Deriving the quadratic formula.

In Summary,1. Derive the quadradic equation, ax^2+bx+c=0, using the fallowing method.2. Divide by a, if a =/ 0.3. Move the constant to the other side of the equation.4. Square half the coeffecient of bx and add it to both sides, fixate the negative sign on -c/a.5. factor the left side, simplify the right side.6. solve for x.
  • #1
XxXCausticxXx
1
0
1.Homework Statement

Derive the quadradic equation, [tex]ax^2 + bx + c = 0[/tex], using the fallowing method.

1. Divide by a, if a =/ 0.
2. Move the constant to the other side of the equation.
3. Square half the coeffecient of bx and add it to both sides, fixate the negative sign on -c/a.
4. factor the left side, simplify the right side.
5. apply the square root property.
6. solve for x.

Homework Equations



[tex]ax^2+bx+c=0[/tex], where a is either 0 or not 0, if it is zero you skip the first step. (duh!)

[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

The Attempt at a Solution



1.[tex]ax^2+bx+c=0[/tex]
2.[tex]x^2+\frac{b}{a}x+\frac{c}{a}=0[/tex]
3.[tex]x^2+\frac{b}{a}x+=\frac{-c}{a}[/tex]
4.[tex]x^2+\frac{b}{a}x+\frac{1}{2}\left (\frac{b}{a} \right )^2=-\frac{c}{a}+\frac{1}{2}\left (\frac{b}{a} \right )^2 [/tex]

5.[tex]\left (x+\frac{b}{a} \right )^2=-\frac{c}{a}+\frac{1}{2}\left (\frac{b}{a} \right )^2 [/tex]

I'm going to split this step up because this is where I run into lots of problems of my understanding of deriving and deriving the quadratic formula.

So, I factored the left side, but can someone explain to me the logic and the why behind this? why is THIS correct? how does it WORK? I do not understand why it factors to [tex]\left (x+\frac{b}{a} \right )^2 [/tex]=[, and therefor I cannot attach it to my intuition, I'm just going off memorization of rules and rote, that an't the way I do things! I need to UNDERSTAND!

EDIT 1: I do not know WHY! the last four tex coded equations aren't working! can someone help me fix that? it would help others in reading the problem I'm presenting, geeze... PLEASE bear with that until I figure out a way to get it fix, I've tried many things over the course of hours and no avail. :(

Mod note: LaTeX fixed[tex]\left(x+\frac{b}{a} \right)^2=-\frac{c}{a}+\frac{b^2}{2a^2}[/tex]

what I don't understand here is why the a has a degree of 2.

[tex]x+\frac{b}{a} =\pm \sqrt{-\frac{c}{a}+\frac{b^2}{2a^2}}[/tex]

ugh...

[tex]x=\frac{b}{a} \pm \sqrt{-\frac{c}{a}+\frac{b^2}{2a^2}}[/tex]
[tex]x=\frac{b}{a} \pm \sqrt{ \frac{b^2}{2a^2}+\frac{c}{a}}[/tex]

These two, are where I get stuck, I don't understand how to form the [tex]b^2-4ac[/tex] in the radicand, I don't know how to form the [tex]2a[/tex] in the denominator, I don't know how the [tex]-b /pm [/tex] ever gets there. I just don't know how to derive the quadratic formula, though, I have tried many times in the past day and a half, I eventually forfeited, and realized that I may need some help figuring this out, so I came to the online community I thought would be best at helping me out with this, and that is physics forums, if you help me understand this, thanks in advance.

If you help me understand my second question, I'll more than appreciate you, I'll remember your name as well! I have been on these forums for a bit of time now, I rarely, rarely RARELY post. but that's going to change starting today, I'm going to hone my skills by asking for help, and I will contribute wherever I can, my goal is to help for every-time I get help, and to help 1000 times, lol, a hefty goal. But I got some bad karma that needs fixing, let's just leave it at that,

so here is the second question, it's about the definition of derivation, and understanding the concept of it, what it's used for, and how it works; and why it was developed. The history of derivation and the common pattern that is understood for every rule of derivation, for now, I only know the power rule of derivation, and I know that Gottfried Wilhelm von Leibniz notation d/dx or dy/dx doesn't indicate division.

at this point I'm trying my best to solve for x, I think one of my problems is that, my expierence and know-how when it comes to rational equations, radicals, rational exponents, and even factoring and complex numbers is kind of bad. With the exception of factoring, my factoring is par enough to do this, as far as working with rational/radical expressions as in this problem, I go off my gut/intuition, what I know, my past experience and what I think logically fits best, and I don't think it succeeds to say the least, lol.

My gut, and my intuition just knows this is wrong, which is why I came here, so I can be told why, and so I can make sense of this problem. It would do wonders for me in my mathematical journey, my fellow thinkers.

This is where I run into problems, now I know what it factors to, what I do not understand, is why it factors to it, it just feels like there is an x that is just sort of being tossed out the window, and as far as simplifying the left side, I struggling with that as well, I go off what I figure is logically correct but it just turns out to be wrong, can someone point out what I'm doing wrong? :( this is a good opportunity for advanced noobs to finally contribute on the physics forums, because we all know you asked more questions then helped solved, and that being a senior here requires that your stats equate to the opposite! ;D

4. Further, questions. WINKWINK. ;D

In both a general and mathematical sense, what is the absolute best definition of derivation? that is to say, what is the best way for a noob like me to understand what derivation is? I live by the philosophy that doing is understanding, I know how to derive a power equation, x^2 = 2x. I understand that, but I only understand it by fallowing rules that I memorized about it, I don't understand exactly why it's important, or why people decided to start deriving and how they used it in their life, I know that to derive to a conclusion, is to be presented with a question, usually a problem, and to come up with a solution. For instance, I'm lost, therefor I derived by logical consequence, a way to find my way home based on what I know and the means to use and work with what I know. but I still do not understand derivation, let alone "integration", aka the INTEGRAL. I understand limits, I've evaluated quite a few. Which is surprising because I'm just now retaking pre-calculus. And this is the very first thing they are teaching the class, is reviewing past concepts, such as working with radicals, rational expressions, complex numbers, and of course, quadratic equations.

But I'm a thirsty person, I thrive on satisfying my curiiosity I always have, can someone teach this humbled musician what it means to derive? and what an integral is? Teach me, CONCEPTUALLY. I beg you!
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
XxXCausticxXx said:
3.[tex]x^2+\frac{b}{a}x+=\frac{-c}{a}[/tex]
4.[tex]x^2+\frac{b}{a}x+\frac{1}{2}\left (\frac{b}{a} \right )^2=-\frac{c}{a}+\frac{1}{2}\left (\frac{b}{a} \right )^2 [/tex]

5.[tex]\left (x+\frac{b}{a} \right )^2=-\frac{c}{a}+\frac{1}{2}\left (\frac{b}{a} \right )^2 [/tex]

I'm going to split this step up because this is where I run into lots of problems of my understanding of deriving and deriving the quadratic formula.

So, I factored the left side, but can someone explain to me the logic and the why behind this? why is THIS correct? how does it WORK? I do not understand why it factors to [tex]\left (x+\frac{b}{a} \right )^2 [/tex]=[, and therefor I cannot attach it to my intuition, I'm just going off memorization of rules and rote, that an't the way I do things! I need to UNDERSTAND!
There's a reason you can't figure out why it factors like that. It's because, on both sides of the equation, you should have added [itex]\left (\frac{b}{2a} \right )^2[/itex] rather than [itex]\frac{1}{2}\left (\frac{b}{a} \right )^2[/itex].

This should help quite a bit. Also, don't forget to express the two terms on the right hand side in terms of a common denominator.

Chet
 
  • #3
When you're completing a square, you square the entire term after dividing by two.
you should get ##(x+\frac{b}{2a})^2 + \frac{c}{a} - \frac{b^2}{4a^2} = 0##
##(x+\frac{b}{2a})^2 = \frac{ab^2-4a^2c}{4a^3}## which should look very familiar after factoring out an a from the numerator and denominator.
 
  • #4
XxXCausticxXx said:

so here is the second question, it's about the definition of derivation, and understanding the concept of it, what it's used for, and how it works; and why it was developed. The history of derivation and the common pattern that is understood for every rule of derivation, for now, I only know the power rule of derivation, and I know that Gottfried Wilhelm von Leibniz notation d/dx or dy/dx doesn't indicate division. Can someone teach this humbled musician what it means to derive? and what an integral is? Teach me, CONCEPTUALLY. I beg you!


Assuming that you actually mean differentiation; the definition of a derivative is below:
##\lim_{Δx \to 0} \frac{f(x+Δx)-f(x)}{Δx}## which is similar to the ##\frac{Δy}{Δx}## that you are probably familiar with. When you're differentiating a function, you're basically finding the slope at a single point of a graph. You're doing this by making the change in x so small (Δx→0) that it is literally the slope of that point on your graph. The amazing thing about it is that instead of using that formula, there are many sets of rules, like the power rule that you know, that allows you to derive the function that will give you the slope at ANY point of a graph of the function that was differentiated. You cannot do this precisely with ##\frac{Δy}{Δx}##. The notation you described above ##\frac{dy}{dx}## simply means the derivative of y with respect to x. There are many notations for derivatives, but the most common are ##y'(x)## (y prime of x) and ##\frac{dy}{dx}## when you're dealing with 1 variable.

Integration on the other hand is the inverse of differentiation, hence why it's called anti-differentiation. The definition of an definite integral (reimann sum) is below:
##\lim_{n \to ∞}\sum\limits_{i=1}^n f(c_i) Δx_i##
What you are doing to taking the sum of a number of rectangles that approaches infinity from a to b, the formula below is the common, and more concise version of the formula above:
##\int_a^b f(x) dx ##
Where dx is very small. This allows the computation of a number of rectangles that approaches infinity under the curve of f(x) from a to b. This is a definite integral. An indefinite is literally an antiderivative, ex. if ##\frac{dy}{dx} = 2x##, then ##y = ∫2x dx = x^2 + C##, you must add a plus C (plus a constant) because there are an infinite number of functions whose derivative is ##2x##, ex.## x^2+9##, ##x^2+648##, etc.

I hope I've answered some of your questions :-)
Also, check out some videos of Khan Academy if you're curious about Calculus topics and concepts, that website is made to satisfy curiosity. You'll be a Calc pro in no time. :cool:
 
Last edited:
  • Like
Likes 1 person
  • #5
xxCaustic, Although it seems reasonable that you would derive something to get its derivative, that's not the usage in English. Instead, we say that you differentiate a function to get its derivative.

In your OP in this thread, you started with a quadratic equation, and from it, derived the Quadratic Formula. That is an appropriate use of "derived."
 
  • #6
XxXCausticxXx said:
4. Further, questions. WINKWINK. ;D

In both a general and mathematical sense, what is the absolute best definition of derivation?

Just to clear something up -- you were discussing a derivation of the quadratic formula. Derivation means "the act of deriving".

Then you started discussing calculus and derivatives. The process of calculating a derivative is called "differentiation", not derivation.

[Edit] Apparently Mark types faster than I do.
 
  • Like
Likes 1 person
  • #7
XxXCausticxXx said:

Homework Equations



[tex]ax^2+bx+c=0[/tex], where a is either 0 or not 0, if it is zero you skip the first step. (duh!)

The first step should just say divide by a. If a is zero you have a linear equation or constant function, not a quadratic. By definition of a quadratic equation, the leading coefficient of the ##x^2## term in non-zero.
 
  • #8
To complete the square it should be:

[tex]x^2 + \frac{b}{a}\ x + (\frac{b}{2a})^2 = (x + \frac{b}{2a})^2 = -\frac{c}{a} + (\frac{b}{2a})^2[/tex]

The next step with the right side

[tex]\frac{b^2}{4a^2} - \frac{c}{a}[/tex]
[tex]\frac{b^2}{4a^2} - \frac{4ac}{4a^2}[/tex]

See if you can continue from here.
 
  • #9
Yes, you should add the two fractions inside the square root.
 
  • #10
XxXCausticxXx said:
But I'm a thirsty person, I thrive on satisfying my curiosity I always have, can someone teach this humbled musician what it means to derive? and what an integral is? Teach me, CONCEPTUALLY. I beg you!

Alright, derivation is just like this. Pretend the math book asked you to: "Derive the quadratic formula from ...(so an' so)..." You just use the equation given and manipulate it in any way that looks promising and keep going. If you get stuck, then you go back to where you think you were wrong. Or, if the question was a word problem or any other indirect information, you just derive an equation from the information given. I hope you understand this.:redface:

Now, an integral is something you will learn in calculus:eek:. Which, if you are in grade eleven or twelve, you will learn it. (at the end of the year.) I wouldn't want to teach it to you now because it would spoil the fun. But what do you mean by conceptually?

P.S. - Musician? I am too.:biggrin:

Maybe you should start a new thread somewhere else where you can ask all the questions you want and people can answer you!
 
Last edited:
  • #11
I'll teach you everything about polynomials if you teach me how to write LaTeX.
 
  • #13
HomogenousCow said:
I'll teach you everything about polynomials if you teach me how to write LaTeX.

Everything about polynomials!? That's a tall order to fill.
 
  • #14
Mentallic said:
Everything about polynomials!? That's a tall order to fill.

Maybe he's a poly-no-it-al. :tongue2:
 

1. What is the quadratic formula?

The quadratic formula is a mathematical formula used to solve quadratic equations, which are equations in the form of ax^2 + bx + c = 0. It gives the values of x that make the equation equal to 0.

2. How is the quadratic formula derived?

The quadratic formula is derived using the process of completing the square, which involves manipulating the quadratic equation until it is in the form of (x + a)^2 = b. This allows us to isolate x and solve for its value.

3. Why is the quadratic formula important?

The quadratic formula is important because it provides a general method for solving any quadratic equation, regardless of its form. It is also used in various fields such as engineering, physics, and finance.

4. Can the quadratic formula be used for all quadratic equations?

Yes, the quadratic formula can be used to solve all quadratic equations. However, it may not always yield real solutions. In such cases, the equation has no real solutions.

5. Are there any other methods for solving quadratic equations?

Yes, there are other methods for solving quadratic equations such as factoring, completing the square, and graphing. However, the quadratic formula is considered the most reliable and efficient method.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
2
Views
863
  • Precalculus Mathematics Homework Help
Replies
5
Views
686
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
13
Views
310
  • Precalculus Mathematics Homework Help
Replies
12
Views
2K
  • Precalculus Mathematics Homework Help
Replies
6
Views
574
  • Precalculus Mathematics Homework Help
Replies
10
Views
617
  • Precalculus Mathematics Homework Help
Replies
4
Views
799
Back
Top