Square Roots of Compound Expressions.

Since the square of $a+b$ is $a^2+2ab+b^2$, the square root of $a^2+2ab+b^2$ is $a+b$.

It is required to find a method of extracting the root $a+b$ when $a^2+2ab+b^2$ is given.

Ex. The first term, $a$, of the root is obviously the square root of the first term, $a^2$, in the expression. $$\begin{gathered} r*3crl\&\&a^2\&+\&2ab\&+\&\mathrm{\backslash multicolumn}1@r|b^2\&a+b \\ \mathrm{\backslash cline}8-8\&\&a^2 \\ \mathrm{\backslash cline}3-32a\&+\&\mathrm{\backslash multicolumn}1@r|b\&\&2ab\&+\&b^2 \\ \&\&\mathrm{\backslash multicolumn}1@r|\&\&2ab\&+\&b^2 \\ \mathrm{\backslash cline}4-7 \end{gathered}$$

If the $a^2$ is subtracted from the given expression, the remainder is $2ab+b^2$. Therefore the second term, $b$, of the root is obtained when the first term of this remainder is divided by $2a$; that is, by double the part of the root already found. Also, since $2ab+b^2=(2a+b)b,$ the divisor is completed by adding to the trial-divisor the new term of the root.

Ex. Find the square root of $25x^2-20x^{3}y+4x^4{y}^2$.

$$\begin{gathered} rcccrll\&\&\&\&25x^2\&\mathrm{\backslash multicolumn}1@r|-20x^{3}y+4x^4{y}^2\&5x-2x^{2}y \\ \mathrm{\backslash cline}7-7\text{{}Here \end{gathered}$$a^{2}$$\begin{gathered} \}\&=\&(5x{)}^2\&=\&25x^2 \\ \mathrm{\backslash cline}5-52a+b\&=\&10x\&-\&\mathrm{\backslash multicolumn}1@r|2x^{2}y\&-20x^{3}y+4x^4{y}^2 \\ \&\&\&\&\mathrm{\backslash multicolumn}1@r|\&-20x^{3}y+4x^4{y}^2 \\ \mathrm{\backslash cline}6-6 \end{gathered}$$

The expression is arranged according to the ascending powers of $x$.

The square root of the first term is $5x$, hence $5x$ is the first term of the root. $(5x{)}^2$ or $25x^2$ is subtracted, and the remainder is $-20x^{3}y+4x^4{y}^2.$

The second term of the root, $-2x^{2}y$, is obtained by dividing $-20x^{3}y$ by $10x$, the double of $5x$, and this new term of the root is also annexed to the divisor, $10x$, to complete the divisor.

The same method will apply to longer expressions, if care be taken to obtain the trial-divisor at each stage of the process, by doubling the part of the root already found, and to obtain the complete divisor by annexing the new term of the root to the trial-divisor.

Ex. Find the square root of $1+10x^2+25x^4+16x^6-24x^5-20x^3-4x.$

$\qquad\makebox [c]{${r*{2}{cr}lll} 16x^{6} &-& 24x^{5} &+& 25x^{4} &-20x^{3} + 10x^{2} &\multicolumn{1}{@{}r|}{- 4x + 1} & 4x^{3} - 3x^{2} + 2x - 1 \ \cline{8-8} 16x^{6} \ \cline{1-1} \multicolumn{1}{@{}r|}{\llap{$8x^3-3x^2$}} &-& 24x^{5} &+& 25x^{4} \ \multicolumn{1}{@{}r|}{}&-& 24x^{5} &+& 9x^{4} \ \cline{2-5} \multicolumn{4}{r|}{8x^{3} - 6x^{2} + 2x} & 16x^{4} &-20x^{3} + 10x^{2} \ & & & \multicolumn{1}{@{}r|}{ }& 16x^{4} &-12x^{3} + \phantom{0}4x^{2} \ \cline{5-6} \multicolumn{5}{r|}{8x^{3} - 6x^{2} + 4x - 1} &-\phantom{0}8x^{3} + \phantom{0}6x^{2} & -4x + 1 \ & & & & \multicolumn{1}{@{}r|}{}&-\phantom{0}8x^{3} + \phantom{0}6x^{2} & -4x + 1 \ \cline{6-7}$\}$

The expression is arranged according to the descending powers of $x$.

It will be noticed that each successive trial-divisor may be obtained by taking the preceding complete divisor with its last term doubled.