Algebra
 ← Completing the Square Quadratic Equation Binomial Theorem →

## 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:

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

In this case, $y = 0$ since we're looking for the root of this function. To solve, first subtract c and divide by a:

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

Take the (plus and minus) square root of both sides to obtain:

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

Subtracting $\frac{b}{2a}$ from both sides:

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

This is the solution but it's in an inconvenient form. Let's rationalize the denominator of the square root:

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

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, Δ. The value of the discriminant tells us some useful information about the roots.

• If Δ > 0, there are two unique real solutions.
• If Δ = 0, there is one unique real solution.
• If Δ < 0, there are two unique, conjugate imaginary solutions.
• If Δ 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