SOLUTION: Solve by completing the square: x^2-4x+6=0 I did x^2-4x+4=-6+4 (x-2)^2=-2 x-2=±√-2 Am I doing it right?

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Question 388773: Solve by completing the square: x^2-4x+6=0
I did x^2-4x+4=-6+4
(x-2)^2=-2
x-2=±√-2

Am I doing it right?
Found 2 solutions by jim_thompson5910, lwsshak3:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!

Start with the given expression.


Take half of the coefficient to get . In other words, .


Now square to get . In other words,


Now add and subtract . Make sure to place this after the "x" term. Notice how . So the expression is not changed.


Group the first three terms.


Factor to get .


Combine like terms.


So after completing the square, transforms to . So .


So is equivalent to .


Now let's solve


Start with the given equation.


Subtract from both sides.


Combine like terms.


Take the square root of both sides.


or Break up the "plus/minus" to form two equations.


or Simplify the square root.


or Add to both sides.


--------------------------------------


Answer:


So the solutions are or .


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

Also, feel free to check out my tutoring website

Jim
Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
what you did so far is correct, but you did not finish the problem
continuing from where you ended,
(x-2)^2=-2
(x-2 = + or -sqrt (-2)
x = 2 + or -sqrt (-2)
at this point you can see there is no real number solution because the sqrt of a negative number is imaginary.
to compete the problem,
x =2 + or -sqrt (2)*sqrt(-1)
sqrt(-1) = i
ans:
x=2 + or -sqrt (2)i
This problem has 2 imaginary number solutions
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