Elements of Algebra

CHAP. IV. OF ALGEBRA. 329 only found in the second power, it is indifferent whether we take $x = -17$, or $x = +17$; either supposition equally gives $2x^2 - 2 = 576 = 24^2$. 53. Question 2. Let the formula $6 + 13x + 6x^2$ be proposed to be transformed into a square. Here, we have $a = 6$, $b = 13$, and $c = 6$, in which neither $a$ nor $c$ is a square. If, therefore, we try whether $b^2 - 4ac$ becomes a square, we obtain 25; so that we are sure the formula may be represented by two factors; and those factors are $(2 + 3x) \times (3 + 2x)$. If $\frac{m(2+3x)}{n}$ is their root, we have $(2 + 3x) \times (3 + 2x) = \frac{m^2(2+3x)^2}{n^2}$, which becomes $3n^2 + 2n^2x = 2m^2 + 3m^2x$, whence we find $x = \frac{2m^2-3n^2}{2n^2-3m^2} = \frac{3n^2-2m^2}{3m^2-2n^2}$. Now, in order that the numerator of this fraction may become positive, $3n^2$ must be greater than $2m^2$; and, consequently, $2m^2$ less than $3n^2$: that is to say, $\frac{m^2}{n^2}$ must be less than $\frac{3}{2}$. With regard to the denominator, if it must be positive, it is evident that $3m^2$ must exceed $2n^2$; and, consequently, $\frac{m^2}{n^2}$ must be greater than $\frac{2}{3}$. If, therefore, we would have the positive values of $x$, we must assume such numbers for $m$ and $n$, that $\frac{m^2}{n^2}$ may be less than $\frac{3}{2}$, and yet greater than $\frac{2}{3}$. For example, let $m = 6$, and $n = 5$; we shall then have $\frac{m^2}{n^2} = \frac{36}{25}$, which is less than $\frac{3}{2}$, and evidently greater than $\frac{2}{3}$, whence $x = \frac{3}{38}$. 54. This third case leads us to consider also a fourth, which occurs whenever the formula $a + bx + cx^2$ may be resolved into two such parts, that the first is a square, and the second the product of two factors: that is to say, in this case, the formula must be represented by a quantity of the form $p^2 + qr$, in which the letters $p$, $q$, and $r$ express quantities of the form $f + gx$. It is evident that the rule for this case will be to make $\sqrt{(p^2 + qr)} = p + \frac{mq}{n}$; for we shall