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

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Answer 1 · 2018-04-23T23:13:24

$x = +-2$

Answer 2 · 2018-04-24T00:19:06

$x=+-2$

$4x^2-16$ 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!

How do you solve 4x^2 - 16 = 04x^2 - 16 = 0?