Question 116770


{{{x^2-2x-1=0}}} Start with the given equation



{{{x^2-2x=1}}} Add 1 to both sides



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

Now square -1 to get 1 (ie {{{(-1)^2=1}}})




{{{x^2-2x+1=1+1}}} Add this result (1) to both sides. Now the expression {{{x^2-2x+1}}} is a perfect square trinomial.





{{{(x-1)^2=1+1}}} Factor {{{x^2-2x+1}}} into {{{(x-1)^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-1)^2=2}}} Combine like terms on the right side


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


{{{x=1+-sqrt(2)}}} Add 1 to both sides to isolate x.


So the expression breaks down to

{{{x=1+sqrt(2)}}} or {{{x=1-sqrt(2)}}}



So our answer is approximately

{{{x=2.41421356237309}}} or {{{x=-0.414213562373095}}}


Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, x^2-2x-1) }}} graph of {{{y=x^2-2x-1}}}



When we use the root finder feature on a calculator, we would find that the x-intercepts are {{{x=2.41421356237309}}} and {{{x=-0.414213562373095}}}, so this verifies our answer.