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22

X + 1

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4 or

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or, clearing of fractions

24 x2 + 148 + 224 25 x + 150 xC + 200; or, transposing and reducing, x2

24; or, changing the sign of each side, then

oca + 20 24; or, completing the square, xo + 2 x + 12 = 24 + 1 = 25.

x2 Taking the square root of each side, we have

+ 5 + 5 1

- 6. Ex. 3. Solve the equation 22 + 6 x + 25 = 0.

We have x + 6 x 25; or, completing the square

och + 6x + 32 = 25 + 9 16; or, extracting the square root of each side

+ N = 16

ii XC 3 + N = 16. As the quantity - 16 has no exact or approximate

v value, the given equation has no real roots. The roots are therefore said to be imaginary.

4. Solution bij breaking into factors.

We have seen (Stage I., Alg., Art. 30) that it is often easy to find by inspection the factors of quadratic expressions. We may make use of this knowledge to solve quadratic equations. Ex. 1. Solve the equation ac +

5 Transposing all to the first side, we have

x2 + 5x 66 = 0. And, resolving the first side into its elementary factors, we get

(x – 6) (20 + 11) = 0. Now, if either of these factors is put equal to 0, the equation is satisfied. Hence, we have, a 6 0, and 2 + 11

2 = 66.

0; 6, and =

11. :: 6 and 11 are the roots required.

or, X

=

Ex. 2. Given - (a + b) 2 + ab = 0, find z.
We have (oc a) (5 6) 0.

Now this equation is satisfied by making either of the factors Hence, os

0, or se = a;
and, as b 0, or s = 6
i. a, b are the roots required

a =

=

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AX

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ab aoba
9.
+

72 = 2.
oca
1

1 10.

= Ca. Va*x* 1

ax + Na2x2 Wat + 2 + ya zc 11.

Vas + " Ta Ocea

a + 1 12.

13. XP 5x + 6 = 0. Nat

1 14. a 72.

15. 3 x2

X = 2. 16. 422

28.
17. 6x + x

35 = 0. 18. 200 + 63 ~ + 8 = 0. 19. ax + bx + c = 0. 14

42 20. 3 ~ 42 – 7.

1. 220 3

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21. 30

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៩.

2

26 22. 2 x + 1

2.2 + 1

= 11.

* See Remark, page 304.

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23. (a + b) (ca + d) = (a + b) (c + dx).
24. (a
b) (202 (a + b) (a - b)

20 6? 25. (a + b) (2 a) (2 6) abx.

6

a? 62 26. b

ab

cd 27. ax? - (c + d) x = bu?

6 28. aʼz– (a - b) x + a3 = 68x2 + (a2 + ab + 6?)x + 69.

23 2 29.

31.

a

+ X

+ a

C

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7 2x + 5 30.

+ 2 x + 3

5 2 x + 9

X 31.

53. 2C

2 x + 9 X + 2

20 10 32. +

0. C + 3

X + 8

7 5x + 3 5 33.

2

7 2 + 4 9 6 x + 5 4 x 1 34.

7 X + 1 2 7

X

2 C 3 2 x + 1

x + 2 7 x + 8 2 + 2 4x + 4 4x + 13

3 x 2 x + 9 36.

2 x + 24
C 2
2 + 2

2x 1
3 x2 2 x + 7 cca + 7 a 3
37.
6 22 4 x + 11

2 2 + 14 a 99 3 22 + 10 aca + 2a + 3 2 cm + 2 x + 10 38. C x + 2

2 + 1 39. 2 + 6 + Nac

C + 10 = 4. 40. 3 x - 4 - 2 2 - 4 = 1

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45. ( - 1)

42. * 12a + ac + 14 a 43. * Vax + b + Vbx + a = (a - b)x + a + b + 2 Vab. b

7

a + b + c . 1).

2. 6 62 46. (a - b) (x + b c) (x + C - a) + a (x + b = c) + 64(x + c a) + c*(x + a 6 o).

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

2c2 48. If, in the equation oc - = a, the quantity be real, show that a cannot be greater than 5.

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20

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,

X + 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 24 + 144 25 2. We have, transposing, ** – 25 c = - 144; or, (sco) - 25 (c) =

144;

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* See Remark, page 304,

=
62 5
4

شد .:

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or, completing the square-
(202)2 – 25 (2%) + (%)?

144 =
20
+;

16 or 9.

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

Ex. 2. Solve 2 + 3 23 88.
We have, (+9) + 3 (2-) = 88; or, transposing,

2c3?
(oc)2 + 3 (204) 88
or, (203 - 8) (28 + 11) = 0.
Hence, the given equation is satisfied by-

acis 8 0, and also by 203 + 11 = 0.
We have then 203 0, and 202 + (11)' = 0;
or, breaking into factors, we have-

oc
(3 - 2) (2 + x2 + 22)
ca

=0.

0;

ac

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- 28

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=

0, and

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From the first of these equations, we get

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

20 + F11 = 0, or a = 11, and ca - 2 711 + 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 cm 9 C + 9 + 40 0. Changing the sign of each side, and transposing, we have6x4- 9 x + 9 = 23 - 75x + 40; x 2 92

oc? adding to each side 13 ac – 92 + 9, then

x2 13.2 - 9 x + 9 + 6 4 * - 9x + 9 = 36 22 - 842 + 49; 9

m2 or, (4 x2 – 9 x + 9) + 2 (38) 14 x2 – 9x + 9 + (3x)?

(6 a) - 2 (6 a) 7 + 7;

x (6 x• ;

75 x

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