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√4x2

=

=

0;

or, ( 14 22 - 9x + 9 + 3x)2 = (6 x — 7)?.

+ :: J4* - 9x + 9 + 3x = + (6 x - 7),............(1.); or, taking the upper sign√4x2 9x + 9 = (6 3 – 7) - 3.2 = 3 – 7.

3 Hence, squaring each side

4 x2 9x + 9 = 9 aca - 42 3 + 49;

33 x + 40
5) (5 x 8) 0.

.. X =
Again, taking the lower sign of (1), we have
4 -9 + 9 = -(

6-7) - 3 = -92 + 7......... (2.); x

x or, squaring

4 22

9 x + 9 81 cca 126 x + 49; from which 77 22 117 oC + 40

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

or, 5 22 or, (

5 or .

0;

.. 3C =

1 or 49.

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Hence, the roots of the given equation are 5, %, 1, 19.

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 xc" 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 — 14 x? 9 x + 9 9 (1) + 7 = - 2,

)

-
and 14 oct - 9 x + 9 = -9 (49) + 7 = 244;

44+ 7 and on substituting either of these values of /4x2 – 9x + 9 along with the corresponding value of x, the given equation is satisfied. Ex. 4. Solve

(a + b + c) (x + b - c)(b + c - x)

(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

- a as a factor of each side (Art. 5).

a

b + c b + c

b + c

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We have

(oc + b + c) (+ b c)
(a + b + c)(a

+ b c)
(a + b)2 – C

(a + b)2 – 0 b+ci
or, taking the difference of numerator and denominator-

(oc + b)2 – (a + b)2
(a + b)2 – ca

b + c

i
(2 a) (x + 26 a)
or,

(a + b) - c
Dividing each side by x - a, we have-

1
(a + b) - co
ca

0, or x = a; Hence also (x + 26 - a) (6 + c – X) = (a + b)2 – c*; an

( ordinary quadratic from which two other roots may be determined.

a

b + c

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a

and a

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a =

b + c

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4 albic 2 p

(a - b)”;
or,
(aba?r*)* – 4 att1 (abx"=>") = (a - b)*;

*
or, completing the square-
(abrer) - 4 ałół (abæ?=?) + (2 a112)*

abse
(a - b)2 + 4 ab = (a + b)”.

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

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Hence, extracting the square root

abx 2 PT.

2 albi

= + (a + b). :: abz"Pi = + (a + b) + 2 alb} = (až +84)or - (at - 1); 2

+63? ***

63

atit 1 1

1 61 Then, raising each side to the (2 pq)th power, we have 1

14 bi Hence, taking the (p = q)th root, we get1

1 33

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c =

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p - 9

+

a2

We have also, since the factor 27 has been struck out

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5. 3 x-6

17

Ex. II. 1. 4 24 11 22 225. 2. 5 206 17 203 184. 3. 2c + 5 -% - 257. 4. (x2 + 3x + 3) + 2 * = 189 -6 2. (2

= 1450. 6 6. och t

= 5.

an 17. W + 12 + N + 12 = 6. 8. hoc 7 9 + 20 = 22 + 9. 12

60 9. - -2)' + = 6 + 5 c.

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.

a? + 2 22

24

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ax

4 1 13. 9 2 + 24 2

+ = 14. XC = Na do

Õnä + a. 1

1 15.

+
(a
Na2

ac*) (a + da2
aca
16. 2

ax + x2

1. 6

b

6 1 . a 72 1107.11 17. 22 P 9

(vē + 27) = 0.
2 a + b^
+ oc

aca 18.

1. 2c2

aa 19. (a4m + 1) (x - 1) 2 (x + 1).

2c2 975 22 9 7 20.

0. oca 9 256 xo

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La

+ x2

+

00 V

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+

= 9.

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9 agc

12 3
5

2 21.

4
+
20 + 1

& 202 22. 6 22 5 x 8 J3 20+ 5 x

4 = 12. 23. 2 202 + 5 x 2 x 12 x 7 x + 1 -- 35. 24. 20 22

8 x 15 x2 3 ax + 2 ao = -7 a?. 25. a*(x-2) + 6 x*(x - 2) Qc 2

= 2430 + 36 5x?. 26. 2 x2 + 20 - 13 x2 2C + 5 = 105. 27. 4" + 12 x - 24 - 2x + 19 = 30,

ac 28. a* + c + 1 = a(a + x + 1). 29. 208 1 = 0.

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c).

30. (2o – a)3 + (22 – 5)2 + (x2 - c)3
= 3 (22 a) (2* - b) (x= – c) + 9 axt a(a + b + c)?.
31. (x - 1) (2— c) = (a - b) (a® - co).

b9
32. (x – a) (x – b) (2 — c) (m a) (m b) (m
33. (x – a) (x – b) (x – c) = (a + b) (a + c) (b + c).
34. (n − 1)» (x2 + ax + a?) as – 23.
35. (ax 5) + (cx - d)3 = (a + c)2 – (6 + d)".
- 03

b +1 alaca + x + 1).

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36. 22" + x2n1

Simultaneous Quadratic Equations.

20.....

xy +

6. The following worked examples are given as specimens of the methods to be employed, but it must be understood that practice alone will give the student complete mastery over equations of this class. Ex. 1. Solve aa + ya

(1.) 2 + y

6........ As we have given the sum of the unknown quantities, we shall work for the difference. From (2), multiplying each side by 2, we have

2 22

+ 2 y 40 and from (1), squaring,

a + 2 20

y = 36 Then, subtracting,

ac 2 xy + y 4; and, taking the square root, we have a

Y + 2........(3). (2) + (3), then 2 x = 8 or

4, and :: 3 =

4 or 2. (2) - (3), then 2 y = 4 or 8, and .. y = 2 or 4.

2 NOTE.-Having found that x = 4, y = 2, we might have told by inspection that the values 2 = 2, y = 4, would also

satisfy the given equations, for x and y are similarly involved in both equations. Ex. 2. Solve ac? yo 5.......

(1.)
wy
6.

..(2.) As we have given here the difference of the squares of the unknown quantities, it will be convenient to work for the sum of the squares,

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