Question 167476

{{{s^2-s-6}}} Start with the left side of the equation.



Take half of the {{{s}}} coefficient {{{-1}}} to get {{{-1/2}}}. In other words, {{{(1/2)(-1)=-1/2}}}.



Now square {{{-1/2}}} to get {{{1/4}}}. In other words, {{{(-1/2)^2=(-1/2)(-1/2)=1/4}}}



{{{s^2-s+highlight(1/4-1/4)-6}}} Now add <font size=4><b>and</b></font> subtract {{{1/4}}}. Make sure to place this after the "s" term. Notice how {{{1/4-1/4=0}}}. So the expression is not changed.



{{{(s^2-s+1/4)-1/4-6}}} Group the first three terms.



{{{(s-1/2)^2-1/4-6}}} Factor {{{s^2-s+1/4}}} to get {{{(s-1/2)^2}}}.



{{{(s-1/2)^2-25/4}}} Combine like terms.



So after completing the square, {{{s^2-s-6}}} transforms to {{{(s-1/2)^2-25/4}}}. So {{{s^2-s-6=(s-1/2)^2-25/4}}}.



So {{{s^2-s-6=0}}} is equivalent to {{{(s-1/2)^2-25/4=0}}}.


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{{{(s-1/2)^2-25/4=0}}} Start with the completed square set to 0.



{{{(s-1/2)^2=0+25/4}}} Add {{{25/4}}} to both sides.



{{{(s-1/2)^2=25/4}}} Combine like terms.



{{{x-1/2=0+-sqrt(25/4)}}} Take the square root of both sides.



{{{s-1/2=sqrt(25/4)}}} or {{{s-1/2=-sqrt(25/4)}}} Break up the "plus/minus" to form two equations.



{{{s-1/2=5/2}}} or {{{s-1/2=-5/2}}}  Take the square root of {{{25/4}}} to get {{{5/2}}}. Note: {{{(5/2)^2=25/4}}} so {{{sqrt(25/4)=5/2}}}



{{{s=1/2+5/2}}} or {{{s=1/2-5/2}}} Add {{{1/2}}} to both sides.



{{{s=(1+5)/2}}} or {{{s=(1-5)/2}}} Combine the fractions



{{{s=6/2}}} or {{{s=-4/2}}} Combine the numerators



{{{s=3}}} or {{{s=-2}}} Reduce.



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Answer:



So the solutions are {{{s=3}}} or {{{s=-2}}}.