Algebra/Quadratic Equation

Derivation
The solutions to the general-form quadratic function $$ax^2+bx+c=0$$ can be given by a simple equation called the quadratic equation. To solve this equation, recall the completed square form of the quadratic equation derived in the previous section:
 * $$y=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}$$

In this case, $$ y = 0 $$ since we're looking for the root of this function. To solve, first subtract c and divide by a:
 * $$\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac{c}{a}$$

Take the (plus and minus) square root of both sides to obtain:
 * $$x+\frac{b}{2a}=\pm\sqrt{\frac{b^2}{4a^2}-\frac{c}{a}}$$

Subtracting $$\frac{b}{2a}$$ from both sides:
 * $$x=-\frac{b}{2a}\pm\sqrt{\frac{b^2}{4a^2}-\frac{c}{a}}$$

This is the solution but it's in an inconvenient form. Let's rationalize the denominator of the square root:
 * $$\sqrt{\frac{b^2}{4a^2}-\frac{c}{a}}=\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\sqrt{b^2-4ac}}{2|a|}=\pm\frac{\sqrt{b^2-4ac}}{2a}$$

Now, adding the fractions, the final version of the quadratic formula is:

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

This formula is very useful, and it is suggested that the students memorize it as soon as they can.

Discriminant
The part under the radical sign, $${b^2-4ac}$$, is called the discriminant, $$\Delta$$. The value of the discriminant tells us some useful information about the roots.
 * If $$\Delta>0$$, there are two unique real solutions.
 * If $$\Delta=0$$, there is one unique real solution.
 * If $$\Delta<0$$, there are two unique, conjugate imaginary solutions.
 * If $$\Delta$$ is a perfect square then the two solutions are rational, otherwise they are irrational conjugates.

Word Problems
Need to pull word problems from http://teachers.yale.edu/curriculum/search/viewer.php?id=initiative_07.06.12_u&skin=h