Question 174719


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



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-1}}}, and {{{c=1}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-1) +- sqrt( (-1)^2-4(1)(1) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-1}}}, and {{{c=1}}}



{{{x = (1 +- sqrt( (-1)^2-4(1)(1) ))/(2(1))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(1)(1) ))/(2(1))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4 ))/(2(1))}}} Multiply {{{4(1)(1)}}} to get {{{4}}}



{{{x = (1 +- sqrt( -3 ))/(2(1))}}} Subtract {{{4}}} from {{{1}}} to get {{{-3}}}



{{{x = (1 +- sqrt( -3 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (1 +- i*sqrt(3))/(2)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (1+i*sqrt(3))/(2)}}} or {{{x = (1-i*sqrt(3))/(2)}}} Break up the expression.  



So the answers are {{{x = (1+i*sqrt(3))/(2)}}} or {{{x = (1-i*sqrt(3))/(2)}}} 



which approximate to {{{x=0.5+0.866i}}} or {{{x=0.5-0.866i}}}