Page images
PDF
EPUB

in any case of this kind, can be resolved into factors, such that one of them shall be a square, it will be sufficient to make the remaining factor a square, in order to render the whole expression so; since a square, multiplied or divided by a square, is still a square.*

EXAMPLES.

1. It is required to find such a value of x as will make 11x3-3x2 a square.

2

Let the given expression 112+3x2= n2x2: agreeably to Case 1.

n2;

Then, by dividing by x2, we shall have 11x+3 = n

and consequently, x =

n 3

11

where n may be any number,

positive or negative, that is greater than √3.

Taking, therefore, n = 2, 3, 4, 5, &c., respectively, we

1 6 13

shall have, in this case, x ==

or 2, the last of

11' 11' 11'

which is the least integral answer that the question admits of. 2. It is required to find such values of x as will make 2x2 + 2x + 1 a square.

Here the last term 1, being a square, let 1+ 2x 2x2 + x13 — (1 + x)2 = 1 + 2x + x2, agreeably to the first part of Case 2.

Then, since the first two terms, on both sides of the equation, destroy each other, we shall have x3

2x2 = x2, or 203 =3x2, and consequently = 3; which, by substitution, makes 1+ 2x 2x2+x3 1+6-18 +27 16 a square, as required.

3

Again, by putting xy + 3, according to Case 3, we shall

=

X2

* The method of determining the factors of which any formula is composed, when it can be done, is to put the given expression = 0, and then find the roots r, ?', &c., of the equation so formed; each of which will give a factor xr, x- 7', and these are generally easily discovered, as we here seek only the rational roots, which are always divisors of the absolute term, or of that which does not contain x. Thus the formula x3- x2-x+1 is resolvable into the factors (1−x)×(1+x)×(1−x), or (1 — x)2 × (1+x); and by putting 1+x= n2, we have 2 = n2-1; where, if n be taken equal to any number whatever, 23 ·C2+1 will be a square; though, by any other mode of solution, it would be difficult to find even two or three values of x. It may here also be observed, there are but few questions in this problem that can be determined in whole numbers. Several of them, likewise, admit only of one answer, and others are totally irresolvable, either in integers or fractions. Thus, if it were required to make 23+1 a square, the only positive value of x that renders this possible, is 2; and the making of 3x2-1 a square, is impossible.

have 12x2x2 + x3 = 1 + 2 (y+3)-2 (y+3)2 + (y+3)3 16+ 17y.+ 7y2 + y3.

And consequently, by making 16+ 17y+7y2+ y3 ̈

17

289

(4 + ——— y)2= 16 + 17y + y2, agreeably to the first part of

8

64

Case 2, by cancelling 16+17y, there will arise 7y2 + y3=

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

Which number, being substituted in the original formula,

[blocks in formation]

3. It is required to find such values of x as will make მეც3

5x2 + 6x + 4 a square.

Here, 4 being a square, let 4 + 6x5x2+3x3 — (2+3x)3 ·4 + 6x + 2x2, as in the first part of Case 2.

9 4

Then, since the first two terms on each side tion destroy each other, we shall have 3x3

3x

5= 9

2, and consequently, in this case,

29

29

45
2

of the equa5x2 = 2x2, or

[blocks in formation]

Whence (2+)2 = (2 + ~~ )2 = (~-~~)2 a square, as

was required.

Or, by the second method of the same Case, let 4 + 6 x

[merged small][merged small][merged small][merged small][ocr errors][merged small]

841

256

x; then, as the first three terms on each side of this

[merged small][merged small][merged small][ocr errors][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small]

of x, that, being substituted in the original formula, will make

it a square.

4. It is required to find such values of x as will make 23+3 a square.

Here, it is evident that the expression is a square when Let, therefore, x= =1+y, and we shall have

х 1.

3 + x3 4 + 3y + 3y2+ y3.

And, as the first part of this is a square, make, according to the first part of Case 2, 4+ 3y + 3y2 + y3 — (2 + 3y)2 4 + 3y+ y2. Then, because the first two terms on each side of the equation destroy each other, we shall have y3 + 3y2 = √y2, or y + 3 = f.

9 16.

[ocr errors]
[blocks in formation]

39 16'

39

[blocks in formation]

; which is a second value of x.

3

Again, let 4+3y+3y2+y3 = (2+y+64y2)2 = 4+

117

1521

39

y,

3y+3y2+12843 +140963, according to the second part of

Case 2.

Then, as the first three terms on each side of the equation

[blocks in formation]

And by proceeding in the same way with either of these new values of x as with the first, other values of it may be obtained; but the resulting fraction will become continually more complicated in each operation.

4

PROBLEM 3.-To find such values of x as will make √ (ax1 + bx3 + cx2 + dx + e) rational, or ax1 + bx3 + cx2 + dx +e= a square.

The resolution of expressions of this kind, in which the indeterminate, or unknown quantity, rises to the fourth power, is the utmost limit of the researches that have hitherto been made on the formula affected by the sign of the square root; and in this problem, as well as in that last given, there are only a few particular cases that admit of answers in rational numbers; the rest being either impossible, or such as afford

one or two simple solutions; which may generally be found as follows:*

RULE 1.—When the last term e, of the given formula, is a square, put it f2, and make f2 + dx + cx2 + bx3 + ax1 d 4cf2 — d2

(f+ x + 2f

8f3

(4cf2 — d2)

[ocr errors]

64f6

-∞2)2 = ƒ2 + dx + cx2 +

d(4cf2 — d2)

8f4

[ocr errors]

Then, by expunging the first three terms, which are common to each side of the equation, there will remain ba3 + d(4cf2 — d2) (4cƒ3 — d2)2

ax1

4

8f4

[ocr errors]

64f8

x2. And consequently, by

dividing by x3, and reducing the result, we shall have x = 64bf8df2 (4cf2-d2)

(4cf-d2)2 -- 64af6

[ocr errors]

which form fails when the coefficients c and d, or b and d, are each 0.

2. When the coefficient a, of the first term of the formula, is a square, put it ÷± g2, and make g2x2 + bx3 + cx2 + dx + e

[merged small][ocr errors][merged small][merged small][ocr errors][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]
[blocks in formation]

(4cg2b)264eg

64dg® — 8bg2 (4cg2 — b); which form likewise

fails under similar circumstances with the former.

3. When the first and last terms of the formula are both squares, put a = g2, and e = f2, and make ƒa + dx + cx2 +

bx3 + g2x2 = (f+x+g2x2)2== f2 + dx + (2fg+

[merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small]

Or, because g enters the given formula only in its second

* As an instance of what is above said, it may be observed, that the only value of x that renders the formula 2x4 3x2+2 a square, is 1; and the formula 24- x21 can never be a square, except when 2+1, or -1,

1.

power, it may be taken either negatively or positively; and 4 de f2 (2fg+c)

consequently, we shall also have x ==

1

f (bf+dg) So that this mode of solution furnishes two different answers. Also, if there be taken for another supposition f2 + dx

b

bf g.

+ cx2 + bx3 + g·2x4 = (ƒ + x + gx2)2 = ƒ2 + x + (2ƒg

b2

2g

+408) x2 + bx2 + g2x2, hence, by cancelling, dx + cx2 = bf x

b2

+ (2ƒg + —) x2; and consequently, x

g (bg — bf) 4.g2 +b2 + g2 (2fg—c)' And because f enters the given formula only in the second power, it may be taken either negatively or positively; and

consequently, we shall also have x

g (bg+bf)
1b2 — g2 (2ƒg + c)°

4

So that this solution likewise furnishes two values for x, which are each different from the former.

But these forms all fail under similar circumstances with those of the second Case.

4. When neither the first nor the last terms are squares, the formula cannot be resolved in any other way than by first endeavouring to discover, by trials, some simple value of the unknown quantity that will answer the conditions of the question, and then finding other values of it according to the methods pointed out in the last two problems.

Thus, let p be a value of a so found, and make ap + bp3 +cp2 + dp + e = q2.

4

4

Then, by putting xy+p, we shall have ap1 + bp3 + cp2 + dp + e = a (y + p)2 + b (y + p)3 + c (y + p)2 + d (y + p) + e = ay1 + (ap + b) y3 + (6ap2 + 3bp + c) y2 + (4ap3 + 3bp2 +2cp + d)y + ap2 + bp3 + cp2 + dp + e, or ax1 + bx2 + cx2 + dx + e = ay1 + (ap + b)y3 + (6ap2 + 3bp + c) y2 + (4ap3 + 3bp2 + 2cp + d) y + q2. From which last formula, the value of y, and consequently that of x, may be found by Case 1.

1

EXAMPLES.

1. It is required to find such a value of x as will make 2x + 3x2 4x3 + 5x4 a square.

Here, the first term 1, being a square, let 1 2x + 3x24x3 +5x=(1 −x + x2)2 = 1

ably to the method in Case 1. Then we shall have 5x⭑

2x + 3x2 - 2x2 + x1, agree

[ocr errors][merged small][merged small]

And consequently, 5x-4x-2; whence x = 23.

4

« PreviousContinue »