Elements of Algebra

CHAP. VIII. OF ALGEBRA. 237 root sought will be $\sqrt{\frac{2+1}{2}} + \sqrt{\frac{2-1}{2}} = \sqrt{\frac{3}{2}} + \sqrt{\frac{1}{2}}$. Let it be required to find the square root of the binomial $11 + 6\sqrt{2}$. Here we shall have $a = 11$, and $\sqrt{b}=6\sqrt{2}$; consequently, $b = 36 \times 2 = 72$, and $a^2 - b = 49$, which gives $c=7$; and hence we conclude, that the square root of $11 + 6\sqrt{2}$ is $\sqrt{9} + \sqrt{2}$, or $3 + \sqrt{2}$. Required the square root of $11 + 2\sqrt{30}$. Here $a=11$, and $\sqrt{b} = 2\sqrt{30}$; consequently, $b = 4 \times 30 = 120$, $a^2 - b = 1$, and $c=1$; therefore the root required is $\sqrt{6} + \sqrt{5}$. 679. This rule also applies, even when the binomial contains imaginary, or impossible quantities. Let there be proposed, for example, the binomial $1 + 4\sqrt{-3}$. First, we shall have $a=1$ and $\sqrt{b} = 4\sqrt{-3}$, that is to say, $b = -48$, and $a^2 - b = 49$; therefore $c=7$, and consequently the square root required is $\sqrt{4} + \sqrt{-3} = 2 + \sqrt{-3}$. Again, let there be given $-\frac{1}{2} + \frac{1}{2}\sqrt{-3}$. First, we have $a = -\frac{1}{2}$; $\sqrt{b} = \frac{1}{2}\sqrt{-3}$, and $b = \frac{1}{4} \times -3 = -\frac{3}{4}$; whence $a^2 - b = \frac{1}{4} + \frac{3}{4} = 1$, and $c=1$; and the result required is $\sqrt{\frac{1}{4}} + \sqrt{-\frac{3}{4}} = \frac{1}{2} + \frac{\sqrt{-3}}{2}$, or $\frac{1}{2} + \frac{1}{2}\sqrt{-3}$. Another remarkable example is that in which it is required to find the square root of $2\sqrt{-1}$. As there is here no rational part, we shall have $a=0$. Now, $\sqrt{b} = 2\sqrt{-1}$, and $b = -4$; wherefore $a^2 - b = 4$, and $c=2$; consequently, the square root required is $\sqrt{1} + \sqrt{-1} = 1 + \sqrt{-1}$; and the square of this quantity is found to be $1 + 2\sqrt{-1} - 1 = 2\sqrt{-1}$. 680. Suppose now we have such an equation as $x^2 = a \pm \sqrt{b}$, and that $a^2 - b = c^2$; we conclude from this, that the value of $x = \sqrt{(\frac{a+c}{2})} \pm \sqrt{(\frac{a-c}{2})}$, which may be useful in many cases. For example, if $x^2 = 17 + 12\sqrt{2}$, we shall have $x = 3 + \sqrt{8} = 3 + 2\sqrt{2}$. 681. This case occurs most frequently in the resolution of equations of the fourth degree, such as $x^4 = 2ax^2 + d$. For, if we suppose $x^2 = y$, we have $x^4 = y^2$, which reduces the given equation to $y^2 = 2ay + d$, and from this we find $y = a \pm \sqrt{(a^2 + d)}$, therefore, $x^2 = a \pm \sqrt{(a^2 + d)}$, and consequently we have another evolution to perform. Now,