Elements of Algebra

238 ELEMENTS SECT. IV. since $\sqrt{b} = \sqrt{(a^2 + d)}$, we have $b = a^2 + d$, and $a^2 - b = -d$; if, therefore, $-d$ is a square, as $c^2$, that is to say, $d = -c^2$, we may assign the root required. Suppose, in reality, that $d = -c^2$; or that the proposed equation of the fourth degree is $x^4 = 2ax^2 - c^2$, we shall then find that $x = \sqrt{(\frac{a+c}{2})} \pm \sqrt{(\frac{a-c}{2})}$. 682. We shall illustrate what we have just said by some examples. 1. Required two numbers, whose product may be 105, and whose squares may together make 274. Let us represent those two numbers by $x$ and $y$; we shall then have the two equations, $xy = 105$ $x^2 + y^2 = 274$. The first gives $y = \frac{105}{x}$, and this value of $y$ being substituted in the second equation, we have $x^2 + \frac{105^2}{x^2} = 274$. Wherefore $x^4 + 105^2 = 274x^2$, or $x^4 = 274x^2 - 105^2$. If we now compare this equation with that in the preceding article, we have $2a = 274$, and $-c^2 = -105^2$; consequently, $c = 105$, and $a = 137$. We therefore find $x = \sqrt{(\frac{137+105}{2})} \pm \sqrt{(\frac{137-105}{2})} = 11 \pm 4$. Whence $x = 15$, or $x = 7$. In the first case, $y=7$, and in the second case, $y=15$; whence the two numbers sought are 15 and 7. 683. It is proper, however, to observe, that this calculation may be performed much more easily in another way. For, since $x^2 + 2xy + y^2$ and $x^2 - 2xy + y^2$ are squares, and since the values of $x^2 + y^2$ and of $xy$ are given, we have only to take the double of this last quantity, and then to add and subtract it from the first, as follows: $x^2 + y^2 = 274$; to which if we add $2xy = 210$, we have $x^2 + 2xy + y^2 = 484$, which gives $x+y=22$. But subtracting $2xy$, there remains $x^2 - 2xy + y^2 = 64$, whence we find $x-y=8$. So that $2x = 30$, and $2y = 14$; consequently, $x = 15$, and $y = 7$. The following general question is resolved by the same method.