• October 9th 2007, 05:12 PM
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.
• October 9th 2007, 05:13 PM
Jhevon
Quote:

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
• October 9th 2007, 05:23 PM
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.
• October 9th 2007, 05:38 PM
Jhevon
Quote:

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
• October 9th 2007, 05:39 PM
facesonfilm
so would the c (which is zero) make the whole 4ac equal zero?
• October 9th 2007, 05:43 PM
Jhevon
Quote:

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
• October 9th 2007, 05:44 PM
facesonfilm
Oh kay, I didnt even catch that. Wow.
that sure makes me feel like a genius.
thanks alot!!
• October 9th 2007, 05:45 PM
SnipedYou
I'm surprised no one has pointed out that B is negative meaning that it should be

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

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

Making it

$-(-5)$ or just $5$
• October 9th 2007, 05:47 PM
Jhevon
Quote:

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

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

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

here $a = 2$, $b = -5$ and $c = 0$
• October 9th 2007, 05:49 PM
facesonfilm
So because I have a negitive with a negitive, it becomes a positive,
even in this case? Correct?
• October 9th 2007, 05:50 PM
Jhevon
Quote:

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
• October 9th 2007, 05:52 PM
facesonfilm
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.
• October 10th 2007, 01:00 PM
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

:confused:
• October 10th 2007, 01:04 PM
Jhevon
Quote:

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

:confused:

i think he was required to use the quadratic formula