How do you solve #x^2=196#?

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Answer 1 · 2016-02-28T23:29:42

#x=14# or #x=-14#

First find the prime factors of #196#

#196# ends with an even digit (#6#), so must be divisible by #2#

#196 = 2 * 98#

#98# ends with an even digit (#8#), so must be divisible by #2#

#196 = 2 * 2 * 49#

The digits of #49# do not add up to a multiple of #3#, so it is not divisible by #3#.

The last digit of #49# is not #5# or #0#, so it is not divisible by #5#.

#49 / 7 = 7#

#196 = 2 * 2 * 7 * 7#

Both distinct prime factors occur twice, so we have a perfect square:

#196 = (2*7)*(2*7) = 14 * 14#

So #14^2 = 196# and #(-14)^2 = 196# too.