Question 299406


{{{2x^2+8x+5}}} Start with the given expression.



{{{2(x^2+4x+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 {{{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}}}



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



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



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



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



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



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



So after completing the square, {{{2x^2+8x+5}}} transforms to {{{2(x+2)^2-3}}}. So {{{2x^2+8x+5=2(x+2)^2-3}}}.



So {{{2x^2+8x+5=0}}} is equivalent to {{{2(x+2)^2-3=0}}}.



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




{{{2(x+2)^2=0+3}}}Add {{{3}}} to both sides.



{{{2(x+2)^2=3}}} Combine like terms.



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



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



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



{{{x+2=sqrt(6)/2}}} or {{{x+2=-sqrt(6)/2}}}  Simplify the square root.



{{{x=-2+sqrt(6)/2}}} or {{{x=-2-sqrt(6)/2}}} Subtract {{{2}}} from both sides.



{{{x=(-4+sqrt(6))/(2)}}} or {{{x=(-4-sqrt(6))/(2)}}} Combine the fractions.



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Answer:



So the solutions are {{{x=(-4+sqrt(6))/(2)}}} or {{{x=(-4-sqrt(6))/(2)}}}.