How do you solve #4x^2 - 16 = 0#?

2 Answers
Apr 23, 2018

Answer:

#x = +-2#

Explanation:

  1. Add #16# to both sides to get #4x^2=16#
  2. Divide both sides by #4# to get #x^2=4#
  3. Find the square root of both sides to get #x = +-2#
Apr 24, 2018

Answer:

#x=+-2#

Explanation:

#4x^2-16# fits the difference of squares property

#a^2-b^2#, which can be factored as #(a+b)(a-b)#

Here, we can take the square root of #4x^2#, which is #2x# and we can take the square root of #16# which is #+-4#.

Thus, #a=2x# and #b=+-4#, so we have

#(2x+4)(2x-4)=0#

We can factor a #2# out of both expressions to get

#2(x+2)(x-2)#

Setting the factors equal to zero, we get

#x=+-2#

Hope this helps!