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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?

Answer:

Instead of

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

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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    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*}".2010-08-19

2 Answers 2

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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 $\frac1{2a}$ and get: $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

Does that make sense?

  • 0
    Looks good now, you've done it right! :)2010-08-19
1

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