Algebra ← Completing the Square Quadratic Equation Binomial Theorem →

## Derivation

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

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

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

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

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

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

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

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

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

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

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

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

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, ${\displaystyle {b^{2}-4ac}}$  , is called the discriminant, ${\displaystyle \Delta }$  . The value of the discriminant tells us some useful information about the roots.

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