Book:Tablet/BM 13901/Examples

Example from BM $\mathit { 13901 }$

I have added up seven times the side of my square and eleven times the area, [getting] $6; 15$ [that is, $6 \dfrac 1 4$ in modern notation].

The numbers here are base $60$.

The solution as given on the tablet is as follows:

You write down $7$ and $11$.

You multiply $6; 15$ by $11$, [getting] $1, 8; 45$.

You break off half of $7$, [getting] $3; 30$ and $3; 30$.

You multiply, [getting] $12; 15$.

You add [this] to $1, 8; 45$ [getting] result $1, 21$.

This is the square of $9$.

You subtract $3; 30$, which you multiplied, from $9$.

Result $5; 30$.

The reciprocal of $11$ cannot be found.

By what must I multiply $11$ to obtain $5; 30$?

[The answer is] $0; 30$, the side of the square is $0; 30$.

In modern notation, we write:

$a = 11, b = 7, c = 6; 25 = 6 \dfrac 1 4$

$a x^2 + b x = c$

with the particular values given.

The technique given is:

Multiply $a$ by $c$ to get $a c$.

Divide $b$ by $2$, to get $\dfrac b 2$.

Square $\dfrac b 2$ to get $\dfrac {b^2} 4$.

Add this to $a c$, which is $a c + \dfrac {b^2} 4$.

Take its square root $\sqrt {a c + \dfrac {b^2} 4}$.

Subtract $\dfrac b 2$, which makes $\sqrt {a c + \dfrac {b^2} 4} - \dfrac b 2$.

Divide this by $a$, to give the answer $x = \dfrac {\sqrt {a c + \frac {b^2} 4} - \frac b 2} a$.

This is equivalent to:

$x = \dfrac {- b + \sqrt {b^2 - 4 a c} } {2 a}$

which is the Quadratic Formula for the equation $a x^2 + b x + c = 0$, substituting $-c$ for $c$.

2008: Ian Stewart: Taming the Infinite ... (previous) ... (next): Chapter $4$: Lure of the Unknown: Equations