Babylonian Mathematics/Examples/Sextic Equation
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Example of Babylonian Mathematics
Simpify the system of simultaneous equations:
| \(\text {(1)}: \quad\) | \(\ds x y\) | \(=\) | \(\ds a\) | |||||||||||
| \(\text {(2)}: \quad\) | \(\ds \dfrac {b x^2} y + \dfrac {c y^2} x + d\) | \(=\) | \(\ds 0\) |
Solution
- $x^3 = \dfrac a {2 b} \paren {-d \pm \sqrt {d^2 - 4 a b c} }$
Proof
| \(\ds y\) | \(=\) | \(\ds \dfrac a x\) | from $(1)$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds b x^2 \dfrac x a + \dfrac {a^2} {x^2} \dfrac c x + d\) | \(=\) | \(\ds 0\) | substituting for $y$ in $(2)$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac {b x^3} a + \dfrac {a^2 c} {x^3} + d\) | \(=\) | \(\ds 0\) | tidying up | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds b x^6 + a d x^3 + a^3 c\) | \(=\) | \(\ds 0\) | multiplying through by $a x^3$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x^3\) | \(=\) | \(\ds \dfrac a {2 b} \paren {-d \pm \sqrt {d^2 - 4 a b c} }\) | simplifying |
$\blacksquare$
Historical Note
David Wells reports in his Curious and Interesting Puzzles of $1992$ that this example (once translated into the modern notation as presented here) comes from a Babylonian clay tablet dating from about $1600$ BCE.
He gives no further details.
Sources
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Dividing a Field