Elements of Algebra

98 ELEMENTS SECT. II. $a+b$ $a+b$ --- $a^2+ab$ $ab+b^2$ --- $a^2+2ab+b^2$ 307. When the root consists of two terms added together, as $a+b$, the square comprehends, 1st, the squares of each term, namely, $a^2$ and $b^2$; and 2dly, twice the product of the two terms, namely, $2ab$: so that the sum $a^2+2ab+b^2$ is the square of $a+b$. Let, for example, $a=10$, and $b=3$; that is to say, let it be required to find the square of $10+3$, or $13$, and we shall have $100+60+9$, or $169$. 308. We may easily find, by means of this formula, the squares of numbers, however great, if we divide them into two parts. Thus, for example, the square of $57$, if we consider that this number is the same as $50+7$, will be found $= 2500+700+49 = 3249$. 309. Hence it is evident, that the square of $a+1$ will be $a^2+2a+1$: for since the square of $a$ is $a^2$, we find the square of $a+1$ by adding to that square $2a+1$; and it must be observed, that this $2a+1$ is the sum of the two roots $a$, and $a+1$. Thus, as the square of $10$ is $100$, that of $11$ will be $100+21$: the square of $57$ being $3249$, that of $58$ is $3249+115=3364$; the square of $59 = 3364+117=3481$; the square of $60 = 3481+119=3600$, &c. 310. The square of a compound quantity, as $a+b$, is represented in this manner $(a+b)^2$. We have therefore $(a+b)^2 = a^2+2ab+b^2$, whence we deduce the following equations: $(a+1)^2=a^2+2a+1$; $(a+2)^2=a^2+4a+4$; $(a+3)^2=a^2+6a+9$; $(a+4)^2=a^2+8a+16$; &c. 311. If the root be $a-b$, the square of it is $a^2-2ab+b^2$, which contains also the squares of the two terms, but in such a manner, that we must take from their sum twice the product of those two terms. Let, for example, $a=10$, and $b=1$, then the square of $9$ will be found equal to $100-20+1=81$. 312. Since we have the equation $(a-b)^2 = a^2-2ab+b^2$, we shall have $(a-1)^2 = a^2-2a+1$. The square of $a-1$ is found, therefore, by subtracting from $a^2$ the sum of the two roots $a$ and $a-1$, namely, $2a-1$. Thus, for