# A ridiculous question

1. Dec 23, 2004

### 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.

2. Dec 23, 2004

### 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..

3. Dec 23, 2004

### Clausius2

Well, now I realise 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!!

Last edited: Dec 23, 2004
4. Dec 23, 2004

### arildno

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

5. Dec 23, 2004

### Clausius2

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

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

6. Dec 23, 2004

### 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:

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

Last edited: Dec 23, 2004
7. Dec 23, 2004

### Clausius2

A bit more elegant, Deathknight.

8. Dec 23, 2004

### arildno

GOD JUL!!!!!!!!!!!!!!1

9. Dec 23, 2004

### tongos

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

10. Dec 23, 2004

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

11. Dec 23, 2004

### kreil

You can work backwards and show it as well:

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

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

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

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

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

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

12. Dec 23, 2004

### Chrono

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

13. Dec 23, 2004

### Ethereal

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

14. Dec 24, 2004

### Muzza

But, um, Deathknight DID complete the square?

15. Dec 24, 2004

### dextercioby

Check this out.
cubic

depressed quartic

full quartic

Daniel.

16. Dec 24, 2004

### Chrono

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

17. Dec 24, 2004

### Hurkyl

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

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