How do you solve #sqrt(-2x) + 64 = 6#?

1 Answer
Alan P.
Jul 31, 2015

There are no solutions to this equation

Explanation:

By convention
#color(white)("XXXX")##sqrt("anything") >= 0#

Therefore for any value of #x#
#color(white)("XXXX")##sqrt(-2x)+64#
#color(white)("XXXX")##color(white)("XXXX")##color(white)("XXXX")##>= 64#
#color(white)("XXXX")##color(white)("XXXX")##color(white)("XXXX")##!= 6#

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