Algebra is a tool to help us solve equations and puzzles. Al Khwarizmi, (750-850) from Persia, was the co-founder of algebra. He found the first systematic solution of the quadratic equation.

In order to uncode a quadratic equation, we must first find what x can be equal to. A quadratic equation usually looks like this:

$a{x}^{2}+bx+c=0$In order to solve for x, lets assume a = 1. The first step is to move "c" to the right side of the equation. This shows:

${x}^{2}+bx=-c$Take a look at the graphic to the upper-right. This uses squares to illustrate the left side of the equation. The x squared is shown by a box that has a length of x on both sides. There is also a length of b divided by 2, which when multiplied by x and added together, results in bx--the middle term.

Use your mouse to hover over the top right corner of the graphic. You will see what "completes" this square--makes it a real square--and will help us solve for x. This step is the beautiful part. By adding a number to both sides of the equation, the left side becomes factorable:

${x}^{2}+bx+\frac{{b}^{2}}{4}=\frac{{b}^{2}}{4}-c$When we factor the left side, we have:

${\left(x+\frac{b}{2}\right)}^{2}=\frac{{b}^{2}}{4}-c$By multiplying the term c on the right side by $\frac{4}{4}$ (which is a legal step because $\frac{4}{4}$ is the same thing as multiplying by one), we can simplify it to:

${\left(x+\frac{b}{2}\right)}^{2}=\frac{{b}^{2}-4c}{4}$Taking the square root of both sides we find:

$x+\frac{b}{2}=\pm \frac{\sqrt{{b}^{2}-4c}}{2}$Moving $\frac{b}{2}$ to the right side of the equation we have:

$x=\frac{-b\pm \sqrt{{b}^{2}-4c}}{2}$With the quadratic equation we can find any solutions, or y-intercepts (where y = 0 on a graph), for x.

DesCartes discovered a special relationship between algebra and geometry. He found you could graph a quadratic and find its roots by seeing where the curve crossed the x-axis.

A cubic can be solved with some special manuevering, making it similar to the quadratic. The cubic: ${x}^{3}+{x}^{2}-12x=0$ can be solved by pulling an x out of the left side of the equation. This makes: $x\left({x}^{2}+x-12\right)=0$ At this point we have a solution of x = 0. After we factor we find two solutions: x = 3 and x = -4.

Some quartics can be solved by substituting z for ${x}^{2}$ , and can result in some imaginary numbers. Other more standard quartics are solved using Pascal's triangle (Weisstein, Eric W. "Pascal's Triangle." From MathWorld--A Wolfram Web Resource) to help us with the coefficients.

This page is based on a lecture by Oliver Knill, Harvard University, on 27 February, 2012.

References: Lecture Handout Lecture Worksheet