32 have, in this case, x order to obtain a rational answer, greater than 6. Whence, by taking m= 5, and n = 2, we shall have a 25-24 1 400 8 14 Jay 7 49 m2 222 m2 n3 m2 8n2 י7 6n2 ; where it appears that, in must be less than 8 and m2 29 m2 n2 +6 25 required. 6. It is required to find such a value of x as will make 2x2 - 2 a square. Ch Here, by comparing this with the general bx + c, as before, we shall have a — 2, And, as neither a nor c are squares, but b2 b 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 2x 2, and x + 1, or 2 (x-1), and (x + 1), which is evident, indeed, in this case, from inspection. m2 •Let, therefore, 2x2 - 2 = 2 (x − 1) × (x + 1) = ~(x+1)2, agreeably to the rule; and there will arise, by division, 2x - 2 (x+1). And consequently, by multiplication, m2 na formula ax2 + 0, and c = 2. 2n2+ m2 and reducing the result, we shall have x = 27 2n2 m2; where, by taking n = 1, and m = 1, we shall have x = 3, and 2x2 2: 18 — 2 = 16 (4)2; or, taking n÷2 and m 3, the result will give x = But as x enters the problem only in its second power, +17 may be taken instead of 17; since either of them give 2x2 2 = 576 = (24)2. 17. 7. It is required to find such a value of x as will make 5x2+36x+7 a square. V 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 - 140 1401156 = 1156 (34) is a square, it can be resolved, as in the last example, into the two factors, 5x + 1, and x + 7. 4ac1296 Whence, putting 5x+36x+7=(5x + 1) × (x + 7) = (x+7)2, there will arise, by dividing, by x+7, 5x + 1 = (x+7). And consequently, by multiplication, and reducing the 7m2 n2 resulting expression, we shall have x = 5n2 m2; where, taking m=2, and n= 1, the substitution will give x = 7 X 4-1 27, which makes 5 X (27)2 + 36 × 27 +7: 5 X 1-4 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 ax2 + bx + c, we have a=6, b = 13, 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) = 169 — 144 144 25 is now a square, this part of the formula can be resolved, by Case 5, into the two factors n 3x+2, and 2x + 3. Whence, by assuming, according to the rule, 6x+13x +10 4 m 2 + (3x + 2) × (2 x + 3) = { 2 + 2 (3∞+2) } 2 = 4 + · * N 4m m2 (3x+2)+(3x+2), we shall have, by cancelling the 4 on each side, and dividing by 3x + 2, 2x + 3 n 4m m2 + (3x+2). n2 And consequently, by multiplying by n2, and transposing the terms, we have 2n2x 3m2x = 4mn + 2m2 3n2, or 4mn + 2m2 3n2 2n2 3m2 Where, putting m = 24+8-27 5 18-12 2, and n = 3, the result will give x ; or, if m be taken 13, and n = 6 4 x 17 x 13 + 2 x (13)2-3 X (17) have x = 2(17)-3(13)2 Which makes 6 × (5)2 + 13 × 5 + 10 = 225 required. 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 = 7. And as neither Daddy = W before, we have a = 13, b = 15, and c 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 12x2 + 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) + m 2m m2 2 (3x+2) } 2 = (1 − x)2 + J (1 − x) × (3x+2)+(3x+ 2)3, we shall have, by cancelling (1 — x)3, and dividing by n n n 2m 3x + 2, 4x + 3 2 m2 (1 − x) + (3x+2); and consequent n n2 ly, by multiplying by n2, and transposing the terms, there will arise 4n2x + 2mnx 3m2x = 2mn + 2m2 · 3n2, or 2mn + 2m2 Chang 00 3n2 4n2 + 2mn 3m2* Where, putting m and n each 1, we shall have. X 2+2 −3 13 15 13 45 63 , which makes + +7= + + 4+2-3 121 9 2 (1)2, as required. 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 1, let, therefore, x = 1+y, according to Case 7, and we X shall have 7x2+2=9+14y+ 7y2; m n Or, putting 9+14y + 7y2 = (3 + y), according to the rule, and squaring the righthand side, 6m m2 9+14y+ 7y2 = 9 + y + may2. Hence, rejecting the 9s and dividing the remaining terms by y, we have 7n3y + 14n2 = 6mn + m2y; and consequently, 6mn 14n2 6mn 14n2 y= where it is 7n2 — m2 7n2 m2 evident that m and n may be any positive or negative numbers whatever. and x = 1+ y 3 3' enters the formula, we X = 1, we shall have Or, since the second power of a only may take, as in a former instance, 7x2 + 2 = 3 + 2 = 3 +38 25 9 a square. Or, if m = 3 and n +.2 = 1, we shall have x = 17, and 7x2 7 × (17)2 + 2 2025 ་ (45)2, a square as before. And by proceeding in this manner, we may obtain as many other values of x as we p ease. PROBLEM 2.-To find such values of x as will make √ (ax3 + bx2 + cx + d) rational, or ax3 + bx2 + cx + d а. 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: 0, put the side of the n2x2. RULE 1.-When the third and fourth terms of the formula are wanting, or c and d are each square sought nx, than ax3 + bx2 And consequently, by dividing each x2, we shall have ax + b n2, or x = be any integral or fractional number whatever. side of the equation by b a 2. When the last term d is a square, put it e2, and assume the side of the required square =e+ x, and the proposed с 2e c2 formula is e2 + cx + bx2 + ax3 e2 + cx + x2. 4e2 Whence, by expunging the terms e2+cx, which are common, and dividing by x2, we shall have 4aex -† 4be2 = c2 c2 ·4be2 and consequently, x= 4ae2 с . Or if, the same case, there be put e+ x + 4be2 €2 x2 2e 8e3 for the side of the required square, we shall have, by squaring, e2 + cx + bx2 + ax3 e2 + cx + bx2 + (4be2 c2)2 64e6 And, as the first three terms (e2 + cx + bx2) are now common, there will arise, by expunging them, and then multiplying them by 64e", 64ae6x38ce2 (4be2 - c2) x2 + (4be2 — c2)2x2. Whence, by dividing each side of this last equation by x3, and reducing the result, we shall have 64ae6 Sce2 (4be2 — c2) e (4be2 - c2) (4be2 — c2)2 which last method gives a new value of x, different from that before obtained. It must be observed, however, that each of these forms fail when the second and third terms of the given formula are wanting, or b and c each 0.* 3. When neither of the above rules can be applied to the question, the formula can be resolved, by first finding, by trial, as in the former problem, some value of the unknown quantity that makes the given expression a square : in which case other values of it may be determined from this, when they are possible, as follows: Thus, let p be a value of x so found, and make 3 ap3 + bp2+cp+d=q2; Then, by putting a=y+p, we shall have ap3 + bp + cp +d=a (y+p)3 + b (y+p)2+c (y+p)+d=ay3 +(3ap+ b) y2 + (3ap2 + 2ap + c) y + ap3 + bp2 + cp + d, or ax3 + bx2 + cx + d = ay3 + (3ap + b) y2 +(3ap2 + 2up + c) y + q2. 3 From which latter form, the value of y, and consequently that of x, may be found by either of the methods given in Case 2. It may also be further remarked, that if the given formula, C * In the first of these methods, the assumed root, et is deterX, 2e mined by first taking it in the form e+nx, and then equating the second term of it, when squared, with the corresponding term of the original formula; when it will be found that n с C Qe 4be2 C2 In like manner the assumed root e+ x+ x2, in the second Ze 8e3 method, is determined by first taking it in the form e+nx+mx2, and then equating the second and third terms of it, when squared, with the corresponding terms of the given formula, when it will be found that C and m 4be2-c2 Ze |