Deriving the quadratic formula question

geramul · 2012-12-01 17:49:26

I have only one question for this.

Let me write this out a little bit into the process already.

Alright so we have

What I want to know is how the common denominator for -c/a is gotten. I thought I had figured it out, but I'm pretty sure I was on the far side of mars in relation to how close I actually was to figuring it out. I THOUGHT, that since a variable with no coefficient has an implied value of 1, that I could turn -c/a into -c/2a that way, therefore gaining the ability to get the common denominator of 4a. This is wrong though I'm pretty sure.

I'm out of ideas at the moment. I need to step away from this and think some more . I would appreciate if someone could explain the process of how -c/a gets the common denominator of 4a.

debjit625 · 2012-12-01 19:40:36

Read this
http://www.mathsisfun.com/algebra/quadr … ation.html

Good Luck

Last edited by debjit625 (2012-12-01 19:41:36)

Bob · 2012-12-01 21:11:09

hi geramul

Welcome to the forum.

I would be happy to explain the steps, but, I cannot improve on the MIF explanation given in the link by debjit625.

After you have had a look, post again if any step is still unclear.

Bob

geramul · 2012-12-01 22:38:50

Thanks guys. I was learning to do this through a different slightly different method however, and even though this method looks a bit shorter I'd prefer to stick with what I'm already familiar with . If you could explain how -c/a gets the common denominator of 4a in my version of the problem bob that'd be great .

Bob · 2012-12-02 01:44:45

hi geramul,

Well I will happily do so, but I have no idea what your method is. Please post what you have so far and I'll try to understand it.

Bob

geramul · 2012-12-02 11:07:59

Alright let me start from the beginning.

So we have

What I do here first is I move the "loose" number over to right.

Now we have

Now I take the coefficient on

and divide it through the entire equation.

Now my method tells me that I take half of the middle term, square it and then add it to both sides.

We end up with this.

This is where I got stuck. I don't know how to get the common denominator of 4a for -c/a. I hope I wrote everything out correctly as I was doing this through memory, and keeping track of exponents and what not can be a little tough when typing it out lol.

Bob · 2012-12-02 11:52:36

OK, so carrying on from there:

Multiply the -c/a term to make a common denominator.

Then put all over this denominator and re-arrange.

Then you can square root everything. Only one +/- sign is needed in the final expression.

Take the b/2a term across to the right hand side, all over the same denominator.

Hopefully that sorts it out for you.

Bob

geramul · 2012-12-02 15:45:12

I actually watched a video that explained it that way. I'm still confused on one thing though, I want to know where the 4a that we multiply -c/a comes from. That's what I'm trying to figure out. The 4a we multiply against -c/a can't just come from nowhere right? Nothing in math can just come from "thin air" right? I mean we can't just say "We're going to multiply -c/a by 4a just because", that 4a has to come from some process and that's what I want to know.

Piman · 2012-12-02 16:30:49

Yout aren't multiplying by 4A but rather multiply by 4A / 4A. Which is the same as multiplying by 1.

geramul · 2012-12-02 17:05:49

Ah, so let me try something here.

So any variable without a coefficient has an implied value of 1 right? So based on that we can look at -c/a as -1c/1a. Now, from 1a we just get 4a/4a? That can't be right though. How can you get 4a/4a from 1a? 1 only has itself as a multiple. Do you understand what I mean?

I'm sorry but when I get stuck on something like this in math I just get fixated on it. I could just take the answer as it is but I NEED to know why it happens.

Bob · 2012-12-02 20:32:25

hi geramul

I actually watched a video that explained it that way. I'm still confused on one thing though, I want to know where the 4a that we multiply -c/a comes from. That's what I'm trying to figure out. The 4a we multiply against -c/a can't just come from nowhere right? Nothing in math can just come from "thin air" right? I mean we can't just say "We're going to multiply -c/a by 4a just because", that 4a has to come from some process and that's what I want to know.

Ok, don't worry. I'll try to sort that out. It is an algbra misunderstanding. When I have trouble with algebra, I go back to some numbers and try the same thing.

Let's say that c = 3, b = 12 and a = 5

Now, if you have two fractions to add together, you have to make the denominators the same. I want both /100

So multiply the fraction by 4 x 5

Now with letters

Bob

geramul · 2012-12-03 09:32:02

Ah I'm sorry bob, but I don't think you're getting what I'm trying to say. I think I have a better way to explain it, check this out.

What I can't figure out is how the common denominator of 4a is gotten. Let me give an example. Lets say we want to add together 1/4 and 5/8. We would look of course at the multiples of 4 and find one that's equal to other denominator. So we find that 8 is the least common denominator for both fractions. We multiply 1/4 by 2/2 to get 2/8, and now we can solve.

My problem is how do we get the common denominator of 4a from -c/a? Like I said earlier, a variable without a coefficient has an implied coefficient of 1. So now we can look at -c/a as -1c/1a. How can we get the common denominator of 4 from 1, when 1 only has itself as a multiple? If both of the bases were the same we could just say it was equal to 1, divide that in half to get 5/10, get 1/2, and THEN we could have gotten the common denominator of 4. I don't see how this works.

Last edited by geramul (2012-12-03 09:40:38)

anonimnystefy · 2012-12-03 09:34:48

Why nog justmultiply the top and the bottom of the fraction by 4 like Bob said?

geramul · 2012-12-03 09:41:27

I get that we're supposed to multiply -c/a by 4a/4a, but I'm not seeing where that 4a/4a is coming from.

Bob · 2012-12-03 09:47:20

Well you could multiply by 7/7 or xyz/xyz or 4pq/4pq

In every case you are just multiplying by 1 so the fraction retains the same value.

The reason for picking 4a/4a is that it then makes the denominator into 4a times a = 4a^2. Now it is the same as the b^2 denominator so we can add the two numerators together, putting them over the same denominator.

Bob

geramul · 2012-12-03 09:56:11

Ahhh... so because -c/a have "no value" or are just equal to 1, we can manipulate them how we want?

Last edited by geramul (2012-12-03 09:56:38)

Bob · 2012-12-03 10:12:53

Follow this link and learn how to complete the square.

[url]http://www.mathsisfun.com/algebra/completing-square.html[/urfl]

There are some good questions at the bottom of the page which will help you to improve your algebra particularly in relation to quadratics.

By the time you have done those questions you will be really good at this.

Bob

Math Is Fun Forum

#1 2012-12-01 17:49:26

Deriving the quadratic formula question

#2 2012-12-01 19:40:36

Re: Deriving the quadratic formula question

#3 2012-12-01 21:11:09

Re: Deriving the quadratic formula question

#4 2012-12-01 22:38:50

Re: Deriving the quadratic formula question

#5 2012-12-02 01:44:45

Re: Deriving the quadratic formula question

#6 2012-12-02 11:07:59

Re: Deriving the quadratic formula question

#7 2012-12-02 11:52:36

Re: Deriving the quadratic formula question

#8 2012-12-02 15:45:12

Re: Deriving the quadratic formula question

#9 2012-12-02 16:30:49

Re: Deriving the quadratic formula question

#10 2012-12-02 17:05:49

Re: Deriving the quadratic formula question

#11 2012-12-02 20:32:25

Re: Deriving the quadratic formula question

#12 2012-12-03 09:32:02

Re: Deriving the quadratic formula question

#13 2012-12-03 09:34:48

Re: Deriving the quadratic formula question

#14 2012-12-03 09:41:27

Re: Deriving the quadratic formula question

#15 2012-12-03 09:47:20

Re: Deriving the quadratic formula question

#16 2012-12-03 09:56:11

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#17 2012-12-03 10:12:53

Re: Deriving the quadratic formula question

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