Solving Quadratics Han Dynasty Style

How did the ancient Chinese solve quadratic equations during the Han dynasty (206 BC - 220 AD)? There is a quadratic equation generated from a geometry problem in the *Jiuzhang Suanshu* (Gougu 20).

今有邑方不知大小，各中開門。出北門二十步有木。出南門十四步，折而西行一千七百七十五步見木。問邑方幾何？

There is a square town of unknown dimensions. There is a gate in the middle of each side. There is a tree located 20 bu outside the North Gate. If one leaves the town by the South Gate, walks 14 bu due south, then walks due west for 1775 bu, the tree will just come into view. What are the dimensions of the town?

Check out the problem in this article here. In this article, I solved the quadratic equation

x
2

34
x
−
# 71000
0

{\displaystyle x^2 + 34x - 71000 = 0 }

using the quadratic formula. However, the ancient Chinese did not solve this problem using the quadratic formula. Rather, the mathematicians of the Far East devised an algorithm called the Horner-Ruffini method in the West (nearly two millennia before its discovery in the West). By the Song dynasty (960 - 1279 AD), Chinese mathematicians extended this algorithm to solve polynomial equations up to the tenth degree.

To solve the above quadratic equation using the Horner-Ruffini method, write it like this

x
2

34
# x

{\displaystyle x^2 + 34x = 71000. }

**Iteration 1**

For the first iteration, the quadratic term should be the dominant term. So search for perfect squares such that

a
2

<
71000
<

b
2

{\displaystyle a^2 < 71000 < b^2 }

.

Notice

200
2

<
71000
<

300
2

{\displaystyle 200^2 < 71000 < 300^2 }

. Let

# x
y
200

{\displaystyle x = y + 200 }

. Then make the transformation.

(
y
200

)
2

34
(
y
200
# )
71000

{\displaystyle (y+200)^2 + 34(y+200) = 71000 }

y
2

400
y
40000
34
y
# 6800
71000

{\displaystyle y^2 + 400y + 40000 + 34y + 6800 = 71000 }

y
2

434
# y
24200

{\displaystyle y^2 + 434y = 24200 }

**Iteration 2**

For the second iteration, the linear term should be the dominant term. So search for values of

a
,
b

{\displaystyle a, b}

 such that

434
a
<
24200
<
434
b

{\displaystyle 434a < 24200 < 434b }

.

Notice

434
(
50
)
<
24200
<
434
(
60
)

{\displaystyle 434(50) < 24200 < 434(60) }

. Let

# y
z
50

{\displaystyle y = z + 50 }

. Then make the transformation.

(
z
50

)
2

434
(
z
50
# )
24200

{\displaystyle (z+50)^2 + 434(z+50) = 24200 }

z
2

100
z
2500
434
z
# 21700
24200

{\displaystyle z^2 + 100z + 2500 + 434z + 21700 = 24200 }

z
2

534
# z
0

{\displaystyle z^2 + 534z = 0 }

Now this is easy to solve by factoring.

z
(
z
534
# )
0

{\displaystyle z(z + 534) = 0 }

Thus

# z
0

{\displaystyle z=0}

 or

# z
−
534

{\displaystyle z = -534 }

. Since

# x
y
200

{\displaystyle x = y + 200 }

 and

# y
z
50

{\displaystyle y = z + 50 }

, it follows that

# x
z
250

{\displaystyle x = z + 250 }

. Therefore,

# x
250

{\displaystyle
x=250
}

 or

# x
−
284

{\displaystyle x = -284 }

.

You might be wondering why bother going through such a labourious process to solve such a simple quadratic equation? Bear in mind that without the electronic calculator, solving square roots and quadratic equation is quite challenging. So during the Han dynasty, this method would be computationally more efficient than the quadratic formula!