# How to solve equatons with slide rule.

The general idea behind solving algebraic (and not only) equations is the following:

Convert your equation to form f(x)+g(x)=c, where both f and g can be represented by some scales of your rule with the same position of the slide.

Then move cursor along this scale until sum of the numbers on both scales will become equal c. Usually you can make just a few readings to determine x. Of course each scale can represent different ranges (orders) and have different signs -- keep this in mind during the operation.

## x2-px+q=0 (CI, D)

1. Divide both parts of equation by x and move p to another side. You get x+q/x=p.
2. Place q on CI scale over D index. You may also place index of CI over q on D with the same effect. D now represents x, CI represents q/x.
3. Move cursor into position where numbers on CI and D scales sum to the value of p.
4. Numbers on CI and D under the cursor are the roots of equation.
Note that you find both roots simultaneously. Instead of CI scale (that is absent on some rules) you can reverse slide and use reverted C scale instead.

### Example: x2-7.1x+5.8=0

• Place left index of CI over 5.8 on D.
• We are going to find values on D and CI that sum to 7.1.
• D has range from 5.8 to 10 (for higher orders the sum will be even more than 10, thus much more then 7.1), CI from 0.58 to 1.
• Try D=5.8 => CI=1, D+CI=6.8 => less than needed
• Try D=7 => CI=0.83, D+CI=7.83 => more than needed
• After one or two more tries find D=6.16, CI=0.94 -- equation roots (D+CI=7.1)

### Square root (CI,D)

The important case of the above equation is p=0. In this case equation becomes x2+q=0 that is x=√-q (q should necessary be negative in this case) and our search condition becomes x=-q/x. This means that you should match the same number on both CI and D scales. More exactly operations become

1. Place q on CI scale over D index. You may also place index of CI over q on D with the same effect.
2. Search the same number on CI and D -- this will be √-q (square root of -q).
This allows extracting square root on rules without A/B scales, but with pair of regular C/D scales (reversing C). Another application -- use A/BI scales instead of CI/D and read result from D to receive ∜-q (4√-q, forth degree root of -q).

#### Avoiding slide reverse (C,D)

Some slide rules (mostly circular, but there are also some "straight" ones) do not allow you to reverse slide. There is workaround still. C/D scale pair has useful property -- they are located close one to another and one may read values from D corresposponding to values of C without aid of cursor.
1. Move cusor over q on scale D.
2. Move slide into such position that number on C under cursor is equal to number on D under C index.
3. This number is √q (square root of q).
Again, this method may be applied to a pair of A/B scales and the result (fourth degree root) may be read from C (under cursor) or D (under index) scales.

One may decide that this method is too awkward and requires too much brain efforts (adding numbers, choosing left or right index, dealing with numbers' orders and signs) to be useful. But it requires only one slide move and you are not involved in calculation of intermediate results of many operations (including addition that cannot be easily performed with rule). With little practice you can solve typical quadratic equation in 5 seconds. (Add 2 seconds to get your rule from your pocket and you will get 7 seconds -- fastest method even today and in nearest future).

What is more important that modification of this method allows to solve cubic and some forth and fifth degree equations with almost the same efforts.

The flaw of this method, however, is that it cannot find complex roots of equation.

## x3-px+q=0 (CI, D, A)

x3-px+q=0 is equivalent to x2+q/x=p

Solving this eqiuation is no more difficult than solving quadratic:

1. Place q on CI scale over D index (or CI index over q on D scale).
2. Match with cursor numbers on CI and A that add up to p
3. Read root from D scale under cursor.
4. Note that value on CI (q/x) is the product of two other roots -- this may help you find them.

### x3-px2+q=0 (CI, A)

x3-px2+q=0 is equivalent to x+q/x2=p

1. Place CI index over q on the A scale.
Uncommon thing here that CI will represent x here while A representing q/x2.
2. Match with cursor numbers on CI and A that add up to p
3. The root is the number under cursor on CI scale.

### Generic cubic equation x3+Bx2+Cx+D=0

Setting x=y-B/3 you will get another equation with the x2 coefficient equal to 0. Solving it with the method described above and subtracting B/3 from its roots you will receive the roots of the original equation.

### Cubic root (D, CI, A)

Again, you can reverse C to get CI scale. When p=0 we get x = 3√-q (third degree root of -q).
• Place index of CI over q on D.
• Search the same number on A and CI with cursor.
• Read cube root of q from D scale under the hairline.
This method allows to calculate qube roots with rules that do not have K scale. (Eg rules with A, B, C, D scales only).
• Place the cursor index over the original number in scale A.
• Adjust the slide so that the number in scale B under X is exactly the same as the number in D opposite 1 (or 10) of C.
• The number so found is the cube root.
Both numbers 6 and 600 should be set on the left half of the A scale. Move slide to the right to find cube ruut from 6 and move slide to the left to find cubic root from 600.

## x4-px+q=0 (D,CI,K)

x4-px+q=0 is equivalent to x3+q/x=p
1. Place q on CI scale over D index (or CI index over q on D scale).
2. Match with cursor numbers on CI and K that add up to p
3. Read root from D scale under cursor.

## x4-px3+q=0 (CI, K)

x4-px3+q=0 is equivalent to x+q/x3=p
1. Place CI index over q on the K scale.
CI will represent x here while K represents q/x3.
2. Match with cursor numbers on CI and K that add up to p
3. The root is the number under cursor on CI scale.

### Fourth degree root (K, CI)

1. Place CI index over the operand on the K scale.
2. Match the same numbers on CI and K with cursor.
3. The root is the number under cursor on CI scale.

### Fourth degree root (A, BI, D)

Actually this method should be described under the quadratic equation chapter, since it calculates square root, using scales A and BI intead of D and CI and then you just have to read the result from D.
• put the slide upside down;
• push the inverted slide to put 1C under your number (for example 256) on A;
• with the hairline search the SAME NUMBER on A and B (here it is 16);
• read the fourth root on D (here it is 4).
Comparing to previous this method gives better precision (because it avoids use of low precision K scale), but requires somewhat irritating slide reversing.

## x5-px2+q=0 (A, D, K, BI)

x5-px2+q=0 is equivalent to x3+q/x2=p

Most probably you do not have BI scale on your rule. You will have to reverse your slide and use reversed B scale instead.

1. Place index of BI under q on A scale.
2. Match with cursor numbers on BI and K that add up to p.
3. Read value of x on D.

### x5-px3+q=0 (A, D, K, BI)

x5-px3+q=0 is equivalent to x2+q/x3=p

### Fifth degree root (A, D, K, BI)

The most useful application of the previous rule is its particular case p=0 that will allow you to calculate fifth degree root.

1. put the slide upside down;
2. push the inverted slide to put 1C under your number (for example 32) on A;
3. with the hairline search the SAME NUMBER on K and B (here it is 8);
4. read the fifth root on D (here it is 2).

I wish to express here my gratitude to Benoit Kammerer who showed me this method. See http://perso.wanadoo.fr/benoit.kammerer/regle-a-calcul/racine5.html for original description and picture.

### Fifth degree root (A,B,C,D)

This ingenious method requres no more scales than are present on classic Mannheim slide rule. John St Clair taught me this method. I never saw it in literature.

First, let's see how a number can be raised to fifth power using only A, B, C, D scales with a single slide movement:
1B->xA; xB=>A(y=x^2); |->yC; |=>A(z=x^5)

To calculate fifth degree root we will try to invert above algorithm. For example, let's calcaulate root of 90.

1) place the cursor over the value "X" (to find the fifth root of X, called r) on A, say 90 for example,.
2) Move index "1" on B to the right until it comes near the required value of about 2plus, say 2.4 (r).
3) look at the value on A of 2.4 on B (Index B1 on A) which is r2 and and compare it to the value on C under the hairline.
4) when the value of r2 (on A) equals to the value under the hairline on C, r is the required fifth root.

## Different choice of f and g

Duane Croft <dcroft@usit.net> noticed that the idea of solving f(x)+g(x)=c could be applied more directly to algebraic equations.

For x^2+bx=c, he put b on the C scale over the index of D. Then he called numbers on the D scale "x", numbers on the C scale "bx" (that is f(x)) and numbers on the A scale "x^2" (that is g(x)). Then he just run the cursor down until he find a place where the A scale plus the C scale ("x^2"+"bx") equals c and read the root off the D scale ("x").

For the other root this procedure should be applied again (or -c should be divided by first root to obtain second root).

Similarly, for x^3+bx=c, the same technique works, except that the K scale is used instead of the A.

## Non algebraic equations

f(x) and g(x) functions mentioned in the description of this method need not be necessary monomials. They can be transcendent functions as well. Common example of such functions expressible on most rules are log10x and ex. See Combinatorics chapter for an example of solving of transcendent equation.
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