Question 124289

{{{y=x^2-6 x-3}}} Start with the given equation



{{{y+3=x^2-6 x}}} Add {{{3}}} to both sides



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}}})





{{{y+3=x^2-6x+9-9}}} Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of {{{9}}} does not change the equation




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




{{{y=(x-3)^2-9-3}}} Now add {{{+3}}} to both sides to isolate y



{{{y=(x-3)^2-12}}} Combine like terms




{{{(x-3)^2-12=0}}} Now to solve for x, let {{{y=0}}}



{{{(x-3)^2=+12}}} Add {{{12}}} to both sides




*[Tex \LARGE x-3=\pm \sqrt{12}] Take the square root of both sides



*[Tex \LARGE x-3=\pm 2\sqrt{3}] Simplify {{{sqrt(12)}}} to get {{{2sqrt(3)}}}



*[Tex \LARGE x=3\pm 2\sqrt{3}] Add {{{3}}} to both sides




So our answers are 


*[Tex \LARGE x=3+ 2\sqrt{3}] or *[Tex \LARGE x=3- 2\sqrt{3}]