Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση
[ocr errors]

a.

(l

44. *

[ocr errors]
[ocr errors]

+

+

[ocr errors]

с

45. C

1).

1).

[ocr errors]

42. * 12 a a + C + 14 a 2 dã
43. * Vax + 6 + 1bx + a = (a - b)x + a + b + 2 Vab.
b

+b
b
a + b + c

2. 6 162 46. (a - b) (x + b + c) (c + c - a) + aʻ(x + 6 c) + 62(x + c a) + c*(c + a b = 0).

1 47. Show that, if x be real, the value of a + cannot lie between 2 and 2.

2c2 48. If, in the equation 2 – 20 = a, the quantity o be real, show that à cannot be greater than 5.

[ocr errors]

Equations which may be solved like Quadratics. 5. Certain equations of a degree higher than the second may be solved like quadratics. It will be seen that, although it is impossible to lay down definite rules for the treatment of every such equation, the object to be attained is either (1.) To throw the equation into the form

X2 + px

9, when X is an expression containing the unknown quantity, and solving when possible this equation for X; or (2.), To strike out from each side a factor containing the unknown quantity, thus reducing the dimensions of the equation, and obtaining a value or values of the unknown quantity by equating this factor to zero.

Ex. 1. Solve the equation 2* + 144 25 2".
We have, transposing, ** – 25 * = - 144;

or, (*) - 25 (2) = - 144;

* See Remark, page 304,

= 6 2 5

4

[ocr errors]

0;

0, and

= 0.

or, completing the square-
(sco) – 25 (2) + (49)

144

= + 7.
:ca 2 + 3 16 or 9.

Hence, ac + 4 or $ 3. The given equation has therefore four roots, being as many roots as the degree of the equation.

Ex. 2. Solve 26 + 3 3 88.
We have, (?) + 3 (c) = 88; or, transposing,

(203)2 + 3 (204) 88
or, (20 - 8) (308 + 11) = 0.
Hence, the given equation is satisfied by-

2013 8 0, and also by x + 11 0.
We have then 203 - 28 0, and 28 + (11)3 = 0;
or, breaking into factors, we have

(2 – 2) (20* + x2 + 2a)
22

711
From the first of these equations, we get-

2 = 0 or a = 2, and x2 + 2 C + 4 = 0, from which two other roots may be found. And from the second equation, we get

+ 911 = 0, or a = 311, and 2c2 Jīl + 7121 0, which gives two other roots.

We have therefore shown how to obtain the six roots of the given equation. Ex. 3. Solve23 22

6 x 14 22 92c + 9 + 40 = 0. Changing the sign of each side, and transposing, we have

6x 14.2 - 9 x + 9 = 23 - 75x + 40; adding to each side 13 – 9 x + 9, then13 22 - 9x + 9 + 6x 74 - 9 x + 9 = 36 22 – 84.C + 49; or, (4 x2 – 9 x + 9) + 2 (3x) 14 x2 – 9x + 9 + (3x)2

= (6 x)2 – 2 (6 x) • 7 + 7*;

75C

or, (14 22 – 9 x + 9 + 3 xc)2

9x + 9 + 3x)2 = (6 x - 7). :: 74 m – 9x + 9 + 3x = + (6 x - 7), ............ (1.); or, taking the upper sign

14xc- 9x + 9 = (6x - 7) - 3x = 3 - 7. Hence, squaring each side

4 oca 9 XC + 9 = - 9 22 42x + 49;

33 x + 40
or, (x - 5) (5x - 8) = 0.

or, 5 22

0;

[ocr errors]

5 or .

Again, taking the lower sign of (1), we have J4 – 9x + 9 = -(6 x - 7) - 3x = -92c + 7........ (2.); or, squaring

4 oca 9x + 9 81 oca 126 x + 49; from which 77 22 117 x + 40 = 0;

or, (0 - 1) (773 - 40) = 0.

=

.. X =

1 or 49.

Hence, the roots of the given equation are 5, 1, 1, 1..

REMARK.—If we proceed to verify these values of x, we shall find that the last two values—viz., 1 and 44—will not satisfy the given equation unless we obtain the value of 14 x

9 x + 9 by means of the equation from which these last roots were found.

Thus from (2) we find, on putting 1 and 44 successively for a

√4x2 9x + 9 9 (1) + 7 = - 2, and J4 9 x + 9 = -9 (44) + 7 = 244; and on substituting either of these values of 14x2 – 9x + 9 along with the corresponding value of 2, the given equation is satisfied. Ex. 4. Solve

(3C + b + c) (2 + b - c)(b + c 2)

(a + b + c)(a + b c) (b + c a). By inspection we see that a is one of the roots. We shall therefore so arrange the equation as to be able to strike out PC – a as a factor of each side (Art. 5).

a

b + c b + c

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

C

[ocr errors]

b + c

ܪ

We have

(a + b + c) (20 + 6 c)
(a + b + c)(a + b

; i
c)
(20 + b)2 ca
or,

(a + b)2 - C
or, taking the difference of numerator and denominator-

(ac + b)2 – (a + b)2
(a + b)2 ca

i
(u a) (x + 2b a)
or,

(a + b)2 ca Dividing each side by a

a, we have 3 + 2 6

1 (a + b)2 - 02

0, or x = a; Hence also (c + 26 a) (b + c – x) = (a + b)– co; an ordinary quadratic from which two other roots may be determined.

[ocr errors]

a

b + c

[ocr errors]

a

and a

a =

b + c

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

* We shall assume in this example that the laws of multiplication and division proved in Algebra, Stage I., Arts. 25, 27, hold for fractional indices,

Hence, extracting the square rootabapp' _ 2 allt

= + (a + b). :, abx ? P9 = + (a + b) + 2 alb} = (aš + b) or – (- 62); ai

63)

;
atot
1

Then, raising each side to the (2 pq)th power, we have

1 14 pq
2P - Q
63

lo
Hence, taking the (p - g)th root, we get-
1 4 pg_ 1 1 4 pq

P-8

[ocr errors]

We have also, since the factor 2o has been struck out

1
XP = 0, and :: 2 = O as another solution,

[ocr errors][ocr errors]

Ex. II. 1. 4 oct

11 cm =

225. 2. 5 206 17 23 184. 3. c + 5 2-2 251. 4. (22 + 3x + 3)2 + 2 x 189 - 6 x 5. 3 - 17 -= 1450.

6 6. c +

5. n. Wat + 12 + N + 12 = 6. 8. N + 9 + 20 2c + 9. 12

60 9.

= 6 + 5 aC. *

20"

« ΠροηγούμενηΣυνέχεια »