SOLUTION: What is an imaginary #, how to solve x^2-4x+8=0 In what way is the i or imaginery # involved.

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Question 944912: What is an imaginary #, how to solve x^2-4x+8=0
In what way is the i or imaginery # involved.
Found 2 solutions by josgarithmetic, josmiceli:
Answer by josgarithmetic(39838) About Me  (Show Source):
You can put this solution on YOUR website!
Try completing the square to help solve the equation. What do you find inside the square root function part of the solution?

As reference, i%5E2=-1, and from this i=-sqrt%28-1%29 or i=sqrt%28-1%29.

x%5E2-4x%2B8=0
x%5E2-4x%2B4%2B8=4, the 4 completes the square.
%28x-2%29%5E2%2B8=4
%28x-2%29%5E2=4-8=-4
x-2=0%2B-+sqrt%28-4%29
x=2%2B-+sqrt%28-4%29
highlight%28x=2%2B-+2sqrt%28-1%29%29


Also for reference, study this:
http://www.algebra.com/my/Completing-the-Square-to-Solve-General-Quadratic-Equation.lesson?content_action=show_dev

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Imaginary numbers allow ways to deal with +sqrt%28-1%29+
+x%5E2+-+4x+%2B+8++=+0+
Complete the square
+x%5E2++-+4x+=+-8+
Take +1%2F2+ of the coefficient of
+x+, square it, and add it to both sides
-------------------------------------
+x%5E2++-+4x+%2B+%28+-4%2F2+%29%5E2+=+-8+%2B+%28+-4%2F2+%29%5E2+
+x%5E2+-+4x+%2B+4+=+-8+%2B+4+
+x%5E2++-+4x+%2B+4+=+-4+
+%28+x+-+2+%29%5E2+=+-4+
Take the square root of both sides
+x+-+2+=+sqrt%28+-4+%29+
+x+-+2+=+sqrt%28+-1%2A4+%29+
+x+-+2+=+sqrt%28+-1+%29+%2A+sqrt%28+4+%29+
+x+-+2+=+2i+
+x+=+2+%2B+2i+
and also,
+x+-+2+=+-2i+
+x+=+2+-+2i+
-----------------
The factors of the equation are:
+%28+x+-+2+-+2i+%29%2A%28+x+-+2+%2B+2i+%29+=+x%5E2+-+4x+%2B+8+
You can prove this just by multiplying it out