Question 147242

{{{x^2+6x+1}}} Start with the given expression



Take half of the x coefficient {{{6}}} to get {{{3}}} (ie {{{(1/2)(6)=3}}}).


Now square {{{3}}} to get {{{9}}} (ie {{{(3)^2=(3)(3)=9}}})





{{{x^2+6x+9-9+1}}} Now add and subtract this value inside the parenthesis. Notice how {{{9-9=0}}}. Since we're adding 0, we're not changing the equation.




{{{(x+3)^2-9+1}}} Now factor {{{x^2+6x+9}}} to get {{{(x+3)^2}}}



{{{(x+3)^2-8}}} Combine like terms



So after completing the square, {{{x^2+6x+1}}} becomes {{{(x+3)^2-8}}}.



In other words, {{{x^2+6x+1=(x+3)^2-8}}}




So {{{x^2+6x+1=0}}} is equivalent to {{{(x+3)^2-8=0}}}



{{{(x+3)^2-8=0}}} Start with the given equation.



{{{(x+3)^2=8}}} Add 8 to both sides.



{{{x+3=0+-sqrt(8)}}} Take the square root of both sides



{{{x=-3+-sqrt(8)}}} Subtract 3 from both sides.



{{{x=-3+sqrt(8)}}} or {{{x=-3-sqrt(8)}}} Break up the expression



{{{x=-3+2*sqrt(2)}}} or {{{x=-3-2*sqrt(2)}}} Simplify the square root.



So the answers are {{{x=-3+2*sqrt(2)}}} or {{{x=-3-2*sqrt(2)}}}



which approximate to {{{x=-0.172}}} or {{{x=-5.828}}}