Elements of Algebra

$x^2 + \frac{bx}{c} + \frac{b^2}{4c^2} - \frac{b^2}{4c^2} + \frac{4ac}{4c^2} = x^2 + \frac{bx}{c} + \frac{a}{c}$; which being multiplied by $c$, gives $cx^2 + bx + a$. We have, therefore, only to multiply one of the factors by $c$, and we obtain the formula in question expressed by the product, $(cx + \frac{b}{2} - \frac{d}{2}) \times (x + \frac{b}{2c} + \frac{d}{2c})$; and it is evident that this solution must be applicable whenever $b^2 - 4ac$ is a square. 51. From this results the third case, in which the formula $a + bx + cx^2$ may be transformed into a square; which we shall add to the other two. 52. This case, as we have already observed, takes place, when the formula may be represented by a product, such as $(f+gx) \times (h+kx)$. Now, in order to make a square of this quantity, let us suppose its root, or $\sqrt{(f+gx) \times (h+kx)} = \frac{m(f+gx)}{n}$; and we shall then have $(f+gx) \times (h+kx) = \frac{m^2(f+gx)^2}{n^2}$; and, dividing this equation by $f+gx$, we have $h+kx = \frac{m^2(f+gx)}{n^2}$; or $hn^2 + kn^2x = fm^2 + gm^2x$; and, consequently, $x = \frac{fm^2-hn^2}{kn^2-gm^2}$. To illustrate this, let the following questions be proposed. Question 1. To find all the numbers, $x$, such, that if 2 be subtracted from twice their square, the remainder may be a square. Since $2x^2 - 2$ is the quantity which is to be a square, we must observe, that this quantity is expressed by the factors, $2(x+1) \times (x-1)$. If, therefore, we suppose its root $= \frac{m(x+1)}{n}$, we have $2(x+1) \times (x-1) = \frac{m^2(x+1)^2}{n^2}$; dividing by $x+1$, and multiplying by $n^2$, we obtain $2n^2x - 2n^2 = m^2x + m^2$, and $x = \frac{m^2+2n^2}{2n^2-m^2}$. If, therefore, we make $m=1$, and $n=1$, we find $x=3$, and $2x^2 - 2 = 16 = 4^2$. If $m=3$ and $n=2$, we have $x = -17$. Now, as $x$ is