Question 115166

{{{x^2-14x+1=0}}} Start with the given equation



{{{x^2-14x=-1}}} Subtract 1 from both sides



Take half of the x coefficient -14 to get -7 (ie {{{-14/2=-7}}})

Now square -7 to get 49 (ie {{{(-7)^2=49}}})




{{{x^2-14x+49=-1+49}}} Add this result (49) to both sides. Now the expression {{{x^2-14x+49}}} is a perfect square trinomial.





{{{(x-7)^2=-1+49}}} Factor {{{x^2-14x+49}}} into {{{(x-7)^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-7)^2=48}}} Combine like terms on the right side


{{{x-7=0+-sqrt(48)}}} Take the square root of both sides


{{{x=7+-sqrt(48)}}} Add 7 to both sides to isolate x.


So the expression breaks down to

{{{x=7+sqrt(48)}}} or {{{x=7-sqrt(48)}}}



So our answer is approximately

{{{x=13.9282032302755}}} or {{{x=0.0717967697244912}}}


Here is visual proof


{{{ graph( 500, 500, -10, 15, -10, 10, x^2-14x+1) }}} graph of {{{y=x^2-14x+1}}}



When we use the root finder feature on a calculator, we would find that the x-intercepts are {{{x=13.9282032302755}}} and {{{x=0.0717967697244912}}}, so this verifies our answer.