Elements of Algebra

CHAP. VI. OF ALGEBRA. 99 example, if $a = 50$, we have $a^2 = 2500$, and $2a - 1 = 99$; therefore $49^2 = 2500 - 99 = 2401$. 313. What we have said here may be also confirmed and illustrated by fractions; for if we take as the root $\frac{3}{5} + \frac{2}{5} = 1$, the square will be, $\frac{9}{25} + \frac{4}{25} + \frac{12}{25} = \frac{25}{25} = 1$. Farther, the square of $\frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ will be $\frac{1}{4} - \frac{1}{3} + \frac{1}{9} = \frac{1}{36}$. 314. When the root consists of a greater number of terms, the method of determining the square is the same. Let us find, for example, the square of $a + b + c$: $a+b+c$ $a+b+c$ --- $a^2+ab+ac$ $ab+b^2+bc$ $ac+bc+c^2$ --- $a^2+2ab+2ac+b^2+2bc+c^2$ We see that it contains, first, the square of each term of the root, and beside that, the double products of those terms multiplied two by two. 315. To illustrate this by an example, let us divide the number 256 into three parts, $200 + 50 + 6$; its square will then be composed of the following parts: $200^2 = 40000$ $50^2 = 2500$ $6^2 = 36$ $2 (50 \times 200) = 20000$ $2 (6 \times 200) = 2400$ $2 (6 \times 50) = 600$ --- $65536 = 256 \times 256$, or $256^2$. 316. When some terms of the root are negative, the square is still found by the same rule; only we must be careful what signs we prefix to the double products. Thus, $(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab - 2ac + 2bc$; and if we represent the number 256 by $300 - 40 - 4$, we shall have, H 2