How do you solve # 3x^2=48#?

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Answer 1 · 2018-03-04T18:29:25

#x=+-4#

We begin by dividing both sides by #3#:

#(cancel3x^2)/cancel3=48/3#

#x^2=16#

Take the square root on both sides (and don't forget the negative root):

#sqrt(x^2)=+-sqrt(16)#

#x=+-4#