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

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Answer 1 · 2018-04-23T23:13:24

$x = \pm 2$ $x = +-2$

Answer 2 · 2018-04-24T00:19:06

$x = \pm 2$ $x=+-2$

$4 x^{2} - 16$ $4x^2-16$ fits the difference of squares property

$a^{2} - b^{2}$ $a^2-b^2$ , which can be factored as $(a + b) (a - b)$ $(a+b)(a-b)$

Here, we can take the square root of $4 x^{2}$ $4x^2$ , which is $2 x$ $2x$ and we can take the square root of $16$ $16$ which is $\pm 4$ $+-4$ .

Thus, $a = 2 x$ $a=2x$ and $b = \pm 4$ $b=+-4$ , so we have

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

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

$2 (x + 2) (x - 2)$ $2(x+2)(x-2)$

Setting the factors equal to zero, we get

$x = \pm 2$ $x=+-2$

Hope this helps!

How do you solve 4x^2 - 16 = 04x2−16=04x^2 - 16 = 0?