quadratic formulaNoobpatty wrote:
Alright now for 14s2- 21s = 0
Do I start by factoring out the GCD for 14s2 and -21s? or do I make the equation 14s2 = 21s ? or am I totally lost
Last edited by Bevo (2009-10-06 16:53:22)
quicker (with a program on your calculator, that is)Bevo wrote:
Yup, factor out what you can.
7s(2s - 3) = 0
s = 3/2, 0
ed: yew don't have to do the quadratic formula.
This. When there is no constant term (ie. it is zero), you can usually just pull one of the variables outside the brackets.. and set that to x = 0 (or s=0 in this case).. and then just solve the part inside the brackets to equal zero, which is easy since it's a linear equation.Bevo wrote:
Yup, factor out what you can.
7s(2s - 3) = 0
s = 3/2, 0
ed: yew don't have to do the quadratic formula.
Heh. We weren't allowed calculators in engineering maths. Wut. yeah.. it's so we could do it by handhaffeysucks wrote:
quicker (with a program on your calculator, that is)
yeah that's what'll eventually happen to memcminty wrote:
mcminty wrote:
This. When there is no constant term (ie. it is zero), you can usually just pull one of the variables outside the brackets.. and set that to x = 0 (or s=0 in this case).. and then just solve the part inside the brackets to equal zero, which is easy since it's a linear equation.Bevo wrote:
Yup, factor out what you can.
7s(2s - 3) = 0
s = 3/2, 0
ed: yew don't have to do the quadratic formula.Heh. We weren't allowed calculators in engineering maths. Wut. yeah.. it's so we could do it by handhaffeysucks wrote:
quicker (with a program on your calculator, that is)
Oh how I miss factoring...mcminty wrote:
This. When there is no constant term (ie. it is zero), you can usually just pull one of the variables outside the brackets.. and set that to x = 0 (or s=0 in this case).. and then just solve the part inside the brackets to equal zero, which is easy since it's a linear equation.Bevo wrote:
Yup, factor out what you can.
7s(2s - 3) = 0
s = 3/2, 0
ed: yew don't have to do the quadratic formula.Heh. We weren't allowed calculators in engineering maths. Wut. yeah.. it's so we could do it by handhaffeysucks wrote:
quicker (with a program on your calculator, that is)
Wait I'm confused how you jumped from 7s(2s - 3) = 0 to knowing s is 3/2 or 0Bevo wrote:
Yup, factor out what you can.
7s(2s - 3) = 0
s = 3/2, 0
ed: yew don't have to do the quadratic formula.
Mhmmm. I was doing calc earlier in the library with my friends and one of them pulled out a calc, I gave them a weird look. Mine's sitting in a box somewhere. I don't even consider using it.mcminty wrote:
Heh. We weren't allowed calculators in engineering maths. Wut. yeah.. it's so we could do it by hand
either 7s or (2s-3) must equal zero for the equation to equal zero.Noobpatty wrote:
Wait I'm confused how you jumped from 7s(2s - 3) = 0 to knowing s is 3/2 or 0Bevo wrote:
Yup, factor out what you can.
7s(2s - 3) = 0
s = 3/2, 0
ed: yew don't have to do the quadratic formula.
When you have those terms multiplied, you can set both of them individually equal to 0.Noobpatty wrote:
Wait I'm confused how you jumped from 7s(2s - 3) = 0 to knowing s is 3/2 or 0Bevo wrote:
Yup, factor out what you can.
7s(2s - 3) = 0
s = 3/2, 0
ed: yew don't have to do the quadratic formula.
oh, duh. Forgot about that.S.Lythberg wrote:
either 7s or (2s-3) must equal zero for the equation to equal zero.Noobpatty wrote:
Wait I'm confused how you jumped from 7s(2s - 3) = 0 to knowing s is 3/2 or 0Bevo wrote:
Yup, factor out what you can.
7s(2s - 3) = 0
s = 3/2, 0
ed: yew don't have to do the quadratic formula.
s=3/2 and s=0 are the only numbers that work
We just gaze blankly at the board in lecture... Wasn't it great when math had numbers in it?Bevo wrote:
Mhmmm. I was doing calc earlier in the library with my friends and one of them pulled out a calc, I gave them a weird look. Mine's sitting in a box somewhere. I don't even consider using it.mcminty wrote:
Heh. We weren't allowed calculators in engineering maths. Wut. yeah.. it's so we could do it by hand
@ Lyth - my friend was doing one of those - my brain did a few spins and then stopped.
I know the feeling. I have 2 more calc classes after this one... not looking forward to those at all.S.Lythberg wrote:
We just gaze blankly at the board in lecture... Wasn't it great when math had numbers in it?
Mathematics without numbers?S.Lythberg wrote:
We just gaze blankly at the board in lecture... Wasn't it great when math had numbers in it?
Last edited by Noobpatty (2009-10-06 17:24:17)
Yep.Noobpatty wrote:
Oh boy. Alright well let's see.... last problem that I've stumbled on. I know 4x2 -20x +25 factors into (2x-5)2... does that mean the only answer is that x= 5/2 ?
(4x2 -20x +25 =0)
Last edited by Bevo (2009-10-06 17:26:28)
yes, double rootNoobpatty wrote:
Oh boy. Alright well let's see.... last problem that I've stumbled on. I know 4x2 -20x +25 factors into (2x-5)2... does that mean the only answer is that x= 5/2 ?
(4x2 -20x +25 =0)
hah well at least you're exercising your brain.Bevo wrote:
Yep.Noobpatty wrote:
Oh boy. Alright well let's see.... last problem that I've stumbled on. I know 4x2 -20x +25 factors into (2x-5)2... does that mean the only answer is that x= 5/2 ?
(4x2 -20x +25 =0)
edit: I thought about it for a bit and then did the quadratic formula in my head to verify my answer. I got it right. FML.
Last edited by Noobpatty (2009-10-06 17:29:45)
haha, so did I, to check for an imaginary rootBevo wrote:
Yep.Noobpatty wrote:
Oh boy. Alright well let's see.... last problem that I've stumbled on. I know 4x2 -20x +25 factors into (2x-5)2... does that mean the only answer is that x= 5/2 ?
(4x2 -20x +25 =0)
edit: I thought about it for a bit and then did the quadratic formula in my head to verify my answer. I got it right. FML.
Oh boy, I guess it's too late to help now, I can't remember mean value theorem for the life of me (shows how valuable it was )Bevo wrote:
Need help with mean value theorem application, class is in an hour :x any help appreciated.
it states f(x) is continuous and differentiable on [a,b] and (a,b) respectively. Which is fine.
then it says there is a #c b/w A and B such that f(b) - f(a)/b - a = f'(c)
which is cool and all. but we get no interval for this problem.
use MVT to prove |sin a - sin b| less than or equal to |a-b| for all a, b. the TA said to check on a<b and b<a, but i dunno what that matters because its absolute value anyway.
so for whatever reason it starts out as f(x) = sin (x). Checked for continuity, and differentiable. Cool. end up with |sin a - sin b| = |cos c|*|a-b|. |cos c| is any number in the interval [0,1].
Okay, so what? How does that prove anything? I seem to be stuck here.
