Need clarity on discriminant (1 Viewer)

[Joshmosh2]

Let me get this clear
If I get an equation, factored somewhere along the line of
2(m-2)x^2+...
When using discriminant or any other formulas in general, do you take into account the inside term, or the entire part?
Thinking about it, technically, the 'a' in ax^2 should include the entire 2(m-2) but it's so tempting just to use the (m-2) hehe

 

[SquareHeartsAdrita]

it is the entire coefficient of x^2, x and the constant

 

[RealiseNothing]

Let me get this clear
If I get an equation, factored somewhere along the line of
2(m-2)x^2+...
When using discriminant or any other formulas in general, do you take into account the inside term, or the entire part?
Thinking about it, technically, the 'a' in ax^2 should include the entire 2(m-2) but it's so tempting just to use the (m-2) hehe

I really don't see how?

 

[braintic]

Let me get this clear
If I get an equation, factored somewhere along the line of
2(m-2)x^2+...
When using discriminant or any other formulas in general, do you take into account the inside term, or the entire part?
Thinking about it, technically, the 'a' in ax^2 should include the entire 2(m-2) but it's so tempting just to use the (m-2) hehe

Your equation is not factored. Only the coefficient of x^2 is factored.

Why would it be remotely tempting to do something you know is wrong?
If you want to divide all terms by 2 .... with regard to determining the nature of the roots (and not actually solving the equation), the logic would still work. But then you can't claim to be using the discriminant ..... unless you want to call it Δ/4.