Elements of Algebra

CHAP. VIII. OF ALGEBRA, 235 squares of such binomials are also binomials of the same kind; in which also one of the terms is always rational. For, if we take the square of $a + \sqrt{b}$, we shall obtain $(a^2+b) + 2a \sqrt{b}$. If therefore it were required reciprocally to take the root of the quantity $(a^2+b) + 2a \sqrt{b}$, we should find it to be $a + \sqrt{b}$; and it is undoubtedly much easier to form an idea of it in this manner, than if we had only put the sign $\sqrt{}$ before that quantity. In the same manner, if we take the square of $\sqrt{a} + \sqrt{b}$, we find it $(a+b)+2\sqrt{ab}$; therefore, reciprocally, the square root of $(a+b)+2\sqrt{ab}$ will be $\sqrt{a}+\sqrt{b}$, which is likewise more easily understood, than if we had been satisfied with putting the sign $\sqrt{}$ before the quantity. 672. It is chiefly required, therefore, to assign a character, which may, in all cases, point out whether such a square root exists or not; for which purpose we shall begin with an easy quantity, requiring whether we can assign, in the sense that we have explained, the square root of the binomial $5+2\sqrt{6}$. Suppose, therefore, that this root is $\sqrt{x}+\sqrt{y}$; the square of it is $(x+y)+2\sqrt{xy}$, which must be equal to the quantity $5+2\sqrt{6}$. Consequently, the rational part $x+y$ must be equal to $5$, and the irrational part $2\sqrt{xy}$ must be equal to $2\sqrt{6}$; which last equality gives $\sqrt{xy}=\sqrt{6}$. Now, since $x+y=5$, we have $y=5-x$, and this value substituted in the equation $xy=6$, produces $5x-x^2=6$, or $x^2=5x-6$; therefore, $x=\frac{5}{2}+\sqrt{(\frac{25}{4}-6)}=\frac{5}{2}+\sqrt{\frac{1}{4}}=\frac{5}{2}+\frac{1}{2}=3$. So that $x=3$, and $y=2$; whence we conclude, that the square root of $5+2\sqrt{6}$ is $\sqrt{3}+\sqrt{2}$. 673. As we have here found the two equations, $x+y=5$, and $xy=6$, we shall give a particular method for obtaining the values of $x$ and $y$. Since $x+y=5$, by squaring, $x^2+2xy+y^2=25$; and as we know that $x^2-2xy+y^2$ is the square of $x-y$, let us subtract from $x^2+2xy+y^2=25$, the equation $xy=6$, taken four times, or $4xy=24$, in order to have $x^2-2xy+y^2=1$; whence by extraction we have $x-y=1$; and as $x+y=5$, we shall easily find $x=3$, and $y=2$: wherefore, the square root of $5+2\sqrt{6}$ is $\sqrt{3}+\sqrt{2}$. 674. Let us now consider the general binomial $a+\sqrt{b}$, and supposing its square root to be $\sqrt{x}+\sqrt{y}$, we shall have the equation $(x+y)+2\sqrt{xy}=a+\sqrt{b}$; so that $x+y=a$, and $2\sqrt{xy}=\sqrt{b}$, or $4xy=b$; subtracting this square from the square of the equation $x+y=a$, that is, from $x^2+2xy+y^2=a^2$, there remains $x^2-2xy+y^2=a^2-b$, the square root of which is $x-y=\sqrt{(a^2-b)}$.