TO describe the Locus of a cubick Equation.
A Cardanick Equation convenient for the purpose, (viz. such as shall have the dioristick limits rational) must have the Coefficient of the roots to be the triple of a square number such is a³-48a=N.
Assume a rank of roots in Arithmetical progression, and raise resolvends thereto a³-48a=N or resolvends.
| R | N | ||
| Such are | 1 | 1 | -48=47 |
| 2 | 8 | -96=88 | |
| 3 | 27 | -144=17 | |
| 4 | 64 | -192=128 | |
| 5 | 125 | -240=115 | |
| 6 | 216 | -288=72 | |
| 7 | 343 | -336=+7 | |
| 8 | 512 | -384=+128 | |
| 9 | 729 | -432=+297 |
Draw a Base line and a perpendicular thereto, and from O in the Base line prick the negative resolvends downwards, and the affirmative ones upwards, and raise their roots upon them as ordinates, a Curve passing through the same is one Moity of the Curve or Locus on the right hand for affirmative roots, and the other moity on the left hand is described in the same manner by assuming a rank of negative roots, and raising resolvends thereunto. The Curve Fig. 4. may give a resemblance of the thing.
And 16 the third part of the Coefficient of the roots cubed is equal to the square of 64 half the resolvend, or dioristick limit. Which in composing of Cardans canon is always substracted from the square of half the absolute, as in the example following.
If I were to find the root belonging to the resolvend 297
The square of half thereof is 22052¼
The square of 64 half the dioristick Limit 4096
The difference is 17956¼
And the rule is 148½ + √17956¼.
148½ + √17956¼.
That is in a quadratick Æquation, if 297 were the sum of the two roots and 64 the root of the Rectangle: then if from the square of half the sum, the rectangle be subducted, there remains the square of half the difference of the
