TO THE RESOLUTION OF PROBLEMS. 103 
2#* -f 24# -f 144 zz 1424; this, ordered, gives # 1 -p 
12# — 640; which by completing the square, becomes 
# 2 4 12# + 36 (=z 640 4 36) —■ 676; whence, extract 
ing the root on both sides, we have x + 6 z: (v/676) 
26; therefore x zz 20, and x 4 12 r 32, are the two 
numbers required. 
For $ 32 ~ 20 = 12 > 
4 - t 32 2 -f* 20’ zz 1424. 
PROBLEM XLIII. 
To divide 36 into three such parts, that the second may 
exceed the first by 4, and that the sum of all their squares 
may be 464. 
Let x be the first part, then the second will be # 44; 
and, the sum of these two being taken from (36) the 
whole, we have 32 — 2#, for the third, or remaining 
part; and so, by the question, x 2 4 x 4 4| x 4- 32 — 2# I 1 
zz 464, that is, 6# 7 — J20# 4 1040 zz 464; whence 
6# 1 — 120# zz — 376, and x 1 — 20# — — 96. Now, 
by completing the square: # a — 20# 4 100 (zz 100 
— 96) zz 4; and, by extracting the root, # — 10 -zz 
4 2. Therefore # zz 10 4 2, that is, # zz 8, or # z= 
12; so that 8, 12, and 16 are the three numbers re 
quired, 
PROBLEM XLIV. 
To divide the number 100 (a) into two such parts, 
that their product and the difference of their squares may 
be equal to each other. 
Let the lesser part be denoted by #, then the greater 
will be a — x, and we shall have a — # x # zz a — #1“ 
— # 2 , that is, ax — # 2 zz a z — 2a#; whence # 1 —3a# 
z: — a x ; and, by completing the square, # 2 — 3a# 
4 zz (— a 2 4 -~—) —~; of which the root being 
extracted, there comes out # — zz 4-1/ ~T L % an( ^ 
2 — y 4 
H 4