Question

The ancient Babylonians knew how to solve quadratic equations. Here is a problem from a cuneiform tablet found in a Babylonian school dating back to about 2000B.C.I have a reed, I know not its length. I broke from it one cubit, and it fittimes along the length of my field. I restored to the reed what I had broken off, and it fit 30times along the width of my field. The area of my field is 375square nindas. What was the original length of the reed? Solve this problem. Use the fact that 1ninda =cubits.

Short answer

The length of the reed is$x=\frac{1+\sqrt{61}}{2}\mathrm{cubits}$

Step-by-step solution

Step-1. Setup the model.

Let x is the length of the reed,w Is the width of the field, l length of the field,A area of the field.

$1\mathrm{ninda}=12 \mathrm{cubits}$

The area of field in cubits is

$\begin{array}{l}A=375 nind{a}^2\\ A=\frac{375 nind{a}^2}{1}\times \frac{144 cubit{s}^2}{1 ninda{s}^2}\\ A=54000 cubit{s}^2\end{array}$

Length of the field yield $l=60(x-1)$

The width of the field is $w=60x$

Step-2. Solve for the area.

Area of the rectangle$=L\times B$

$\begin{array}{l}A=w\times l\\ A=60(x-1)\times 60x\\ 54000=60x-60\times \left(60x\right)\\ 54000=3600x^2-3600x\\ 3600x^2-3600x-54000=0\end{array}$

Step-3. Solve for x.

Using quadratic formula, we get:

$\begin{array}{l}x=\frac{-(-3600)\pm \sqrt{{(-3600)}^2-4\left(3600\right)(-5400)}}{2\left(3600\right)}\\ x=\frac{3600\pm \sqrt{12960000+777600000}}{7200}\\ x=\frac{3600\pm \sqrt{790560000}}{7200}\\ x=\frac{3600\pm 3600\sqrt{61}}{7200}\\ x=\frac{1\pm \sqrt{61}}{2}\end{array}$

Hence, the length of the reed is $x=\frac{1\pm \sqrt{61}}{2}$.