but by subtraction, 2y= ±32, or 40,
and y=16, or 20.

41. Given (x2 + y2)× (x + y) = 2336)

and (x2—y2). (x − y) = 576) to find the

values of x and y.

From the first equation, ́x3+x2y+xy2+y3=2336; and from the second, x3- x2y-xy2+y3 = 576;

.. by subtraction,

adding this to the first equation,

3

2x2y+2x y2 = 1760;

x3+3x2y + 3x y2+y3=4096;

.. extracting the cube root, x+y= 16,

and 2xy.(x+y)=1760,

2

4xy = 220;

.. by subtraction, x2 -2xy + y2 = 36,

and therefore x − y = ±6;

Dividing the second equation by xy, x+y=4;

3

:. x3 + 3 x3y + 3x y2+y3=64.

But from the first equation, 3x2y - xy2+y3 = 0;

.. by subtraction, 4x2y+4xy=64;

.. (x+y).xy= 16,

and xy=4.

But x2+2xy + y2 = 16,

Again, dividing the first equation by this last,

44. Given (x2-y3) × (x − y) = 3xy and (x*—y1) × (x2 — y2)=45 x2 y2

values of x and y.

Dividing the second equation by the first,

(x2+ y2)×(x+y)=15xy,

or x3+x2y+xy2+y3 = 15xy;

but from the first, x3-x2y-xy2+y3 = 3xy;

.. by addition, 2x3

+2y3 = 18xy;

and

y3=9xy;

but by subtraction, 2x2y+2xy= 12xy;

.. dividing by 2xy, x+y=6;

whence x3+3x2y+3xy2+y3=216;

but 3

.. by subtraction,

+y3=9xy;

3x2y+3xy2 =216-9xy,

or 3.(x+y).xy 18 xy=216-9xy;

=

.. by subtraction, x2-2xy+y2=4;

and extracting the root, x-y= ±2;

72

SECT. IV.

Solution of Adfected Quadratics, involving only one unknown Quantity.

(28.) LET the terms be arranged on one side of the equation, according to the dimensions of the unknown quantity, beginning with the highest; and (17) the known quantities be transposed to the other side; then, if the square of the unknown quantity has any coefficient, either positive or negative, let all the terms be divided by this coefficient (13). If the square of half the coefficient of the second term be now added (11) to both sides of the equation*, that side which involves the unknown quantity will become a complete square; and (19) extracting the square root on both sides of the equation, a simple equation will be obtained, from which the values of the unknown quantity may be determined.

* This is called completing the square; and that a complete square is thus obtained may be easily proved.

Let x2+2ax be the proposed quantity on one side, when the terms are arranged according to the form prescribed above; and suppose y2=the quantity requisite to complete the square. Now the square of xd = x2 + 2 dx + d2, where it is evident that four times the product of the extreme terms is equal to the square of the middle term; and therefore, in order that r2+2ax + y2 may be a square, 4x2 y2 must be equal to 4 a2x2; therefore y2=a2=the square of half the coefficient of the middle term.

It may be observed, that all equations may be solved as quadratics, by completing the squares, in which there are two terms involving the unknown quantity or any function of it, and the index of one is double that of the other. Thus,

70

2

x+px3=q, x2′′-px"=q, x2+ =q, x2 + x2 = a, a2x2 + ax =

3n

2

2n

b,

r3n+ax = b, pr1 pr = d, x2 + px + q +(x2+px + q) = r‚ x2. (x2+ax)2 + bx. (x2 + a x) = d, are of the same form as quadratics, and the value of the unknown quantity may be determined in the same manner. Many equations also, in which more than one unknown quantity are involved, may in a similar manner be reduced to lower dimensions by completing the square, x2y2 + pxy = q, x3 + y3]2 + p . (x2 + y3) = r.

32

of this kind occur in the following

EXAMPLES.

as

Instances

1. Given 2+8x=33, to find the values of x.

2

Completing the square, x2+8x+16=49 ;

and extracting the root, x+4= ±7;

whence, by transposition, r = 3, or 11.

2. Given x2+6x+4=59, to find the values of x.
By transposition, x2+6x=55;

and completing the square, x+6x+9=64;

.. extracting the root, x+3= ±8;

3. Given -8x+10=19, to find the values of r.

x

By transposition, x-8x=9;

and completing the square, x2-8x+16 = 25 ;