Question 99093


{{{x^2-6x-3=0}}} Start with the given equation



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



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

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




{{{x^2-6x+9=3+9}}} Add this result (9) to both sides. Now the expression {{{x^2-6x+9}}} is a perfect square trinomial.





{{{(x-3)^2=3+9}}} Factor {{{x^2-6x+9}}} into {{{(x-3)^2}}}  (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




{{{(x-3)^2=12}}} Combine like terms on the right side


{{{x-3=0+-sqrt(12)}}} Take the square root of both sides


{{{x=3+-sqrt(12)}}} Add 3 to both sides to isolate x.


So the expression breaks down to

{{{x=3+sqrt(12)}}} or {{{x=3-sqrt(12)}}}



So our answer is approximately

{{{x=6.46410161513775}}} or {{{x=-0.464101615137754}}}


Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, x^2-6x-3) }}} graph of {{{y=x^2-6x-3}}}



When we use the root finder feature on a calculator, we would find that the x-intercepts are {{{x=6.46410161513775}}} and {{{x=-0.464101615137754}}}, so this verifies our answer.