MATHEMATICS AND ASTRONOMY. 
37 
Let AB (flg. 2) be the given line, and suppose K to be the required 
point; let AB be denoted by a, and AKby x; then by the given condition 
a 2 
x (a — x) = 
and by the ordinary process of solution 
According to Carnot, the appearance of the imaginary indicates that there 
is no such point as is required between A and B, but that it is outside AB 
B 
K 
A 
Fig. 2. 
E 
A 
G 
D 
fig. 3. 
on the line prolonged. If it is supposed to be beyond B on the line pro 
duced, the equation takes the modified form x(x — a) = h a 2 , giving 
1 . / 3 « 2 
x = à« ± V -j- 
Of these two roots he considers 
only to be a true solution of the question ; while 
is the solution on the hypothesis that the point is on the line produced, but 
on the side of A. Buee views these answers as the solutions of connected 
equations, not of the given equation. His solution is represented (flg. 3) 
by drawing two mutual perpendiculars KC and KB to represent \/—1 ~ 
and their opposites KB and KG to represent — |/— 1 \; C and D or E and 
G are the points required. But Buee does not show how the square of 
± -|- i/—1 is to be represented? If the one component of the line is 
perpendicular to the other, ought not the square of the sum to be equal to 
the sum of the squares? But this does not agree with the principles of 
algebra, for 
0« + V— 1 VY — x * — 2/ 2 + 2]/—1 xy-