Elements of Algebra

CHAP. VIII. OF ALGEBRA. 105 $1+\sqrt{2}$ $4+2\sqrt{2}$ $1+\sqrt{2}$ $2-\sqrt{2}$ $1+\sqrt{2}$ $8+4\sqrt{2}$ $\sqrt{2}+2$ $-4\sqrt{2}-4$ $1+2\sqrt{2}+2=3+2\sqrt{2}$. $8-4=4$. 329. What we have said applies also to imaginary quantities; we shall only observe farther, that $\sqrt{-a}$ multiplied by $\sqrt{-a}$ produces $-a$. If it were required to find the cube of $-1+\sqrt{-3}$, we should take the square of that number, and then multiply that square by the same number; as in the following operation: $-1+\sqrt{-3}$ $-1+\sqrt{-3}$ $1-\sqrt{-3}$ $-\sqrt{-3}-3$ $1-2\sqrt{-3}-3=-2-2\sqrt{-3}$ $-1+\sqrt{-3}$ $2+2\sqrt{-3}$ $-2\sqrt{-3}+6$ $2+6=8$. 330. In the division of surds, we have only to express the proposed quantities in the form of a fraction; which may be then changed into another expression having a rational denominator; for if the denominator be $a+\sqrt{b}$, for example, and we multiply both this and the numerator by $a-\sqrt{b}$, the new denominator will be $a^2-b$, in which there is no radical sign. Let it be proposed, for example, to divide $3+2\sqrt{2}$ by $1+\sqrt{2}$: we shall first have $\frac{3+2\sqrt{2}}{1+\sqrt{2}}$; then multiplying the two terms of the fraction by $1-\sqrt{2}$, we shall have for the numerator: $3+2\sqrt{2}$ $1-\sqrt{2}$ $3+2\sqrt{2}$ $-3\sqrt{2}-4$ $3-\sqrt{2}-4=-\sqrt{2}-1$;