Question 109388


{{{x^2-4x-1=0}}} Start with the given equation



{{{x^2-4x=1}}} Add 1 to both sides



Take half of the x coefficient -4 to get -2 (ie {{{-4/2=-2}}})

Now square -2 to get 4 (ie {{{(-2)^2=4}}})




{{{x^2-4x+4=1+4}}} Add this result (4) to both sides. Now the expression {{{x^2-4x+4}}} is a perfect square trinomial.





{{{(x-2)^2=1+4}}} Factor {{{x^2-4x+4}}} into {{{(x-2)^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-2)^2=5}}} Combine like terms on the right side


{{{x-2=0+-sqrt(5)}}} Take the square root of both sides


{{{x=2+-sqrt(5)}}} Add 2 to both sides to isolate x.


So the expression breaks down to

{{{x=2+sqrt(5)}}} or {{{x=2-sqrt(5)}}}



So our answer is approximately

{{{x=4.23606797749979}}} or {{{x=-0.23606797749979}}}


Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, x^2-4x-1) }}} graph of {{{y=x^2-4x-1}}}



When we use the root finder feature on a calculator, we would find that the x-intercepts are {{{x=4.23606797749979}}} and {{{x=-0.23606797749979}}}, so this verifies our answer.