Originally Posted by
topsquark By "quadratic" do you mean you need to derive the quadratic formula?
This method uses "Completing the squares." If you are not comfortable with this method then I'd suggest you look at some examples first.
$\displaystyle ax^2 + bx + c = 0$
$\displaystyle ax^2 + bx = -c$
$\displaystyle a \left ( x^2 + \frac{b}{a}x \right ) = -c$
Now complete the square:
$\displaystyle a \left ( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} \right ) = -c$
$\displaystyle a \left ( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} \right ) - a \cdot \frac{b^2}{4a^2} = -c$
$\displaystyle a \left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2}{4a} - c$
$\displaystyle a \left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2 - 4ac}{4a}$
$\displaystyle \left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2 - 4ac}{4a^2}$
$\displaystyle x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$
$\displaystyle x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$
$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
There might be a "cooler" way to derive this, but I think this is the most "transparent" method. (The one that clearly shows where everything comes from.)
-Dan