ma as m? 6n have, in this case, x = where it appears that, in 8n2 ma order to obtain a rational answer, must be less than 8 and n greater than 6. Whence, by taking m=5, and n=2, we shall have x = 25 24 1 8 14 400 which makes + +6= 32 25 7 49 7 49 required. 6. It is required to find such a value of x as will make 2002 - 2 a square. Here, by comparing this with the general formula ax? + bx + c, as before, we shall have a=2, b == 0, and c =- 2. And, as neither a nor c are squares, but b2 4ac 4ac 4 (2 X-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. ma ·Let, therefore, 2x2 -2=2(- 1) x(x + 1) = 8 agreeably to the rule; and there will arise, by division, ma 2x - 2 n? 2n + ma and reducing the result, we shall have x = ? by taking n= 1, and m= 1, we shall have x = 3, and 2002 2 = 18 -2 = 16 2 = 16 = (4)”; or, taking n= 2 and m= result will give x = 17. But as enters the problem only in its second power, +17 may be taken instead of - 17; since either of them -2=576 (24) 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= 36, and c And as neither a nor c are squares, but he 4ас 1296 - 140 = 1156 = (34) is a square, it can be resolved, as in the last example, into the two factors, 5x + 1, and + 7. Whence, putting 5x3 + 36x + 7 = (5x + 1) x (x +7)= ma (x + 7), there will arise, by dividing, by x + 7,5x + 1 = ma (x + 7). 3, the give 2x2 5, 6 na 22 m т 10. And consequently, by multiplication, and reducing the 7ma resulting expression, we shall have a = where, 5na taking m=2, and n=l, the substitution will give a = 7 X 4-1 = 27, which makes 5 X (27) + 36 X 27+7= 5 X 1-4 4624 = (68)?, as required. 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 axa + bx + c, we have a'= 6,b=13, and c= And as neither a, c, nor 62 - 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 X 6)= 169 — 144 = 25 is now a square, this part of the formula can be resolved, by Case 5, into the two factors 3x + 2, and 2x + 3. Whence, by assuming, according to the rule, 6x2 + 13x + 10 = 4 + (3x + 2) x (2x+3)=32 +-(3x + 2)2=4+ + . 4m m2 (3x + 2)+ m2 (3x + 2), we shall have, by cancelling the 4 on each side, and dividing by 3x +2, 2x + 3 = 4m m? + (3x + 2). And consequently, by multiplying by no, and transposing the terms, we have 2nox 3m-x -= 4mn + 2ma 3n2, or 4mn + 2ma 3n? 2na Зm? ? Where, putting m= = 2, and n= 3, the result will give x = 24 +8 -- 27 or, if m be taken = 13, and n=17, we shall 18 -- 1 2 6 -5, 2(17) – 3(13) 71 Which makes 6 X (5)2 + 13 X 5 + 10 = 225 = = 225 = (15), as required. 9. It is required to find such a value of x as will make 1323 + 15x + 7 a square. m N n n 른 5 have a Here, by comparing this with the general formula, as before, we have a = 13,b= 15, and c=7. = 7. And as neither a, b, nor 12 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)or 1 2x + 2a be subtracted from the given expression, and there will remain 1202 + 1700 + 6. And as (17) 4 (6 X 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 – )2 ++(4x+3) > (3x+2) ={(1 — «) + 2m ma 2) X (3x + 2), we shall have, by cancelling (1 - x), and dividing by 2m 3x + 2, 4x + 3 = (1 - x) + (3x + 2); and consequent na ly, by multiplying by n", and transposing the terms, there will arise 4n-x + 2mnx Зm°х 2mn + 2m2 - 3n, 3no, or 2mn + 2ma 472 + 2mn Зm?" Where, putting m and n each =l, we shall have. 2+2-3 13 15 13 45 63 }, which makes + +7 + 3 + 4+2 - 3 9 9 9 9 121 », as required. 9 3 n 22 ma n 3n? 10. It is required to find such a value of x as will make 7002 4-2 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 -ty, according to Case 7, and we shall have 7x2 + 2 = 9 + 14y + 7ya; Or, putting 9 + 14y + 7yo = (3 + y)", according to the rule, and squaring the righthand side, 9 + 14y + 7y=9+ m 6т n y + maya. Hence, rejecting the 9s and dividing the remaining terms by y, we have 7noy + 14n* = 6mn + mʻy; and consequently, and X 18 25. 9 a 14n? Omn 14n2 Y = and x =lt where it is 7na ma 7n? ma evident that m and n may be any positive or negative numbers whatever. If, for instance, m and n be each taken = 1, we shall have 4 1 y Or, since the second power of x only 3 3 enters the formula, we may take, as in a former instance, x = }, which value makes 7x2 + 2 7+2= + a square. Or, if m= 3 and n= 1, we shall have x = 17, and 7202 + 2 = X (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 fi ease. PROBLEM 2.---To find such values of X as will make v (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 + bx* = nox2. And consequently, by dividing each side of the equation by na b 2*, we shall have ax +b=n?, or x = be any integral or fractional number whatever. 2. When the last term d is a square, put it = f', and assume the side of the required square = e+ ®, and the proposed 2e . where n may a с c2 formula is ea + cx + bxc2 + ax = e + cx + 2. . 4e2 Whence, by expunging the terms es + cx, which are common, and dividing by a?, we shall have 4ae*x + 4bea = c; ca 4be2 and consequently, x = 4ae 4be ca Or if, the same, case, there be put et xt 2e 8e3 for the side of the required square, we shall have, by squaring, с e (4be ca) ez + cx + baca to ax3 et cx + baca + 8e4 (4be ca) 24. And, as the first three terms (es + cx + bxo) 64€ are now common, there will arise, by expunging them, and then multiplying them by 640, 64aea = 8ce2 (4be? --co) + (4be - ca). Whence, by dividing each side of this last equation by 3*, and reducing the result, we shall have 64ce® – 8ceo (4be? – (*) (46e* - ) 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 1 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 ap3 + bpa + cp + d=qo; Then, by putting x =ý tp, we shall have ap + bpa + cp +d=ay + p)3 + bly + p + cy + p) +d=ay +(3ap+ b) ya + (3apa + 2ap + c).y + ap3 + bp + cp + d, or ax: + bx2 + cx + d = ay + (3ap+by + (3apa + 2ap + c)y + q* 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, 2e2, is deter C * In the first of these methods, the assumed root, et mined by first taking it in the form e tnx, 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 2e 4602-02 In like manner the assumed root ett 22, in the second 8e3 method, is determined by first taking it in the form etnutmx2, 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 4be2 c C2 n and m |