- x
^{2}-px+q=0 (CI, D) - x
^{3}-px+q=0 (CI, D, A) - x
^{4}-px+q=0 (D,CI,K) - x
^{4}-px^{3}+q=0 (CI, K) - x
^{5}-px^{2}+q=0 (A, D, K, BI) - Different choice of f and g
- Non algebraic equations

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.

- Divide both parts of equation by x and move p to another side. You get x+q/x=p.
- 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.
- Move cursor into position where numbers on CI and D scales sum to the value of p.
- Numbers on CI and D under the cursor are the roots of equation.

- 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)

The important case of the above equation is p=0. In this case equation
becomes x^{2}+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

- Place q on CI scale over D index. You may also place index of CI over q on D with the same effect.
- Search the same number on CI and D -- this will be √-q (square root of -q).

- Move cusor over q on scale D.
- Move
**slide**into such position that number on C under cursor is equal to number on D under C index. - This number is √q (square root of q).

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.

And completely another method is described in trigonometric section.

x^{3}-px+q=0 is equivalent to x^{2}+q/x=p

Solving this eqiuation is no more difficult than solving quadratic:

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

x^{3}-px^{2}+q=0 is equivalent to x+q/x^{2}=p

- Place CI index over q on the A scale.

Uncommon thing here that CI will represent x here while A representing q/x^{2}. - Match with cursor numbers on CI and A that add up to p
- The root is the number under cursor on CI scale.

- 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.

"Search with slide" technique, which was successfully used to calculate square root, will allow to use C instead of CI. (See http://www.sliderules.clara.net/a-to-z/tys/tys05.htm for details.)

- 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.

- Place q on CI scale over D index (or CI index over q on D scale).
- Match with cursor numbers on CI and K that add up to p
- Read root from D scale under cursor.

- Place CI index over
*q*on the K scale.

CI will represent x here while K represents*q/x*.^{3} - Match with cursor numbers on CI and K that add up to p
- The root is the number under cursor on CI scale.

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

- 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).

x^{5}-px^{2}+q=0 is equivalent to x^{3}+q/x^{2}=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.

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

x^{5}-px^{3}+q=0 is equivalent to x^{2}+q/x^{3}=p

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

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

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 r^{2}and and compare it to the value on C under the hairline.

4) when the value of r^{2}(on A) equals to the value under the hairline on C, r is the required fifth root.

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.