Help in getting the Quadratic Equation

I'm starting a chapter on Functions and they had the steps shown to reach the p-q equation.

$$x_{1,2} = -\frac{p}{2} \pm\sqrt{\left(\frac{p}{2}\right)^2 - q}$$

So I wanted to do the same with the Quadratic Equation. I'm using the base linear equation

$$ax+by+c = 0.$$

The solution I have so far is as follows:

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

$$x^2 + \frac{b}{a}x = -\frac{c}{a}$$

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

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

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

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

My problem comes from trying to solve the insides of the square root:

$$\sqrt{(\frac{b}{2a})^2 - \frac{c}{a}} = \sqrt{\frac{b^2}{4a2} - \frac{c}{a}}$$

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

$$= \sqrt{\frac{b^2 - 4ac}{\left(2a\right)^2}}$$

Then:

$$x_{1,2} = \frac{-\left(\frac{b}{2a}\right) \pm\sqrt{b^2 -4ac}}{2a}$$

but there is still the problem of the -(b/2a) outside of the sqrt.

What am I doing wrong? Also, Tex is awesome; is there a better way to do the 1,2 subscripts than _1,_2?

$$x_{1,2} = \frac{-\left(\frac{b}{2a}\right) \pm\sqrt{b^2 -4ac}}{2a}.$$

The solution goes

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

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

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A note on the TeX subscripts: try "x_{1,2}". As for the derivation, are you already familiar with "completing the square"? Otherwise, one thing you can try is to make the substitution $x=u-\frac{b}{2a}$, solve for u, and then reexpress the whole mess in terms of x. Good luck! – J. M. Aug 19 '10 at 13:20
The typical TeX way to do the 1,2 subscripts would be x_{1,2}, which yields $x_{1,2}$ as desired. In other words, you use curly braces, like so: "_{subscripts go here}". – Alex Basson Aug 19 '10 at 13:57
@Mangaldan: Thanks for the TeX tip! Yes, "completing the square" is what I had to use to get the quadratic equation, but I'm not sure what you mean. After all, I've already applied it earlier in the solution. – IAE Aug 19 '10 at 14:23
SB, your dividing of the depression term $-\frac{b}{2a}$ again with $2a$ was what threw me off. :) At least WWright has already pointed you in the proper direction. – J. M. Aug 19 '10 at 14:39

Sorry I don't know how to do tex on websites, but I'm trying to learn.

You just made a small mistake on the final step. In the second to last step, we actually have our full equation as:

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

Now we can collect the common factor of 1/2a and get: $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

Does that make sense?

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I learned TeX more or less by looking at the Tex that other users wrote. View the source code for this page and then you can see how I wrote the math in TeX. It's quite easy actually, and you can display it by starting and ending with . – IAE Aug 19 '10 at 13:52
WWright, this may be a useful crutch: codecogs.com/latex/eqneditor.php – J. M. Aug 19 '10 at 13:53
thanks for the help, give me a few minutes and it'll look right, hopefully :) – WWright Aug 19 '10 at 14:08
Looks good now, you've done it right! :) – J. M. Aug 19 '10 at 14:21

You factored 2a out of the square root and put it in the denominator without factoring it out of -(b/2a).

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