Elements of Algebra

288 ELEMENTS SECT. IV. But we have already seen, (Art. 675. and 676.), that the square root of $a \pm \sqrt{b}$, when $\sqrt{(a^2 - b)} = c$, is expressed by $\sqrt{(a \pm \sqrt{b})} = \sqrt{(\frac{a+c}{2})} \pm \sqrt{(\frac{a-c}{2})}$: so that, as in the present case, $a = 8$, and $\sqrt{b} = 2\sqrt{15}$; and consequently, $b = 60$, and $c = \sqrt{(a^2 - b)} = 2$, we have $\sqrt{(8 + 2\sqrt{15})} = \sqrt{5} + \sqrt{3}$, and $\sqrt{(8 - 2\sqrt{15})} \ldots = \sqrt{5} - \sqrt{3}$. Hence, we have $\sqrt{p} = 1$, $\sqrt{q} = \frac{\sqrt{5}+\sqrt{3}}{2}$, and $\sqrt{r} = \frac{\sqrt{5}-\sqrt{3}}{2}$; wherefore, since we also know that the product of these quantities is positive, the four values of $x$ will be: 1. $x = \sqrt{p} + \sqrt{q} + \sqrt{r} = 1 + \frac{\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}}{2} \ldots$ $= 1 + \sqrt{5}$, 2. $x = \sqrt{p} - \sqrt{q} - \sqrt{r} = 1 + \frac{-\sqrt{5}-\sqrt{3}-\sqrt{5}+\sqrt{3}}{2} \ldots$ $= 1 + \sqrt{5}$, 3. $x = -\sqrt{p} + \sqrt{q} - \sqrt{r} = -1 + \frac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{2} \ldots$ $= -1 + \sqrt{3}$, 4. $x = -\sqrt{p} - \sqrt{q} + \sqrt{r} = -1 + \frac{-\sqrt{5}-\sqrt{3}+\sqrt{5}-\sqrt{3}}{2}$ $= -1 - \sqrt{3}$. Lastly, as we have $y = x + 2$, the four roots of the given equation are: 1. $y = 3 + \sqrt{5}$, 2. $y = 3 - \sqrt{5}$, 3. $y = 1 + \sqrt{3}$, 4. $y = 1 - \sqrt{3}$. QUESTIONS FOR PRACTICE. 1. Given $z^4 - 4z^3 - 8z + 32 = 0$, to find the values of $z$. Ans. $4, 2, -1 + \sqrt{-3}, -1 - \sqrt{-3}$. 2. Given $y^4 - 4y^3 - 3y^2 - 4y + 1 = 0$, to find the values of $y$. Ans. $\frac{-1 \pm \sqrt{-3}}{2}$, and $\frac{5 \pm \sqrt{21}}{2}$. 3. Given $x^4 - 3x^2 - 4x = 3$, to find the values of $x$. Ans. $\frac{1 \pm \sqrt{13}}{2}$, and $\frac{-1 \pm \sqrt{-3}}{2}$.