How do you solve #x+sqrt(6+sqrt(x^2)) = 0# ?

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Answer 1 · 2017-06-18T12:16:39

#x=-3#

Given:

#x+sqrt(6+sqrt(x^2)) = 0#

Subtract #x# from both sides to get:

#sqrt(6+sqrt(x^2)) = -x#

Note that since #sqrt(...) >= 0# we must have #x <= 0#.

Square both sides to get:

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

Since #x <= 0#, we have #sqrt(x^2) = -x# and hence this simplifies to:

#6-x = x^2#

Subtract #6-x# from both sides to get:

#0 = x^2+x-6#

#color(white)(0) = (x+3)(x-2)#

Hence:

#x = -3" "# or #" "x = 2#

We need to discard the extraneous solution #x = 2# as we require #x < 0#.

So the only solution is:

#x = -3#