V Algebra

Algebra

History

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.

Completing the Square

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} + b x + 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} + b x = - 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} + b x + \frac{b^{2}}{4} = \frac{b^{2}}{4} - c

When we factor the left side, we have:

{(x + \frac{b}{2})}^{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:

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

Taking the square root of both sides we find:

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

Moving $\frac{b}{2}$ to the right side of the equation we have:

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

With the quadratic equation we can find any solutions, or y-intercepts (where y = 0 on a graph), for x.

More Discoveries

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: $x^{3} + x^{2} - 12 x = 0$ can be solved by pulling an x out of the left side of the equation. This makes: $x (x^{2} + x - 12) = 0$ 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