How do you solve #2x^2 = 16#?

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Answer 1 · 2015-08-02T00:32:54

#x=2sqrt2, -2sqrt 2#

#2x^2=16#

Divide both sides of the equation by #2#.

#x^2=16/2# =

#x^2=8#

Take the square root of both sides.

#x=+-sqrt8#

#x=+-sqrt(2xx2xx2)#

#x=+-sqrt(2^2xx2)#

Apply the rule #sqrt (a^2)=a#.

#x=+-2sqrt2#

Solve for #x#.

#x=2sqrt 2#

#x=-2sqrt2#