Question 3196: how do you find the complex solutions to X^2+2=2x
Answer by mixerimaythyme(7)   (Show Source):
You can put this solution on YOUR website! if you mean the solutions to x^2+2=2x , then...
put the equation in standard form (ax^2+bx+c), which is x^2-2x+2=0
then since factoring does not work with this problem you can either use the quadradic formula [ -b plus or minus the square root of (b^2 minus 4ac) all over 2a] which turns out like...
-(-2) plus or minus the sqrt of [(-2)^2-4(1)(2)] / 2(1)
= 2 plus or minus the sqrt of [4-8] / 2
= 2 plus or minus the sqrt of [-4] / 2
= 2 plus or minus 2i / 2
which simplifies to...
1 plus or minus i so x= 1+i or x=1-i
but if you want to solve this by completing the square...
put equation in standard form...then
subtract 2 from both sides leaving an open space on both sides for when you do complete the square... you end up getting... x^2-2x+__=-2+__
Now to get the last term, you must square have of the miidle term (-2)... which is 1,... add that to both sides and you should get...
x^2-2x+1=-2+1...factor the left side and simplify the right side.
so far you should have... (x-1)^2=-1...
take the sqrt of both sides...
x-1= plus or minus i ...
add 1 to both sides...
x=1+i or x=1-i
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