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?

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> 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?      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E2-4x%2B6 Start with the given expression.


Take half of the x coefficient -4 to get -2. In other words, %281%2F2%29%28-4%29=-2.


Now square -2 to get 4. In other words, %28-2%29%5E2=%28-2%29%28-2%29=4


x%5E2-4x%2Bhighlight%284-4%29%2B6 Now add and subtract 4. Make sure to place this after the "x" term. Notice how 4-4=0. So the expression is not changed.


%28x%5E2-4x%2B4%29-4%2B6 Group the first three terms.


%28x-2%29%5E2-4%2B6 Factor x%5E2-4x%2B4 to get %28x-2%29%5E2.


%28x-2%29%5E2%2B2 Combine like terms.


So after completing the square, x%5E2-4x%2B6 transforms to %28x-2%29%5E2%2B2. So x%5E2-4x%2B6=%28x-2%29%5E2%2B2.


So x%5E2-4x%2B6=0 is equivalent to %28x-2%29%5E2%2B2=0.


Now let's solve %28x-2%29%5E2%2B2=0


%28x-2%29%5E2%2B2=0 Start with the given equation.


%28x-2%29%5E2=0-2Subtract 2 from both sides.


%28x-2%29%5E2=-2 Combine like terms.


x-2=%22%22%2B-sqrt%28-2%29 Take the square root of both sides.


x-2=sqrt%28-2%29 or x-2=-sqrt%28-2%29 Break up the "plus/minus" to form two equations.


x-2=i%2Asqrt%282%29 or x-2=-i%2Asqrt%282%29 Simplify the square root.


x=2%2Bi%2Asqrt%282%29 or x=2-i%2Asqrt%282%29 Add 2 to both sides.


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Answer:


So the solutions are x=2%2Bi%2Asqrt%282%29 or x=2-i%2Asqrt%282%29.


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) About Me  (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