Question 169929


{{{3x^2+6x-4}}} Start with the right side of the equation.



{{{3(x^2+2x-4/3)}}} Factor out the {{{x^2}}} coefficient {{{3}}}. 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 {{{2}}} to get {{{1}}}. In other words, {{{(1/2)(2)=1}}}.



Now square {{{1}}} to get {{{1}}}. In other words, {{{(1)^2=(1)(1)=1}}}



{{{3(x^2+2x+highlight(1-1)-4/3)}}} Now add <font size=4><b>and</b></font> subtract {{{1}}} inside the parenthesis. Make sure to place this after the "x" term. Notice how {{{1-1=0}}}. So the expression is not changed.



{{{3((x^2+2x+1)-1-4/3)}}} Group the first three terms.



{{{3((x+1)^2-1-4/3)}}} Factor {{{x^2+2x+1}}} to get {{{(x+1)^2}}}.



{{{3((x+1)^2-7/3)}}} Combine like terms.



{{{3(x+1)^2+3(-7/3)}}} Distribute.



{{{3(x+1)^2-7}}} Multiply.



So after completing the square, {{{3x^2+6x-4}}} transforms to {{{3(x+1)^2-7}}}. So {{{3x^2+6x-4=3(x+1)^2-7}}}.



So {{{3x^2+6x-4=0}}} is equivalent to {{{3(x+1)^2-7=0}}}.



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{{{3(x+1)^2-7=0}}} Start with the given equation.



{{{3(x+1)^2=0+7}}}Add {{{7}}} to both sides.



{{{3(x+1)^2=7}}} Combine like terms.



{{{(x+1)^2=(7)/(3)}}} Divide both sides by {{{3}}}.



{{{x+1=0+-sqrt(7/3)}}} Take the square root of both sides.



{{{x+1=sqrt(7/3)}}} or {{{x+1=-sqrt(7/3)}}} Break up the "plus/minus" to form two equations.



{{{x+1=sqrt(21)/3}}} or {{{x+1=-sqrt(21)/3}}}  Simplify the square root.



{{{x=-1+sqrt(21)/3}}} or {{{x=-1-sqrt(21)/3}}} Subtract {{{1}}} from both sides.



{{{x=(-3+sqrt(21))/(3)}}} or {{{x=(-3-sqrt(21))/(3)}}} Combine the fractions.



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Answer:



So the solutions are {{{x=(-3+sqrt(21))/(3)}}} or {{{x=(-3-sqrt(21))/(3)}}}.