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.

2. Originally Posted by facesonfilm
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.
that is wrong. c is zero here

3. 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.

4. Originally Posted by facesonfilm
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.
you would not have the -4(2) under the square root. since that is -4ac and c is zero

5. so would the c (which is zero) make the whole 4ac equal zero?

6. Originally Posted by facesonfilm
so would the c (which is zero) make the whole 4ac equal zero?
umm, yes. zero times any number is zero

7. Oh kay, I didnt even catch that. Wow.
that sure makes me feel like a genius.
thanks alot!!

8. I'm surprised no one has pointed out that B is negative meaning that it should be

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

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

Making it

$\displaystyle -(-5)$ or just $\displaystyle 5$

9. Originally Posted by SnipedYou
I'm surprised no one has pointed out that B is negative meaning that it should be

$\displaystyle \frac{5 \pm \sqrt{25}}{4}$
we would have gotten to that

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 $\displaystyle c = 0$

10. So because I have a negitive with a negitive, it becomes a positive,
even in this case? Correct?

11. Originally Posted by facesonfilm
So because I have a negitive with a negitive, it becomes a positive,
even in this case? Correct?
yes, the negative of a negative is a positive

12. 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.

13. 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

14. Originally Posted by shuggyboi
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

i think he was required to use the quadratic formula