logosomatw

2022-01-21

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

Kaitlynn Noble

Step 1
$4{x}^{2}-16$
fits the difference of squares property
${a}^{2}-{b}^{2}$, which can be factored as $\left(a+b\right)\left(a-b\right)$
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 $±4$
Thus, $a=2x$ and $b=±4$, so we have
$\left(2x+4\right)\left(2x-4\right)=0$
We can factor a 2 out of both expressions to get
$2\left(x+2\right)\left(x-2\right)$
Setting the factors equal to zero, we get
$x=±2$

lilwhitelieyc

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
$x=±2$

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