# The Quadratic Equation Process, is this wrong?

1. Feb 9, 2005

### mathzeroh

Hellooooo everybody. i have a very distinct feeling that there's something very wrong here...

OK, what i'm trying to do is to get to the quadratic equation through the standard form of a polynomial equation in the second degree.

watch this, and go to this website. it lists the process very briefly http://mathworld.wolfram.com/QuadraticEquation.html...in this picture http://mathworld.wolfram.com/qimg558.gif isn't it wrong to put the x+b/2a inside the parenthesis with the WHOLE thing raised to the second power?? i mean doesn't that mean that the WHOLE THING INCLUDING the 'x' gets squared?? that's wrong!! aha!! that's wrong!! i knew it!! i knew it!! Nah I'm just kidding there but i really do want to know if that's a typo or not. I think that when completing the square, you should only raise the "b" term in (the standard form) to the second power. like this:

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

$$(\frac{b}{2})^2$$

subtract the 'c' and bring it over to the other side, then complete the square:
$$ax^2+bx+(\frac{b}{2})^2=-c+(\frac{b}{2})^2$$

is this right, as opposed to the ones shown on that website? :uhh: :uhh:

Last edited: Feb 9, 2005
2. Feb 9, 2005

### StatusX

The site is right, but I'm not sure what you're trying to do. What is your next step?

3. Feb 9, 2005

### dextercioby

I'm sure what's on that site is not wrong.How about u finish your calculations...???

Daniel.

4. Feb 9, 2005

### mathzeroh

Hey, how's it going? i'm gonna go do that, i'll get back at you all.

5. Feb 9, 2005

### mathzeroh

okay guys im in a pot-hole.

here's my work:

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

$$ax^2+bx+(\frac{b}{2})^2=-c+(\frac{b}{2})^2$$

$$ax^2+bx+(\frac{b^2}{4})=-c+(\frac{b^2}{4})$$

$$\left(x-\frac{b}{2a}\right)^2=\left(\frac{-c+\frac{b^2}{4}}{a}\right)$$

Then i took the square root of both sides and got this:

(x-(b/2a))=\sqrt(-c+(b^2/4)/ \sqrt(a)

then i did that thing where you can't have a square root on the bottom (forgot what its called), and to do that, i multiplied the top and bottom of the right side of the equation with the square root of a over the square root of a (1 in essence).

then it wasn't looking like it was going in the right direction...help. :uhh: :uhh:

6. Feb 9, 2005

### StatusX

After you implicity divided by a between the last two (latex) steps, you no longer had a perfect square on the left. You need that last term to be b^2/4a^2. That's why they divided by a first. Once you correct that, your method becomes identical to theirs.

7. Feb 9, 2005

### dextercioby

Do you agree that the RHS of your last equation (written in tex) can be written
$$\frac{b^{2}-4ac}{4a}$$

Now extract square root...

Daniel.

P.S.Yes,an "a" is missing in the numerator...

8. Feb 9, 2005

### mathzeroh

i looked and looked and looked at it, but it didn't click.

how on earth did they go from here:
http://mathworld.wolfram.com/qimg557.gif

to here:
http://mathworld.wolfram.com/qimg558.gif

what happened to the x in the left in the second step?? i mean, it was, (b/a)x

9. Feb 9, 2005

### StatusX

expand the squared term and you'll see

10. Feb 9, 2005

### mathzeroh

11. Feb 9, 2005

### mathzeroh

hahah yeah, thanks man, i just saw that.

12. Feb 9, 2005

### dextercioby

As an advice for future "events",ALWAYS TRUST MATHWORLD from WOLFRAM.COM.

I haven't seen one single mistake on their site.

Daniel.

13. Feb 9, 2005

### mathzeroh

HOLY MOLY!!

its like it all just revealed itself right before my eyes!! IT WORKS!! BY GOD ALMIGHTY IT WORKS!! thanks for ur help! :)

thanks for the tip Daniel. i'll keep that in mind. mathworld rules.