Question 92972
Let's use the quadratic formula to find the roots:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{x^2+x+1=0}}} ( notice {{{a=1}}}, {{{b=1}}}, and {{{c=1}}})


{{{x = (-1 +- sqrt( (1)^2-4*1*1 ))/(2*1)}}} Plug in a=1, b=1, and c=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)}}} Combine like terms in the radicand (everything under the square root)




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




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




After simplifying, the quadratic has roots of


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


Notice if we graph the quadratic {{{y=x^2+x+1}}}, we get


{{{ graph( 500, 500, -15.5, 14.5, -14.25, 15.75, x^2+x+1) }}} graph of {{{y=x^2+x+1}}}


And we can see that there are no real roots


To visually verify the answer, check out <a href=http://www.math.hmc.edu/funfacts/ffiles/10005.1.shtml>this page</a> to see a visual representation of imaginary roots