Making sense of William Jones's solution of quadratic equations and notation

Question

In his History of Mathematical notation, Cajori (1993) writes about Jones's approach to the solution of a quadratic equation as follows:

William Jones, when discussing quadratic equations, says: "Therefore if $ \vee $ be put for the Sign of any Term, and $ \wedge $ for the contrary, all Forms of Quadratics with their Solutions, will be reduc'd to this one. If $ x x \vee a x \vee b = 0 $ then $ \wedge \frac 1 2 a \overline { \pm a a \wedge b } \! \, | ^ { \frac 1 2 } $."

Could you please help me make sense:

(a) What is meant with "the contrary" of "any sign"?

(b) Do I understand correctly that the "then" part supposed to be a solution to the preceding quadratic equation? If yes, how does this work exactly? How does it align with how we would symbolize the solution today?

Answer 1

By "contrary", it looks like he just means "opposite sign" - so if $\lor$ is "+" then $\land$ is "-" and vice versa.

In other words, this is saying that the solutions to $x^2 + ax + b = 0$ are $-\frac{1}{2}a + \sqrt{a^2 - b}$ and similarly the solutions to $x^2 - ax - b = 0$ are $+\frac{1}{2}a + \sqrt{a^2 + b}$, although it looks like there's a factor missing somewhere that would actually make the expression correct (there might also be some grouping involved in the way the notation works that I haven't picked up on).

Answer 2

The "$\vee$" and "$\wedge$" appear to play the role of modern "$\pm$" and "$\mp$", assigned independently to $a$ and $b$. The "contrary" of a sign is simply its opposite.

So, I would translate the example equation like this:

$$xx\vee a x \vee b = 0 \qquad\to\qquad x^2 \pm_1 ax \pm_2 b = 0$$

(I've placed subscripts on the "$\pm$"s to indicate that they're independent signs.)

By the Quadratic Formula, we "know" the solution is

$$ x \quad=\quad \frac12\left(-(\pm_1 a) \pm \sqrt{(\pm_1a)^2-4(\pm_2b)}\right) \quad=\quad \frac12\left(\mp_1 a \pm \sqrt{a^2\mp_24b}\right) \tag1 $$ Translating this back to the other form is a little tricky (there may be a typo, or I may be misinterpreting the notation), but even here we notice that the original "$\pm$"s have become "$\mp$"s (except for the squared sign that just disappears); thus, "$\vee$"s have become "$\wedge$"s. Expanding $(1)$, we get

$$ \mp_1 \frac12a \pm \frac12\sqrt{a^2\mp_24b} \quad=\quad \mp_1 \frac12a \pm \sqrt{\frac14\left(a^2\mp_24b\right)}\quad=\quad \mp_1 \frac12a \pm \sqrt{\frac14a^2\mp_2b} \tag2$$

This almost matches the notation $$\wedge \frac12 a\;\pm\; (aa\wedge b)^{1/2} \tag3$$ except the $\frac14$ seems to have gone missing from the $a^2$ (or $\frac12$ is missing from the $a$ that's being squared).