1. ## Can someone look this over please?

So, I was given the task of proving the quadratic formula for homework... I think I've pretty much got it, but I'd like to be sure, so if someone can look over my working it'd be much appreciated.

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

Completing the square gives;

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

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

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

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

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

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

2. Originally Posted by SuperCalculus
So, I was given the task of proving the quadratic formula for homework... I think I've pretty much got it, but I'd like to be sure, so if someone can look over my working it'd be much appreciated.

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

Completing the square gives;

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

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

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

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

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

$\displaystyle x = - \frac{b}{2a} \pm \frac{\sqrt {b^2 - 4ac}}{2a} = \frac{-b\pm \sqrt {b^2 - 4ac}}{2a}$
Pretty much but the $\displaystyle \pm$ sign should also be in the second last step

interestingly it's different to how I derive it still correct ^_^

3. Originally Posted by e^(i*pi)
Pretty much but the $\displaystyle \pm$ sign should also be in the second last step

interestingly it's different to how I derive it still correct ^_^
Thank you ^.^. How do you derive it?

4. Originally Posted by SuperCalculus
Thank you ^.^. How do you derive it?
I divide by a initially

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

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

$\displaystyle \left(x+\frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} = 0$

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

Then as you did it