Aside from the greater numbers, the biquadratic root algorithm is probably better behaved than the cubic, and therefore easier.

We use the same rationale as before except, of course we use:

[1.1]

Answer | a+b | Answer so far | |

a^{4}+4a^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4} | Take the 4th root from the first group. This is a | a | |

a^{4} | Subtract to form a remainder | ||

4a^{3}b | 4a^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4} | We seek b, such that (4a ^{3}+6a^{2}b+4ab^{2}+4b^{3})b is equal to the remainder. If b is relatively small compared with a, then we can use 4a^{3} to divide into the remainder to find b | a+b |

4a^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4} | We compute (4a^{3}+6a^{2}b+4ab^{2}+4b^{3})b, and subtract it to get the new remainder. | ||

0 | No more remainders, we are done. The quartic root of the expression is (a+b) |

Observations

- First we find the quartic root of the first group (a)
- We seek b, dividing 4a
^{3}into the remainder. - We subtract 4a
^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4}from the remainder - We continue from step 2.

1+x/4-3/3x^{2}x2+7/128x^{3}... | Answer so far | ||

1+x | a^{4}=1 | 1 | |

1 | Subtract 1 | ||

4*x/4 | x | Remainder | |

x+3/8x^{2}+x^{3}/16+x^{4}/256 | Find a b such that 4b=x b=x/4 4a ^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4}=4·1 ^{3}·(x/4)+6·1^{2}(x/4)^{2}+4·1·(x/4)^{3}+(x/4)^{4}=x+3/8x ^{2}+x^{3}/16+x^{4}/256Subtract this from the remainder above to create the new remainder below. | 1+x/4 | |

4·(1+x/4)^{3}·(-3/8x^{2}) | -3/8x^{2}-x^{3}/16-x^{4}/256 | 4·(1+x/4)^{3}b=-3/8x^{2}...b=-3/32x ^{2} | 1+x/4-3/32x^{2} |

-3/8x^{2}-9/32x^{3}... | a=(1+x/4) b=-3/32x ^{2}(Working only as far as x ^{3})4a ^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4}=4(1+3/4x+3x ^{2}/16+x^{3}/64)(-3/32x^{2})+...-3/8x ^{2}-9/32x^{3}...Subtract this from our remainder above to create a new remainder below. | 1+x/4-3/32x^{2} +7/128x³ | |

4·(1+x/4-3/32x^{2})^{3}·7/128x^{3} | 7/32x^{3}... |

To the tenth term (1+x)

[2.1]

[1.2]

Also, we divide the number into groups of 4 from the units, because a number between 0 and 10 raised to the fourth power is less than 10000 (four digits).

Example: Find the quartic root of 694837277761 by hand. (As the ancients used tables, we can (without cheating) use a calculator, if we wish, for the arithmetic only).

Answer | 913 | Answer so far | |

6948 3727 7761 | Divide the number in grous of 4 | ||

6561 | We need to find the quartic root of 6948, for our variable a This means two rough square roots. √6948≅80 √80≅9 9 ^{4}=6561This must be right because a cannot be 10 (a number between 0 and 10 to the fourth power is less than 10000), and 9 ^{4} is less than the number.So a=9 Subtract and bring down the next group of 4. | 9 | |

387 3727 | 4000a^{3}b≤3 873 7274000(9 ^{3})=2 916 000b=1 | 91 | |

2964961 | 4000a^{3}b+600a^{2}b^{2}+40ab^{3}+b^{4}=4000(729)+600(81)+40(9)+1 ^{4}=2964961 Subtract to get the new remainder: | ||

3014284000·3 | 9087667761 | 4000a^{3}=30142840003014284000b=9087667761 b=3 | 913 |

9087667761 | 4000(91)^{3}·3+600(91)^{2}·9+40(91)·27+3^{4}=9087667761 | ||

0 | We are done. The fourth root is 913 |

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