# Math Help - [SOLVED] Basic Functions Question

1. ## [SOLVED] Basic Functions Question

I've solved the defined functions to this answer

$x^2-2x+2kx-4k+36=0$

Then I've converted it into the $ax^2+bx+c$ format like this:

For bx: $(-2x+2kx) = -x(2-2k)$

For c: $(-4k+36)$

So now I have a= 1 b= (2-2k) c= (-4k+36)

Am I doing it right? I'm not getting the correct discriminant/roots with the above.

Any help fellas?

Thanks!

2. Originally Posted by unstopabl3
I've solved the defined functions to this answer

$x^2-2x+2kx-4k+36=0$

Then I've converted it into the $ax^2+bx+c$ format like this:

For bx: $(-2x+2kx) = -x(2-2k)$

For c: $(-4k+36)$

So now I have a= 1 b= (2-2k) c= (-4k+36)

Am I doing it right? I'm not getting the correct discriminant/roots with the above.

Any help fellas?

Thanks!
Dear unstopabl3,

If you want to convert your equation into $ax^2+bx+c$ form, b should be, $b=2k-2$ not $b=2-2k$

Did you get the idea?

3. Originally Posted by unstopabl3
I've solved the defined functions to this answer

$x^2-2x+2kx-4k+36=0$

Then I've converted it into the $ax^2+bx+c$ format like this:

For bx: $(-2x+2kx) = -x(2-2k)$

For c: $(-4k+36)$

So now I have a= 1 b= (2-2k) c= (-4k+36)

Am I doing it right? I'm not getting the correct discriminant/roots with the above.

Any help fellas?

Thanks!
well according to $ax^{2}+bx+c$ your $a=1$ is correct. However your $b$ should be $b = -2 + 2k$ and your $c = -4k + 36$ is also correct.

4. Thanks for the quick replies, got the correct answer now