Algebra/Quadratic Equation

Derivation

The solutions to the general-form quadratic function $a{x}^2+bx+c=0$ 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:

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

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.