How do you solve #(x-2)^2=-3#?

1 Answer
George C.
May 1, 2016

#x = 2+-sqrt(3)i#

Explanation:

To solve this we want to take the square root of both sides, but the right hand side is negative. If #a# is any Real number then #a^2 >= 0#, so we need a Complex number, in this case #sqrt(3)i#.

If #a < 0# then #sqrt(a) = sqrt(-a)i#

So given:

#(x-2)^2 = -3 = (sqrt(3)i)^2#

We must have:

#x-2 = +-sqrt(3)i#

Note the #+-#, since if #a^2 = b# then #(-a)^2 = b# too.

Then adding #2# to both sides we get:

#x = 2+-sqrt(3)i#

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