Algebra/Chapter 9/Quadratic Equation

Algebra

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