How do you solve #11^ { 2} + x ^ { 2} = 23^ { 2}#?

1 Answer
Tom M.
Dec 11, 2017

#x=+-2sqrt102#

Explanation:

First, let's subtract #11^2# from both sides.

#x^2=23^2-11^2#
#x^2=529-121#
#x^2=408#

Take the square root of both sides

#x=+-sqrt408#

#x=+-2sqrt102#

Why do we have plus or minus? Well, what is #(sqrt408)^2#? #408#.
How about #(-sqrt408)^2#. This is #408# as well. So #x# can take both the positive and negative values.

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