Math Help - Finding points where equation at a certain slope

So when you have an equation such as x^2 + 4x + 9 = 0, and you're looking for the two x values where y = 0, you can use the quadratic equation or you can factor it out or simplify it... So forgetting the equation mentioned above and going to this one

-3x^2 + 12x = 0, I want to find the x values where this is true. I could... multiply the whole equation by (-1/3) which would give me...
x^2 -4 = 0. Then we can do x^2 = 4, and then square both sides which gives us... x = 2 and/or x = (-2). So this means that the two x points where y = 0 are x = 2 or x= -2...

What doesnt make sense to me is that you can also do...

-3x^2 + 12x = 0... and you can factor out the -3... which makes it... -3x(x - 4) = 0.. then you can do the thing where you split it and do
-3x = 0 and x - 4 = 0... so x = 0 and/or x = 4...

Mathematically I think I'm doing correct steps for both but I get different answers for x.

When should you do which method?

Originally Posted by brianscherms

So when you have an equation such as x^2 + 4x + 9 = 0, and you're looking for the two x values where y = 0, you can use the quadratic equation or you can factor it out or simplify it... So forgetting the equation mentioned above and going to this one

-3x^2 + 12x = 0, I want to find the x values where this is true. I could... multiply the whole equation by (-1/3) which would give me...
x^2 -4 = 0. Then we can do x^2 = 4, and then square both sides which gives us... x = 2 and/or x = (-2). So this means that the two x points where y = 0 are x = 2 or x= -2...

What doesnt make sense to me is that you can also do...

-3x^2 + 12x = 0... and you can factor out the -3... which makes it... -3x(x - 4) = 0.. then you can do the thing where you split it and do
-3x = 0 and x - 4 = 0... so x = 0 and/or x = 4...

Mathematically I think I'm doing correct steps for both but I get different answers for x.

When should you do which method?

You made a careless error.

$-3x^2 + 12x = 0 \implies - \dfrac{1}{3} (-3x^2 + 12x) = - \dfrac{1}{3} * 0 \implies x^2 - 4x = 0.$

You got, in error, $-3x^2 + 12x = 0 \implies - \dfrac{1}{3} (-3x^2 + 12x) = - \dfrac{1}{3} * 0 \implies x^2 - 4 = 0.$ Should be - 4x not - 4.

Had you checked your answer, you would have seen your error.

$-3(-2)^2 + 12(- 2) = - 3 * 4 - 12 * 2 = - 12 - 24 = - 36 \ne 0.$

$-3(2)^2 + 12(2) = - 3 * 4 + 12 * 2 = - 12 + 24 = 12 \ne 0.$

Whereas $-3(0)^2 + 12(0) = 0 + 0 = 0\ and\ - 3(4)^2 + 12(4) = - 3 * 16 + 48 = - 48 + 48 = 0.$

Originally Posted by brianscherms

So when you have an equation such as x^2 + 4x + 9 = 0, and you're looking for the two x values where y = 0, you can use the quadratic equation or you can factor it out or simplify it... So forgetting the equation mentioned above and going to this one

-3x^2 + 12x = 0, I want to find the x values where this is true. I could... multiply the whole equation by (-1/3) which would give me...
x^2 -4 = 0. Then we can do x^2 = 4, and then square both sides which gives us... x = 2 and/or x = (-2). So this means that the two x points where y = 0 are x = 2 or x= -2...

What doesnt make sense to me is that you can also do...

-3x^2 + 12x = 0... and you can factor out the -3... which makes it... -3x(x - 4) = 0.. then you can do the thing where you split it and do
-3x = 0 and x - 4 = 0... so x = 0 and/or x = 4...

Mathematically I think I'm doing correct steps for both but I get different answers for x.

When should you do which method?

read carefully what you wrote and think about whether you need to ask this.

(hint: in first equation you magically change from $-3x^2+12x$ to $x^2-4$ )