Algebra/Chapter 9/Quadratic Equation

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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:

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

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:

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, $\mathrm{\Delta }$ . The value of the discriminant tells us some useful information about the roots.

• If $\mathrm{\Delta }>0$ , there are two unique real solutions.
• If $\mathrm{\Delta }=0$ , there is one unique real solution.
• If $\mathrm{\Delta }<0$ , there are two unique, conjugate imaginary solutions.
• If $\mathrm{\Delta }$ 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