Elements of Algebra

CHAP. IV. OF ALGEBRA. 331 $2x^2 - 1 = 1$; or if we make $m = 1$, and $n = 2$, we find $x = \frac{5}{3}$, and $2x^2 - 1 = \frac{41}{9}$; lastly, if we suppose $m = 1$, and $n = -2$, we find $x = -5$, or $x = +5$, and $2x^2 - 1 = 49$. 56. Question 4. To find numbers whose squares doubled and increased by 2, may likewise be squares. Such a number, for instance, is 7, since the double of its square is 98, and if we add 2 to it, we have the square 100. We must, therefore, have $2x^2 + 2$ a square; and as $a = 2$, $b = 0$, and $c = 2$, so that neither $a$ nor $c$, nor $b^2 - 4ac$, (the last being $= -16$), are squares, we must, therefore, have recourse to the fourth rule. Let us suppose the first part to be 4, then the second will be $2x^2 - 2 = 2(x+1) \times (x-1)$, which presents the quantity proposed in the form $4 + 2(x+1) \times (x-1)$. Now, let $2 + \frac{m(x+1)}{n}$ be its root, and we shall have the equation $4 + 2(x+1) \times (x-1) = 4 + \frac{4m(x+1)}{n} + \frac{m^2(x+1)^2}{n^2}$, in which the squares 4, are destroyed; so that after having divided the other terms by $x+1$, we have $2n^2x - 2n^2 = 4mn + m^2x + m^2$; and, consequently, $x = \frac{4mn+m^2+2n^2}{2n^2-m^2}$. If, in this value, we make $m = 1$, and $n = 1$, we find $x = 7$, and $2x^2 + 2 = 100$. But if $m = 0$, and $n = 1$, we have $x = 1$, and $2x^2 + 2 = 4$. 57. It frequently happens, also, when none of the first three rules applies, that we are still able to resolve the formula into such parts as the fourth rule requires, though not so readily as in the foregoing examples. Thus, if the question comprises the formula $7 + 15x + 13x^2$, the resolution we speak of is possible, but the method of performing it does not readily occur to the mind. It requires us to suppose the first part to be $(1-x)^2$ or $1 - 2x + x^2$, so that the other may be $6 + 17x + 12x^2$: and we perceive that this part has two factors, because $17^2 - (4 \times 6 \times 12), = 1$, is a square. The two factors therefore are $(2+3x) \times (3+4x)$; so that the formula becomes $(1-x)^2 + (2+3x) \times (3+4x)$, which we may now resolve by the fourth rule.