Elements of Algebra

236 ELEMENTS SECT. IV. Now, $x + y = a$; we have therefore $x = \frac{a + \sqrt{(a^2 - b)}}{2}$, and $y = \frac{a - \sqrt{(a^2 - b)}}{2}$; consequently, the square root required of $a + \sqrt{b}$ is $\sqrt{(\frac{a + \sqrt{(a^2 - b)}}{2})} + \sqrt{(\frac{a - \sqrt{(a^2 - b)}}{2})}$. 675. We admit that this expression is more complicated than if we had simply put the radical sign $\sqrt{}$ before the given binomial $a + \sqrt{b}$, and written it $\sqrt{(a + \sqrt{b})}$: but the above expression may be greatly simplified when the numbers $a$ and $b$ are such, that $a^2 - b$ is a square; since then the sign $\sqrt{}$, which is under the radical, disappears. We see also, at the same time, that the square root of the binomial $a + \sqrt{b}$ cannot be conveniently extracted, except when $a^2 - b = c^2$; in this case, the square root required is $\sqrt{(\frac{a + c}{2})} + \sqrt{(\frac{a - c}{2})}$; but if $a^2 - b$ be not a perfect square, we cannot express the square root of $a + \sqrt{b}$ more simply, than by putting the radical sign $\sqrt{}$ before it. 676. The condition, therefore, which is requisite, in order that we may express the square root of a binomial $a + \sqrt{b}$ in a more convenient form, is, that $a^2 - b$ be a square; and if we represent that square by $c^2$, we shall have for the square root in question $\sqrt{(\frac{a + c}{2})} + \sqrt{(\frac{a - c}{2})}$. We must farther remark, that the square root of $a - \sqrt{b}$ will be $\sqrt{(\frac{a + c}{2})} - \sqrt{(\frac{a - c}{2})}$; for, by squaring this formula, we get $a - 2\sqrt{(\frac{a^2 - c^2}{4})}$; now, since $c^2 = a^2 - b$, or $a^2 - c^2 = b$, the same square is found $= a - 2\sqrt{\frac{b}{4}} = a - \frac{2\sqrt{b}}{2} = a - \sqrt{b}$. 677. When it is required, therefore, to extract the square root of a binomial, as $a \pm \sqrt{b}$, the rule is, Subtract from the square ($a^2$) of the rational part the square ($b$) of the irrational part, take the square root of the remainder, and calling that root $c$, write for the root required, $\sqrt{(\frac{a+c}{2})} \pm \sqrt{(\frac{a-c}{2})}$. 678. If the square root of $2 + \sqrt{3}$ were required, we should have $a = 2$ and $\sqrt{b} = \sqrt{3}$; wherefore $a^2 - b = 4 - 3 = 1$; so that, by the formula just given, the