Elements of Algebra

CHAP. V. OF ALGEBRA. 221 Consequently, our equation will assume the following form, $\frac{9 \times 169}{17 \times 49}x^2 = 4 \times 17 \times 169$, which, divided by 169, is reduced to $\frac{9}{17 \times 49}x^2 = 4 \times 17$; multiplying also by $17 \times 49$, and dividing by 9, we have $x^2 = \frac{4 \times 289 \times 49}{9}$, in which all the factors are squares; whence we have, without any further calculation, the root $x = \frac{2 \times 17 \times 7}{3} = \frac{238}{3} = 79\frac{1}{3}$, as before. 637. Question 5. A company of merchants appoint a factor at Archangel. Each of them contributes for the trade, which they have in view, ten times as many crowns as there are partners; and the profit of the factor is fixed at twice as many crowns, per cent, as there are partners. Also, if $\frac{1}{100}$ part of his total gain be multiplied by $2\frac{1}{2}$, it will give the number of partners. That number is required. Let it be $x$; and since, each partner has contributed $10x$, the whole capital is $10x^2$. Now, for every hundred crowns, the factor gains $2x$, so that with the capital of $10x^2$ his profit will be $\frac{1}{5}x^3$. The $\frac{1}{100}$ part of his gain is $\frac{1}{500}x^3$; multiplying by $2\frac{1}{2}$, or by $\frac{5}{2}$, we have $\frac{5}{1000}x^3$, or $\frac{1}{200}x^3$, and this must be equal to the number of partners, or $x$. We have, therefore, the equation $\frac{1}{200}x^3 = x$, or $x^3 = 225x$; which appears, at first, to be of the third degree; but as we may divide by $x$, it is reduced to the quadratic $x^2 = 225$; whence $x = 15$. So that there are fifteen partners, and each contributed 150 crowns. QUESTIONS FOR PRACTICE. 1. To find a number, to which 20 being added, and from which 10 being subtracted, the square of the sum, added to twice the square of the remainder, shall be 17475. Ans. 75. 2. What two numbers are those, which are to one another in the ratio of 3 to 5, and whose squares, added together, make 1666? Ans. 21 and 35. 3. The sum $2a$, and the sum of the squares $2b$, of two numbers being given; to find the numbers. Ans. $a - \sqrt{(b - a^2)}$, and $a + \sqrt{(b - a^2)}$.