Hey there. I am having difficulties with the Quadratic formula,

I know what goes where in the cases, but its just solving from there.

like in this equasion.

2x^2-5x=0

I know how to get it into the quadratic formula

this is what I got.

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- Oct 9th 2007, 05:12 PMfacesonfilmThe Quadratic Formula.
Hey there. I am having difficulties with the Quadratic formula,

I know what goes where in the cases, but its just solving from there.

like in this equasion.

2x^2-5x=0

I know how to get it into the quadratic formula

this is what I got. - Oct 9th 2007, 05:13 PMJhevon
- Oct 9th 2007, 05:23 PMfacesonfilm
I know C is zero there.. but isnt it in the form of Ax^2 + Bx + C?

so then when you plug it into the quadratic formula, you dont input a C. - Oct 9th 2007, 05:38 PMJhevon
- Oct 9th 2007, 05:39 PMfacesonfilm
so would the c (which is zero) make the whole 4ac equal zero?

- Oct 9th 2007, 05:43 PMJhevon
- Oct 9th 2007, 05:44 PMfacesonfilm
Oh kay, I didnt even catch that. Wow.

that sure makes me feel like a genius.

thanks alot!! - Oct 9th 2007, 05:45 PMSnipedYou
I'm surprised no one has pointed out that B is negative meaning that it should be

$\displaystyle \frac{5 \pm \sqrt{25}}{4}$

Since the quadratic is:

$\displaystyle \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

Making it

$\displaystyle -(-5)$ or just $\displaystyle 5 $ - Oct 9th 2007, 05:47 PMJhevon
we would have gotten to that :D

we just had to deal with a more fundamental error first. anyway, so that we are all clear and on the same page:

the formula is: $\displaystyle x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$

here $\displaystyle a = 2$, $\displaystyle b = -5$ and $\displaystyle c = 0$ - Oct 9th 2007, 05:49 PMfacesonfilm
So because I have a negitive with a negitive, it becomes a positive,

even in this case? Correct? - Oct 9th 2007, 05:50 PMJhevon
- Oct 9th 2007, 05:52 PMfacesonfilm
okay, I wasn't 100 % on that because our teacher just gave us the formula today like 5 minutes before the bell so I was kinda foggy on that.

- Oct 10th 2007, 01:00 PMshuggyboi
why dont you just factorise "2x^2 -5x = 0"

to "x(2x-5) = 0"

then x would have the values:

x = 0 and 2x-5 = 0

2x = 5

x =5/2

:confused: - Oct 10th 2007, 01:04 PMJhevon