Find the solution of #x^2-x+1=0# in the form #a+bi#?

Question

1597 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-14T03:33:32

#x=1/2+sqrt(3)/2i,1/2-sqrt(3)/2i#

Solve:

#x^2-x+1=0# is a quadratic equation in standard form:

#ax^2+bx+c=0#,

where:

#a=1#, #b=-1#, and #c=1#.

Use the quadratic formula to solve for #x#.

#x=(-b+-sqrt((b^2)-4ac))/(2a)#

Plug in the known values.

#x=(-(-1)+-sqrt((-1)^2-4*1*1))/(2*1)#

Simplify.

#x=(1+-sqrt(1-4))/2#

#x=(1+-sqrt(-3))/2#

Solutions for #x#.

#x=1/2+sqrt(3)/2i,1/2-sqrt(3)/2i#

graph{y=x^2-x+1 [-14.47, 14, -2.79, 11.45]}