Elements of Algebra

CHAP. V. OF ALGEBRA. 219 of $x$ will be not only $+7$, but also $-7$, which is expressed by $x = \pm 7$. So that all those questions admit of a double answer; but it will be easily perceived that in several cases, as those which relate to a certain number of men, the negative value cannot exist. 632. In such equations, also, as $ax^2 = bx$, where the known quantity $c$ is wanting, there may be two values of $x$, though we find only one if we divide by $x$. In the equation $x^2 = 3x$, for example, in which it is required to assign such a value of $x$, that $x^2$ may become equal to $3x$, this is done by supposing $x = 3$, a value which is found by dividing the equation by $x$; but, beside this value, there is also another, which is equally satisfactory, namely, $x = 0$; for then $x^2 = 0$, and $3x = 0$. Equations therefore of the second degree, in general, admit of two solutions, whilst simple equations admit only of one. We shall now illustrate what we have said with regard to pure equations of the second degree by some examples. 633. Question 1. Required a number, the half of which multiplied by the third, may produce 24. Let this number be $x$; then by the question $\frac{1}{2}x$, multiplied by $\frac{1}{3}x$, must give 24; we shall therefore have the equation $\frac{1}{6}x^2 = 24$. Multiplying by 6, we have $x^2 = 144$; and the extraction of the root gives $x = \pm 12$. We put $\pm$; for if $x = +12$, we have $\frac{1}{2}x = 6$, and $\frac{1}{3}x = 4$: now, the product of these two numbers is 24; and if $x = -12$, we have $\frac{1}{2}x = -6$, and $\frac{1}{3}x = -4$, the product of which is likewise 24. 634. Question 2. Required a number such, that being increased by 5, and diminished by 5, the product of the sum by the difference may be 96. Let this number be $x$, then $x+5$, multiplied by $x-5$, must give 96; whence results the equation, $x^2 - 25 = 96$. Adding 25, we have $x^2 = 121$; and extracting the root, we have $x = 11$. Thus $x+5 = 16$, also $x-5 = 6$; and, lastly, $6 \times 16 = 96$. 635. Question 3. Required a number such, that by adding it to 10, and subtracting it from 10, the sum, multiplied by the difference, will give 51. Let $x$ be this number; then $10+x$, multiplied by $10-x$, must make 51, so that $100 - x^2 = 51$. Adding $x^2$, and subtracting 51, we have $x^2 = 49$, the square root of which gives $x=7$. 636. Question 4. Three persons, who had been playing, leave off; the first, with as many times 7 crowns, as the second has three crowns; and the second, with as many