2((x+5/4)^2-25/16-1/2)}}} Factor {{{x^2+(5/2)x+25/16}}} to get {{{(x+5/4)^2}}}.



{{{2((x+5/4)^2-33/16)}}} Combine like terms.



{{{2(x+5/4)^2+2(-33/16)}}} Distribute.



{{{2(x+5/4)^2-33/8}}} Multiply.



So after completing the square, {{{2x^2+5x-1}}} transforms to {{{2(x+5/4)^2-33/8}}}. So {{{2x^2+5x-1=2(x+5/4)^2-33/8}}}.



So {{{2x^2+5x-1=0}}} is equivalent to {{{2(x+5/4)^2-33/8=0}}}.