Early Forms of Solution to Quadratic Equation - History Essay Example

 

 

Need

essay sample on "Early Forms of Solution to Quadratic Equation"

? We will write a cheap essay sample on "Early Forms of Solution to Quadratic Equation" specifically for you for only $12 - Early Forms of Solution to Quadratic Equation introduction.90/page

More History Essay Topics.

 

Quadratic Equation

It is claimed that the Babylonians were the first people to solve a quadratic equation. But during the times of the ancient Babylonians, there was no such thing as quadratic equation but they have problems that are solved using the approach of quadratic equation (O’Connor & E F Robertson, 1996).  This method is very much used for finding the sides of a rectangle when only the difference between the sides and the area of the rectangle are given. To illustrate how the Babylonians method works, consider a rectangle with its length d greater than the width and an area A. the solutions to this problem are as follows:

Let L be the length and W be the width of the rectangle. Using the conditions given,

L=W+d               (1)

A=LW                 (2)

Substituting L to (2) and collecting terms to the left side of the equation, we arrive at

W2+Wd-A=0

Using the quadratic formula and rejecting the negative root, we have

W=(-d+√(d2+4A))/2

L=W+d=(d+√(d2+4A))/2

The Babylonians also arrive at this answer using a series of steps. First, they take half the difference between the sides, d/2. Then they square the result, giving d2/4. They add this to the area and take the square root of the sum, arriving at √(d2/4+A) or √(d2+4A)/2. They subtract half the difference to the result to find the width and add half the difference to the result to get the length. The answers are

W=(-d+√(d2+4A))/2

L=W+d=(d+√(d2+4A))/2,

which are the same answer we get using quadratic formula.

Aside from the Babylonians, the Arabs also find a way of solving quadratic equations. But unlike the Babylonians, who solve for the sides of the rectangle, they use the sides and areas of squares and rectangles to solve a quadratic equation.  The mathematician who introduced this approach is al-Khwarizmi. For example, to solve the root of the quadratic equation x2+20x=300, consider a square having side of length x. Putting four 5 by x rectangles on the square, we arrive at a figure below, having a total area of x2+5x+5x+5x+5x or x2+20x or 300.

By adding 5 by 5 squares on each side, we will have a square that looks like the figures below.

The area of the resulting square is 300+25+25+25+25 or 400. The side of the bigger square is √ 400 or 20. From the figure, we could see that this side is 10 longer than x, and so we have x=10. This method is similar to the modern method of completing the square.

Exercises

A rectangular plate is 5 inches longer than it is wide and has an area of 24 square inches. Use the Babylonian method to find the length and the width, indicating each step as you go.
Here’s a historical quadratic equation that has a geometric source. A straight rod is 1 unit long. You have to cut the rod at a distance x from the left end as in the picture

where x is the longer of the two pieces. The cut must be made so that the ratio of the piece of length x to the entire length is the same as the ratio of the shorter piece to the longer piece. At what distance x should you make the cut? Have you seen this number before?

Answer

The difference between the sides of the rectangle is 5. Half of 5 is 2.5. Adding its square, 0.0625, to 24 and getting the square root, we get 4.5. Subtracting 2.5 from 4.5, we get the width, which is 3, and by adding 2.5 to the 4.5 we get the length, which is 8. Therefore, the sides of the rectangle are 3 inches and 8 inches.
It is given that the ratio between the ratio of the piece of length x to the entire length is the same as the ratio of the shorter piece to the longer piece. Because the total length of the rod is 1, the shorter piece should be 1-x. using ratio and proportion, we have
x/1=(1-x)/x

Multiplying both sides of the equation by (1)(1-x),

(x)(x)=(1-x)(1)

x2+x=1

1/4
x
1/4
1/4
Area=x2
1/4
1/4
x
1/4
Consider a square with sides of length x. By adding four ¼ by x rectangles, we get a figure that have an area of x2+4(x)(1/4) or x2+x or 1. Adding ¼ by ¼ square on each side, we get a square that have a total area of 1+4(1/4)2 or 5/4. The side of the larger square is √(5/4) or (√5)/2. Because (√5)/2 is ½ longer than x, x must be ½-(√5)/2. This number is popularly known as phi, or the golden ratio and has many applications to architecture, arts, mathematics and other fields of science (Knott, 2007).

 

References

Knott, R. (2007). The Golden section ratio: Phi. Retrieved December 13, 2007 from http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/phi.html#golden.

O’Connor, J.J., Robertson, E.F. (1996). Quadratic etc equations. Retrieved December 13, 2007 from http://www-groups.dcs.st-and.ac.uk/~history/HistTopics /Quadratic_etc_equations.html.

 

 

Haven't found the Essay You Want?

Get your custom essay sample

For Only $13/page