29 = 225 = = (15)3, which is a square number, as was required. 4. It is required to find such a value of x as will make 7x2-5x+1 a square. Here the last term 1 being a square, let there be taken, according to Case 4, Then, by rejecting the 1 on each side of the equation, m m 2m 1)2= x + 1. n m2 2m and dividing by x, we shall have 7x-5 ing n = 3, and m = 8, we shall have x= , or by tak 72, as required. 5. It is required to find such a value of x as will make 8x2+14x+6 a square. Here, by comparing this expression with the general formula, ax2 + bx + c, we shall have a = 8, b 14, and c = 6. And, as neither a nor c, in the present instance, are squares, but b2 — 4ac 196 — 1924 is a square, the given expression can be resolved, by Case 5, into the two following factors, 8x+6 and x + 1. Let, therefore, 8x2 + 14x + 6 = (8x + 6) (x + 1) = +1)2, agreeably to the rule there laid down. = m2 (x Then there will arise, by dividing each side by x+1, And consequently, by multiplication and reduction, we shall greater than 6. m2 n must be less than 8 and Whence, by taking m = 5, and n = 2, we shall have x= 8 14 400 which makes + +6= = (~2-9)2, 49 7 49 as 6. It is required to find such a value of x as will make 2x22 a square. = Here, by comparing this with the general formula ar3 + bx + c, as before, we shall have a = = 2, b O, and c =- 2. And, as neither a nor c are squares, but b2 4ac- 4ac 4 (2 × −2) = 16 is a square, the root of which is 4, the given expression can be resolved, by Case 5, into the two factors 2x2, and x + 1, or 2 (x − 1), and (x + 1), which is evident, indeed, in this case, from inspection. Let, therefore, 2x2-2 = 2 (x − 1) × (x + 1) = agreeably to the rule; and there will arise, by division, m2 2x-2=(x+1). And consequently, by multiplication, and reducing the result, we shall have x = by taking n = 1, and m = 1, we shall have x = 3, and 2x2 2 = 16 (4)2; or, taking n = 2 and m= 3, the But as a enters the problem only in its second power, +17 may be taken instead of - 17; since either of them 7. It is required to find such a value of x as will make 5x2+36x+7 a square. Here, by comparing the expression with the general formula, we shall have a = 5, b = = 36, and c = 7. And as neither a nor c are squares, but b2 - 4ac = 1296 140 = 1156 = (34)2 is a square, it can be resolved, as in the last example, into the two factors, 5x + 1, and x + 7. Whence, putting 5x+36x+7=(5x + 1) X (x+7)= m2 n2 (x+7), there will arise, by dividing, by x+7, 5x+1= n2 (x+7). m2 And consequently, by multiplication, and reducing the resulting expression, we shall have x = 7m2 5n2 mai where, taking m=2, and n= 1, the substitution will give x= 27, which makes 5 x (27) +36 × 27 +7= 8. It is required to find such a value of x as will make 6x2+13x + 10 a square. Here, by comparing the given expression with the general formula ax + bx + c, we have a = 6, b13, and c = 10. And as neither a, c, nor b2 4ac, are squares, the question, if possible, can only be resolved by the method pointed out in Case 6. In order, therefore, to try it in this way, let the first simple square 4, be subtracted from it, and there will remain, in that case, 6x2+13x + 6. = Then since (13)2 - 4 (6 × 6) 169144 25 is now a square, this part of the formula can be resolved, by Case 5, into the two factors 3+2, and 2x + 3. Whence, by assuming, according to the rule, 6x2 + 13æ + 10 = 4 + (3x + 2) × (2x + 3) = { 4m n (3x+2)+(3x+2), we shall have, by cancelling the 4 on each side, and dividing by 3x+2, 2x + 3 4m m2 n + (3x+2).- = And consequently, by multiplying by n2, and transposing the terms, we have 2n2x 2, and n = 3, the result will give x= ; or, if m be taken 13, and n = 17, we shall 4 x 17 x 13 + 2 x (13) — 3 × (17)2 2(17) 3 (13) - Which makes 6 x (5)2 + 13 x 5+10=225 required. 355 = 5, 71 (15), as 9. It is required to find such a value of x as will make 13x2+15x+7 a square. = Here, by comparing this with the general formula, as before, we have a 13, b = 15, and c = 7. And as neither a, b, nor b2. 4ac, are squares, the answer to the question, if it be resolvable, can only be obtained by Case 6. In order, therefore, to try it in that way, let (1-x)2 or 1 2x + x2 be subtracted from the given expression, and there will remain 12x+17x+6. = And as (17)2 - 4 (6 × 12), which is 1, is now a square, this part of the formula can be resolved by Case 5, into the two factors 4x+3 and 3x+2. Whence, assuming 13x2+15x+7= (1 − x)2 +(4x+3) × (3x+2) = {(1 − x) + 2m m2 3x+2) } 2 = 2 = (1 − x )2 + 2TM (1 − x) × (3x+2)+7 7 (3x + n n 2)3, we shall have, by cancelling (1 - x)3, and dividing by ly, by multiplying by n2, and transposing the terms, there will arise 4n2x + 2mnx 3m2x =2mn + 2m2 3n2, or x = Where, putting m and n each = 1, we shall have x = 13 45 63 1, which makes + +7= + + == 13 15 9 9 9 as required. 9 10. It is required to find such a value of x as will make 7x2+2 a square. Here it is easy to perceive that neither of the former rules will apply. But as the expression evidently becomes a square when x= 1, let, therefore, x = 1 + y, according to Case 7, and we shall have 7x2+2=9+14y+ 7y3 ; m Or, putting 9 +14y+ 7y3 (3 + y), according to the rule, Hence, rejecting the 9s and dividing the remaining terms by y, we have 7n2y + 14n2 = 6mn + m3y; and consequently, y= and x = 1 + 14n2 7n2 m2 ; where it is 7n2 m2 evident that m and n may be any positive or negative numbers whatever. y may take, as in a former instance, 7x2 + 2 = 3 + 2 = 3 +12 = 25 1, we shall have x Or, if m = 3 and n = 17, and 7x2 +2 7 × (17)2 + 2 = 2025 = (45), a square as before. And by proceeding in this manner, we may obtain as many other values of x as we p ease. = a PROBLEM 2.-To find such values of x as will make √ (ax3 + bx2 + cx + d) rational, or ax3 + bx2 + cx + d = square. This problem is much more limited and difficult to be resolved than the former; as there are but a few cases of it that admit of answers in rational numbers, and in these the rules for obtaining them are of a very confined nature, being mostly such as are subject to certain limitations, or that admit only of a few simple answers, which, in the instances here mentioned, may be found as follows: RULE 1.-When the third and fourth terms of the formula are wanting, or c and d are each = 0, put the side of the square sought nx, than ax3 + bx2 = n2x2. And consequently, by dividing each side of the equation by x2, we shall have ax + b = n2, or x = n2 = b α be any integral or fractional number whatever. 2. When the last term d is a square, put it the side of the required square =e+ с 2e where n may = e2, and assume = x, and the proposed c2 formula is e2 + ex + bx2 + ax3 = e2 + cx + x2. 4e2 Whence, by expunging the terms e2 + cx, which are common, and dividing by x2, we shall have 4ae2x + 4be2 = c2; and consequently, x = c2-4be2 4ae2 C Or if, the same case, there be put e+ ·x+ 2e 4be2 -C 8e3 for the side of the required square, we shall have, by squaring, |