July 16th 1670.

Worthy SrSir

I sometimes thought to have altered & enlarged Kinkhuysedn his discourse upon surds but judging those examples I added would in some measure supply his defects I contented my selfe wthwith doing that onely. But since you would have it more fully done, if the booke goe not immediately into yethe presse I desire you'le send it back wthwith those notes I have made (since you are resolved to print them also) & I will doe something more to it or if you please to send all but the first sheete or two, while ytthat other is in the press is printing, Ile reveiw the rest & not only supply yethe wants about surds but that about Æquations soluble by trisection, & somthing more I would say in the chapter [Quomodò quæstio aliqua ad æquationem redigatur.] that being the most requisite & desirable doctrine to a Tyro & scarce touched upon by any writer unles in generall circumstances bidding them onely Nota ab ignotis non discernere & adhibere debitum ratiocionium.

As to Fergusons rendering the roots of Æquations soluble by trisection, his defect will appeare by example. Let us take his 2d ${\mathrm{x}}^3=6\mathrm{x}+4$, in pag 12. In order to solve this hee bidds extract yethe cubick root of these binomiums $2+\sqrt{}-4$, & $2-\sqrt{}-4$ To doe this his rule pag 4 is: " Multiply yethe binomium by $1000$, put in pure numbers &c: Now $2+\sqrt{}-4in1000$ makes $2000+\sqrt{}-4000000$, but to put this in pure numbers is impossible for $\sqrt{}-4000000$ is an impossibele quantity & hath noe pure number answering to it. His rule therefore failes & The like difficulty is in his 3d example & in all other such cases. In generall I see not wtwhat hee hath done more then in Cardans rules. For in this instance Cardans rule will give you $\mathrm{x}=$$=\sqrt{}c:\overline{2+\sqrt{}-4}+\sqrt{}c:\overline{2-\sqrt{}-4}$. in wchwhich yethe only difficulty as before is to extract yethe rootes of yethe binomiums $2+\sqrt{}-4$ & $2-\sqrt{}-4$. Which roots indeed are $-1+\sqrt{}-1$ & $-1-\sqrt{}-1$, as he assignes them, but tells not how to extract them. Nor doe I see wtwhat hee hath done more then Descartes in his Solution of biquadratick Equations: for both goe yethe same way to worke in reducing them first to Cubick & then to quadratick æquation. Lastly I see not in what case his rules will render rothe roots of cubick or biquadratick Æquations in proprio genere where those of Cardan or Descartes will not. But in hast I must take my leave remaining