How do you solve #(x+2)^2 = 20#?

1 Answer
Azimet
Jan 16, 2017

#x = -2 +- sqrt20#

Explanation:

There are many approaches to this type of problem. One option is to expand the square and use the quadratic formula to solve the equation. However in this specific case, we have a simpler way of solving this:

#(x+2)^2 = 20#. Since both #(x+2)^2# and #20# are positive, we can take the square roots of both:

#sqrt((x+2)^2) = sqrt20#

#abs(x+2) = sqrt20#.

We use an absolute value, since we do not know if #x# is greater or less than #-2#, so we don't know if #x + 2# is positive or negative. We can, however, get rid of the absolute value, since if #absa = b#, where #b >= 0#, then #a = +-b#.

#x+2=+-sqrt20 => x = -2 +- sqrt20#.

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