Question 171916


{{{3x^2+5x-1}}} Start with the left side of the equation.



{{{3(x^2+(5/3)x-1/3)}}} 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 {{{5/3}}} to get {{{5/6}}}. In other words, {{{(1/2)(5/3)=5/6}}}.



Now square {{{5/6}}} to get {{{25/36}}}. In other words, {{{(5/6)^2=(5/6)(5/6)=25/36}}}



{{{3(x^2+(5/3)x+highlight(25/36-25/36)-1/3)}}} Now add <font size=4><b>and</b></font> subtract {{{25/36}}} inside the parenthesis. Make sure to place this after the "x" term. Notice how {{{25/36-25/36=0}}}. So the expression is not changed.



{{{3((x^2+(5/3)x+25/36)-25/36-1/3)}}} Group the first three terms.



{{{3((x+5/6)^2-25/36-1/3)}}} Factor {{{x^2+(5/3)x+25/36}}} to get {{{(x+5/6)^2}}}.



{{{3((x+5/6)^2-37/36)}}} Combine like terms.



{{{3(x+5/6)^2+3(-37/36)}}} Distribute.



{{{3(x+5/6)^2-37/12}}} Multiply.



So after completing the square, {{{3x^2+5x-1}}} transforms to {{{3(x+5/6)^2-37/12}}}. So {{{3x^2+5x-1=3(x+5/6)^2-37/12}}}.



So {{{3x^2+5x-1=0}}} is equivalent to {{{3(x+5/6)^2-37/12=0}}}.



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{{{3(x+5/6)^2-37/12=0}}} Start with the given equation.



{{{3(x+5/6)^2=0+37/12}}}Add {{{37/12}}} to both sides.



{{{3(x+5/6)^2=37/12}}} Combine like terms.



{{{(x+5/6)^2=(37/12)/(3)}}} Divide both sides by {{{3}}}.



{{{(x+5/6)^2=37/36}}} Reduce.



{{{x+5/6=0+-sqrt(37/36)}}} Take the square root of both sides.



{{{x+5/6=sqrt(37/36)}}} or {{{x+5/6=-sqrt(37/36)}}} Break up the "plus/minus" to form two equations.



{{{x+5/6=sqrt(37)/6}}} or {{{x+5/6=-sqrt(37)/6}}}  Simplify the square root.



{{{x=-5/6+sqrt(37)/6}}} or {{{x=-5/6-sqrt(37)/6}}} Subtract {{{5/6}}} from both sides.



{{{x=(-5+sqrt(37))/(6)}}} or {{{x=(-5-sqrt(37))/(6)}}} Combine the fractions.



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Answer:



So the solutions are {{{x=(-5+sqrt(37))/(6)}}} or {{{x=(-5-sqrt(37))/(6)}}}.