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Arithmetic Algebra Biquadratic Roots (Determining the 4th Root of a Number)

Main Arithmetic Algebra Page

Page Contents

  1. Roots of the Fourth Degree
  2. Rationale
  3. Application to Algebra
  4. Application to Arithmetic

Roots of the Fourth Degree

The fourth degree is sometimes called biquadratic or quartic. Having understood the principle with square roots and cube roots, the principle with quartic roots and above becomes clear.

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:
quarticRoots(a+b)pow4.gif [1.1]

Rationale

To understand how to extract biquadratic or quartic roots, and to give us a vocabulary in the ensuing sections, we can study the following. The answers are obvious and evident, but the method is new to those who do not know how to extract quartic roots.
Answera+bAnswer so far
a4+4a3b+6a2b2+4ab3+b4Take the 4th root from the first group.
This is a
a
a4Subtract to form a remainder
4a3b4a3b+6a2b2+4ab3+b4We seek b, such that
(4a3+6a2b+4ab2+4b3)b is equal to the remainder. If b is relatively small compared with a, then we can use 4a3 to divide into the remainder to find b
a+b
4a3b+6a2b2+4ab3+b4We compute (4a3+6a2b+4ab2+4b3)b, and subtract it to get the new remainder.
0No more remainders, we are done. The quartic root of the expression is (a+b)

Observations
  1. First we find the quartic root of the first group (a)
  2. We seek b, dividing 4a3 into the remainder.
  3. We subtract 4a3b+6a2b2+4ab3+b4 from the remainder
  4. We continue from step 2.
The above is true for algebra, but slightly different for arithmetic.

Application to Algebra


1+x/4-3/3x2x2+7/128x3...Answer so far
1+xa4=11
1Subtract 1
4*x/4xRemainder
x+3/8x2+x3/16+x4/256Find a b such that 4b=x
b=x/4
4a3b+6a2b2+4ab3+b4=
413(x/4)+612(x/4)2+41(x/4)3+(x/4)4=
x+3/8x2+x3/16+x4/256
Subtract this from the remainder above to create the new remainder below.
1+x/4
4(1+x/4)3(-3/8x2)-3/8x2-x3/16-x4/2564(1+x/4)3b=-3/8x2...
b=-3/32x2
1+x/4-3/32x2
-3/8x2-9/32x3...a=(1+x/4)
b=-3/32x2
(Working only as far as x3)
4a3b+6a2b2+4ab3+b4=
4(1+3/4x+3x2/16+x3/64)(-3/32x2)+...
-3/8x2-9/32x3...
Subtract this from our remainder above to create a new remainder below.
1+x/4-3/32x2 +7/128x
4(1+x/4-3/32x2)37/128x37/32x3...

To the tenth term (1+x)1/4 is:
quarticRootsBinomial.gif [2.1]

Application to Arithmetic

The difference between this and the other roots is that we use this basic equation
quarticRoots1.gif [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). 

Answer913Answer so far
6948 3727 7761Divide the number in grous of 4
6561We need to find the quartic root of 6948, for our variable a
This means two rough square roots.
√694880
√809
94=6561
This must be right because a cannot be 10 (a number between 0 and 10 to the fourth power is less than 10000), and 94 is less than the number.
So a=9
Subtract and bring down the next group of 4.
9
387 37274000a3b≤3 873 727
4000(93)=2 916 000
b=1
91
29649614000a3b+600a2b2+40ab3+b4=
4000(729)+600(81)+40(9)+14=
2964961
Subtract to get the new remainder:
3014284000390876677614000a3=3014284000
3014284000b=9087667761
b=3
913
90876677614000(91)33+600(91)29+40(91)27+34=
9087667761
0We are done. The fourth root is 913









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