SOLUTION: how would i solve equations over complex numbers? for example, x^2-2x+2=0

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson  -> Lesson -> SOLUTION: how would i solve equations over complex numbers? for example, x^2-2x+2=0      Log On


   

Question 554326: how would i solve equations over complex numbers? for example, x^2-2x+2=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You would use the quadratic formula.


x%5E2-2x%2B2=0 Start with the given equation.


Notice that the quadratic x%5E2-2x%2B2 is in the form of Ax%5E2%2BBx%2BC where A=1, B=-2, and C=2


Let's use the quadratic formula to solve for "x":


x+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


x+=+%28-%28-2%29+%2B-+sqrt%28+%28-2%29%5E2-4%281%29%282%29+%29%29%2F%282%281%29%29 Plug in A=1, B=-2, and C=2


x+=+%282+%2B-+sqrt%28+%28-2%29%5E2-4%281%29%282%29+%29%29%2F%282%281%29%29 Negate -2 to get 2.


x+=+%282+%2B-+sqrt%28+4-4%281%29%282%29+%29%29%2F%282%281%29%29 Square -2 to get 4.


x+=+%282+%2B-+sqrt%28+4-8+%29%29%2F%282%281%29%29 Multiply 4%281%29%282%29 to get 8


x+=+%282+%2B-+sqrt%28+-4+%29%29%2F%282%281%29%29 Subtract 8 from 4 to get -4


x+=+%282+%2B-+sqrt%28+-4+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%282+%2B-+2%2Ai%29%2F%282%29 Take the square root of -4 to get 2%2Ai.


x+=+%282+%2B+2%2Ai%29%2F%282%29 or x+=+%282+-+2%2Ai%29%2F%282%29 Break up the expression.


x+=+%282%29%2F%282%29+%2B+%282%2Ai%29%2F%282%29 or x+=++%282%29%2F%282%29+-+%282%2Ai%29%2F%282%29 Break up the fraction for each case.


x+=+1%2Bi or x+=++1-i Reduce.


So the solutions are x+=+1%2Bi or x+=++1-i


If you need more help, email me at jim_thompson5910@hotmail.com

Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you

Jim