How is the quadratic formula derived and why is it not always applicable?

[Clausius2]

I think all people here are going to laugh at me, but I'm going to be strong enough to formulate this question which should be made when I was a schoolboy.

What is the demonstration of the 2nd order equation solution?

$$ax^2+bx+c=0$$

then all teachers said:

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

Where does it come from? How could I demonstrate that formula? We memorized it so well that we haven't ask ourselves how is proved.

 

[arildno]

It's all about completion of squares:
1) Multiply your equation with "4a":
$$(2ax)^{2}+2*(2ax)b+4ac=0$$
2) I'll leave to you the next steps..
Hint: adding 0 in the form of $$b^{2}-b^{2}$$ might work nicely..

 

[Clausius2]

It's all about completion of squares:
1) Multiply your equation with "4a":
$$(2ax)^{2}+2*(2ax)b+4ac=0$$
2) I'll leave to you the next steps..
Hint: adding 0 in the form of $$b^{2}-b^{2}$$ might work nicely..

Well, now I realize I will never win the Fields Medal I think I'm going to delete this thread. :yuck:

Or doing it better, I will leave this here to show the people what kind of guy is an engineer guru here. :rofl: :rofl:

EDIT: MERRY CHRISTMAS!

 

[arildno]

BUENO NATIVITAD?
Does that mean "Merry Christmas" in Spanish?
That's what I intended, at least..

 

[Clausius2]

BUENO NATIVITAD?
Does that mean "Merry Christmas" in Spanish?
That's what I intended, at least..

My question was horrible, but your is spanish is worst.

Merry Christmas!=FELIZ NAVIDAD! in spanish

Sorry, but I haven't found any norwegian dictionary over here... :uhh:

 

[DeathKnight]

It can be also done like that:
By dividing both sides by 'a' we get:
$$x^2+\frac{b}{a}x+\frac{c}{a}=0$$
Taking the constant term to the right side of the equation and then completing the square. This will give us the following equation:
$$x^2+\frac{b}{a}x+(\frac{b}{2a})^2=-\frac{c}{a}+(\frac{b}{2a})^2$$
This is equal to:
$$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2)}$$
By making x the subject of the formula we get:

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

 

[Clausius2]

A bit more elegant, Deathknight.

 

[arildno]

My question was horrible, but your is spanish is worst.

Merry Christmas!=FELIZ NAVIDAD! in spanish

Sorry, but I haven't found any norwegian dictionary over here... :uhh:

GOD JUL!1

 

[tongos]

find the vertex by the use of the derivative and apply the symmetric shape to find the zeros.

 

[DeadWolfe]

This makes me kinda glad my high school teacher showed us derivations of everything.

 

[kreil]

You can work backwards and show it as well:

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

$$(x+\frac{b+\sqrt{b^2-4ac}}{2a})(x+\frac{b-\sqrt{b^2-4ac}}{2a})=0$$

$$x^2+{\frac{bx-x\sqrt{b^2-4ac}}{2a}}+{\frac{bx+x\sqrt{b^2-4ac}}{2a}}+\frac{b^2-b^2+4ac}{4a^2}=0$$

$$x^2+\frac{2bx}{2a}+\frac{4ac}{4a^2}=0$$

$$x^2+\frac{bx}{a}+\frac{c}{a}=0$$

$$ax^2+bx+c=0$$

 

[Chrono]

A bit more elegant, Deathknight.

Yeah, I like that way of doing it a lot better than completing the squares.

 

[Ethereal]

Can someone show the derivations for the solutions of cubic and quartic equations as well?

 

[Muzza]

Yeah, I like that way of doing it a lot better than completing the squares.

But, um, Deathknight DID complete the square?

 

[dextercioby]

Can someone show the derivations for the solutions of cubic and quartic equations as well?

Check this out.
cubic

depressed quartic

full quartic

Daniel.

 

[Chrono]

But, um, Deathknight DID complete the square?

I leaning towards his version of it other than the way we're all taught to do it.

 

[Hurkyl]

I leaning towards his version of it other than the way we're all taught to do it.

The problem is that it isn't useful in other situations where you want to complete the square... such as when doing this integral:

$$\int \frac{1}{3x^2 + 6x - 7} \, dx$$