THE APPLICATION OF ALGEBRA 
therefore x — 
3 a 
± v 4 
ban 
But x, by the nature 
2 — r 4 
of the problem, being less than a, the upper sign (f) 
gives x too great; so that x— —1/ ^^- — 38,19658, 
2 v 4 
&c. must be the true value required. 
PROBLEM XLV. 
The sum, and the sum of the squares, of two numbers 
being given; to fnd the numbers. 
Let half the sum of the two numbeus be denoted by a, 
half the sum of their squares by b, and half the difference 
of the numbers by x; then will the numbers themselves 
be represented by a — x, and <z f x, and their squares 
by a* — 2ax ¡- x\ and a 1 f 2ax f x*; and so we 
have a z — Sax f x 7 4- a 2 4- 2ax \- x 1 — <2b, by the ques 
tion. Which equation, contracted and divided by 2 gives 
a 1 4- x 1 — b; whence x 1 — b — a\- and consequently 
x — %/b — a 1 » Therefore the numbers sought are 
a — \/b — a% and a + \/h — a\ 
PROBLEM XLVI. 
•' • - • 
The sum, and the sum of the cubes of two numbers 
being given; to find the numbers. 
Let the two numbers be expressed as in the preceding 
problem, and let the sum of their cubes be denoted by 
Therefore will a — x] 1 + a -f xV ~ c, that is, by 
involution and reduction, 2a 3 -f- 6ax r — c; whence 
c — 2 a 3 c a 1 , 
— , and x i= 
3