Question

How do you solve the equation `3(x-2)^2-12=0`

Answer 1

Lets simplify this scary thing 1st

This qill be `3( x^2 + 4 - 4x) - 12 = 3x^2 - 12 x`

so this is rewritten as`3x( x - 4) = 0`

The zero rule here has to be either `x- 4 = 0 or 3x = 0`

Hence the roots of these equations are `x =4 , 0`

Answer 2

Another path you could take to solving this equation would be to pull out a GCF. As you can see, all your terms on the left are divisible by 3. And since you just have 0 on the right, you can just divide everything by 3.

Then you get:

`(x-2)^2 - 4 = 0`

Now you could expand the binomial here, but since 4 is a perfect square, it's just much easier to move it over to the other side (add 4 to both sides), and then take the square root of both sides of the equation.

You'd get:

`x-2 = +-2`

Make sure you pay attention to the fact that since you took a square root, it's plus OR minus 2, and not just 2

Now, just add 2 to both sides, and you're done.

`x = 2+-2 = 4 or 0`

Hope that helped :)