ALGEBRA. 
appears, that A has won three shillings ; but 
had he ventured to bet five shillings to two, 
and lost one game more out of the same num- 
ber, he would have lost thirty shillings: how 
many games did they play ? 
, (be the number of games A 
Letx 1 
( won, 
y the number B won, 
then 2 a; is what A won of B, 
and 3 y what B won of A. 
tix — 3 y == 3, by the question ; 
( A would win on the 
, x . } ' ( 2 d supposition, 
y -j- 1 . 5, B would win, 
5y + 5— 2 a: + 2 = 30, by 
the question. 
or 5 y — 2 x — 50 — 3 — 2 — 23 
therefore, 5 y — 2 x — 23 
and 9.x — 3 y = 3 
by addition, 5 y — 3 y — 26 
2» = 26 
y — 13 
2 x = 3 -j- 3 y = 3 -f 39 = 42 
a: = 21 
x 4- y = 34, the num. of games. 
ON QUADRATIC EQUATIONS. 
When the terms of an equation involve the 
square of an unknown quantity, but the first 
power does not appear, the value of the 
square is obtained by the preceding rules ; 
and by extracting the square root on both 
sides, the quantity itself is found. 
Ex. 1. Let 5 a? — 45 = 0 ; to find a;. 
By trans. 5 a? = 45 
a? = 9 
therefore, x — a/ 9 = -j- 3. 
The signs + and — are both prefixed to 
the root, because the square root of a quan- 
tity may be either positive or negative. The 
sign of x may also be negative ; but still x 
will be either equal to -j- 3 or — 3. 
Ex. 2. Let ax 1 — bed ; to find x. 
a 
If both the first and second powers of the 
unknown quantity be found in an equation : 
Arrange the terms according to the dimen- 
sions of the unknown quantity, beginning with 
the highest, and transpose the known quan- 
tities to the other side ; then, if the square 
of the unknown quantity be affected with a 
co-efficient, divide all the terms by this co- 
efficient, and if its sign be negative, change 
the signs of all the terms, that the equation 
may be reduced to this form, x 2 -4- P x — 
+ q. Then add to both sides the square of 
half the co-efficient of the first power of the 
unkwown quantity, by which means the first 
side of the equation is made a complete 
square, and the other consists of known quan- 
tities; and by extracting the square root 
on both sides, a simple equation, is obtained, 
from which the value of the unkwown quan- 
tity may be found. 
Ex. 1 . Let x 2 -j— p x xz. q ; now, we know 
that a? + 4 is tlie s fi uare °f x -J-|, 
V 1 
add therefore, —• to both sides, and we have 
4 
j p2. 
x 2 -J-p x -(-4 = ? + --; then by extract- 
ing the square root on both sides, 
x -\- P - = ± v / q + ~, and by transp. 
x — — fi \/ ?+f- 
In the same manner, if a? — px — q,x is 
found to be ^4 \/ r ] +-7 • 
4 
Ex. 2. Let x 2 — 1 2 a: 35 = 0 ; to find x. 
By transposition, x 2 — 12 x = — 35, and 
adding the square of 6 to both sides of the 
equation, 
x 2 — 12 a; 4- 36 = 36 — 35 = 1 ; 
then extracting the square root on both sides, 
x — 6=41 
.T = 6 4 1 — ^ or 5; either of which, 
substituted for x in the original equation, 
answers the condition, that is, makes the 
whole equal to nothing. 
Ex. 3. Let x -j-\/ 5 re — 10 = 8 j to find x. 
By transposition, y/ 5 x 4 - 10 = 8 — x 
squar. both sides, 5 x 4- 10 = 64 — 1 6a: 4 - a? 
' a 2 — 21a: = 10 — 64= — 54 
441 441 
eompl. the sq. x 2 — 21 x 4- 54 
441 — 216 , . 441 225 
•, or x 2 — 21 x 4 - - 
4 
4 
4 
21 . 15 
extracting the square root, x — -—= 4 
21 + 15 „ 
x — — — — = 3 or 18. 
2 
By this process two values of x are found, 
but on trial it appears, that 18 does not an- 
swer the condition of the equation, if we sup- 
pose that v+5 x + 10 represents the posi- 
