How do you solve #w^2/24-w/2+13/6=0# by completing the square?

1 Answer
Jul 18, 2016

Answer:

The solutions will be #w = 6 +- 4i#.

Explanation:

We can start by removing fractions from the mix by multiplying both sides by #24#:
#w^2 - 12w + 52 = 0#

Now observing that we need an equation looking like #w + b# where #2b = -12# it is clear that the squared term will be #w - 6#.

Since #(w-6)^2 = w^2 - 12w + 36# we can take #36# out of #52#, this gives us:
#(w-6)^2 + 16 = 0#

we can manipulate this:
#(w-6)^2 = -16#

And take the square root of both sides:
#w-6 = +- 4i#
#w = 6 +- 4i#

You can check this answer by inputting the coefficients into the quadratic equation as well.