logosomatw

2022-01-21

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

Kaitlynn Noble

Beginner2022-01-22Added 12 answers

Step 1

$4{x}^{2}-16$

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

Thus,$a=2x$ and $b=\pm 4$ , 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

$x=\pm 2$

fits the difference of squares property

Here, we can take the square root of

Thus,

We can factor a 2 out of both expressions to get

Setting the factors equal to zero, we get

lilwhitelieyc

Beginner2022-01-23Added 10 answers

Step 1

1) Add 16 to both side to get

$4{x}^{2}=16$

2) Divide both sides by 4 to get

${x}^{2}=4$

3) Find the square root of both sides to get

$x=\pm 2$

1) Add 16 to both side to get

2) Divide both sides by 4 to get

3) Find the square root of both sides to get