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MATHEMATICS.

SECOND STAGE.

SEOTION II.

ALGEBRA.

CHAPTER I.

QUADRATIC EQUATIONS.

1. Equations of this class, when reduced to a rational integral form, contain the square of the unknown quantity, but no higher powers.

When the equation contains the square only of the unknown quantity, and not the first power, it is called a pure quadratic. Thus, ac 25 0, 4 a + 10 19, 5х2

180, are pure quadratics. When the equation contains the square of the unknown quantity, as well as the first power, it is called an adfected quadratic. Thus, ca 5 x

30 = 0, 2 x2 + 3 + 3 6, are adfected quadratics.

=

6, oc?

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Pure Quadratics. 2. To solve these, we treat them exactly as we do simple equations, until we obtain the value of the square of the unknown quantity; then, taking the square root of each side, we obtain the value of the unknown quantity. It will be

seen (Stage I, Alg.; Art. 35) that the unknown quantity in a pure quadratic has always two values, equal in magnitude, but of opposite sign. Ex. 1. Given 33" + 12 = 687, find s. We have 33 687 - 12 675, or = 225.

is = 15. 22 +7 22 - 7

56 Ex. 2. Giren

find ... 23 - 77 2F+TE Er Bringing the fractions on the first side to a con non do nominator, we have(2 + 7- (22-7

56 z (43 - 49)

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[ocr errors]

or,

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2 (43

63-93 1

1 49

€ 59 clearing of frustions; ; then, 6 - 99 = 47 - 49, from which

* = 25

• 5.

Adiected narratics. 3. Schutin by rk jsting the opste

Suppose we are greatest 1 2 3 = 34, to find a It will be resouzaina 7

**, is a perfet square, viz, the sparen 2-4, Isis erit, 24., a to first side of raw a serius size by tia addition on a a doiros

Adding, there was a test runn, we have

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4**. Taking in the risk an Bitte meters

24

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MATHEMATICS.

SECOND STAGE.

SECTION II.

ALGEBRA.

CHAPTER I.

QUADRATIC EQUATIONS.

1. Equations of this class, when reduced to a rational integral form, contain the square of the unknown quantity, but no higher powers.

When the equation contains the square only of the unknown quantity, and not the first power, it is called a pure quadratic. Thus, ac 25 0, 42cm + 10

19, 5 cl quadratics. When the equation contains the square of the unknown quantity, as well as the first power, it is called an adfected quadratic. Thus, ac

6, oca

30 0, 2 x2 + 2 + 3 6, are adfected quadratics.

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180, are pure

5 x =

[ocr errors]

=

Pure Quadratics. 2. To solve these, we treat them exactly as we do simple equations, until we obtain the value of the square of the unknown quantity; then, taking the square root of each side, we obtain the value of the unknown quantity. It will be

.. X =

[ocr errors]

find x.

seen (Stage I., Alg.; Art. 35) that the unknown quantity in a pure quadratic has always two 'values, equal in magnitude, but of opposite sign. Ex. 1. Given 3 x + 12 687, find ac. We have 3 c2 = 687 12

675, or ac* = 225.

† 15. 2x + 7 2x 7

56 Ex. 2. Given

2* - 730 2x2 + 7 x 62 - 99' Bringing the fractions on the first side to a common de nominator, we have

(2 sc + 7)2 – (2 x – 7) 56
2 (4 x2 – 49)

6 cm 99'

56 XC (4 x2 49) 6

99'
1

1
49 6x2 – 99, clearing of fractions;
then, 6 m2

99 4 x2 49, from which
oca 25

† 5.

56 3

or,

or, 4

=

X =

find x.

Adfected Quadratics, 3. Solution by completing the square. Suppose we have given the equation c* + 2 ax = 3 a', to

It will be remembered that (x2 + 2 ax + a?) is a perfect square, viz., the square of (x + a). It is evident, then, that the first side of our equation will become a perfect square by the addition of a? as a third term.

Adding, then, a' to each side of the given equation, we haveacea + 2 ax + a?

4 aʻ, or

(2c + a)2 Taking now the square root of each side, we have

+ 2 a. + 2 a

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ac + a =

..

a.

= a or

3 a.

We may remark that the quantity a’, added to the expression 2 + 2 ax in order to make it a perfect square, is the square of half the coefficient of x.

The operation itself is called completing the square.

An adfected quadratic may therefore be solved as follows:

1. Reduce and arrange it until all the terms involving x are on the first side, and the coefficient of xo is unity.

2. COMPLETE THE SQUARE by adding to each side the SQUARE OF HALF the coefficient of x.

3. Take the square root of each side, put a double sign to the second side, and transpose the term of the first side not involving x.

Ex. 1. Solve the equation 3 m2 + 18x + 4 52.
We have 3 x2 + 18 * 52 4

48; or, dividing each side by 3,

2a + 6 x 16. Here, the coefficient of x is 6, the half of which is 3. Adding then the square of 3 to each side, to complete the square, we have

wca + 6 x + 32 = 16 + 9 25. Taking the square root of each side, we have

x + 3 = + 5. .. x = $ 5

3 2 or - 8.

4x + 9 12 x + 17 Ex. 2. Given

x + 4
3 + 2

2 x + 7 6x + 16'

2 + 3

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find x.

We have (1-1)-(1-1)-(2-2)

;

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X + 4

15
2

16.
1
1
15

5
or, simplifying,
x + 2

6x + 16 2 x + 7 (oC + 4) (2c + 2) 15 (2 x + 7) – 5 (6 x + 16); or, (3 + 2) (2 + 4) (6 x + 16) (2 2 + 7) 2

25 or, simplifying, x2 + 6x + 8

12 x + 74 x + 112

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