Elements of Algebra

CHAP. VIII. OF ALGEBRA. 239 2. Required two numbers, whose product may be $m$, and the sum of the squares $n$. If those numbers are represented by $x$ and $y$, we have the two following equations: $xy = m$ $x^2 + y^2 = n$. Now, $2xy = 2m$ being added to $x^2 + y^2 = n$, we have $x^2 + 2xy + y^2 = n + 2m$, and consequently, $x + y = \sqrt{(n + 2m)}$. But subtracting $2xy$, there remains $x^2 - 2xy + y^2 = n - 2m$, whence we get $x - y = \sqrt{(n - 2m)}$; we have, therefore, $x = \frac{1}{2}\sqrt{(n + 2m)} + \frac{1}{2}\sqrt{(n - 2m)}$; and $y = \frac{1}{2}\sqrt{(n + 2m)} - \frac{1}{2}\sqrt{(n - 2m)}$. 684. 3. Required two numbers, such, that their product may be 35, and the difference of their squares 24. Let the greater of the two numbers be $x$, and the less $y$: then we shall have the two equations $xy = 35$, $x^2 - y^2 = 24$; and as we have not the same advantages here, we shall proceed in the usual manner. Here, the first equation gives $y = \frac{35}{x}$, and, substituting this value of $y$ in the second, we have $x^2 - \frac{1225}{x^2} = 24$. Multiplying by $x^2$, we have $x^4 - 1225 = 24x^2$; or $x^4 = 24x^2 + 1225$. Now, the second member of this equation being affected by the sign +, we cannot make use of the formula already given, because having $c^2 = -1225$, $c$ would become imaginary. Let us therefore make $x^2 = z$; we shall then have $z^2 = 24z + 1225$, whence we obtain $z = 12 \pm \sqrt{(144 + 1225)}$ or $z = 12 \pm 37$; consequently, $x^2 = 12 \pm 37$; that is to say, either $= 49$, or $= -25$. If we adopt the first value, we have $x = 7$, and $y = 5$. The second value gives $x = \sqrt{-25}$; and, since $xy=35$, we have $y = \frac{35}{\sqrt{-25}} = \sqrt{\frac{1225}{-25}} = \sqrt{-49}$. 685. We shall conclude this chapter with the following question. 4. Required two numbers, such, that their sum, their product, and the difference of their squares, may be all equal.