How do you solve the equation #2x^2-512=0#?

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Answer 1 · 2017-04-28T00:24:32

#x=+-16#

Add #512# to both sides:

#2x^2-512+512=0+512#

This becomes:

#2x^2=512#

Divide both sides by #2#:

#(2x^2)/2=512/2#

This becomes:

#x^2=256#

Square root both sides:

#sqrt(x^2)=+-sqrt(256)#

This becomes:

#x=+-sqrt(256)#

Break down #256# into its factors:

#x=+-sqrt(2*2*2*2*2*2*2*2)#

Take out pairs of the same number:

#x=+-(2*2*2*2)#

This leaves you with:

#x=+-16#