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Substituting now in these values of x, y, and 2, the values of a, b, and c, the times in which A, B, and C could perform the work when acting alone, will be found thus:

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1

4

x2-4x-45, by transposition.

x2-4x+4=49, completing square, (Art. 82.) x-2=7, extracting root.

.. x=27=9, or -5.

It may also be solved at once, by substituting in the general formula, (Art. 80), the proper values of a, b, and c, found in (1) above, which gives

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Art. 81

2

2x2+2x=24, by subtracting 6

from both.

16x2+16x+4=196.

Extracting root 3 4x+2=14.

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9

x+1 x+2

(1)×x(x+1)(x+2) 2 30x2+90x+60—30x2—60x—

9x2+9x.

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(12.) Given

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(2)Transposed Art. 81.

-3 x+4

(1)×8(x−3)(x+4) 2 8x+32+8x-24-9x2+9x—

Extracting root

108.

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+

9.

X- -4

2

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Extracting root 5 x-9-3.

6 ..x=93=12, or 6.

SOLUTIONS OF QUESTIONS PRODUCING QUADRATIC

EQUATIONS.

1. Let x the less, then x+15 will be the greater; and,

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2. Let x=

the less, then 100-x will be the greater; and, 1 | x(100-x)=2059.

by the question,

changing signs of (2) Art. 82. Extracting root

2

100x-x2-2059.

3 x2-100x=-2059.
4x-100x+2500-441.

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3. Let the less, then x+8 will be the greater; and,

by the question,

Art. 82

1x(x+8)=240.

2x2+8x=240.

3x2+8x+16=256.

Extracting the root 4x+4=16.

5 ..x=16—4—12, or -20. 6x+8=20, or -12.

4. Let the prime cost of the piece; then by the principles of "Loss and Gain," we have the following

proportion :

1

x: 100 :: 24: 100+x.

Art. 68

2

x2

Art. 82

Extracting root

5 ..

+100x=2400.
3x2+100x+2500-4900.
4x+50-70.

x=70-50-20, or -120.

5. Let x the number of oxen bought; ..

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price in pounds paid for each; and since he retained six to himself, the number sold was (x-6), and the price for

480

which each was sold was ; but by the question,

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2 480x-2880+4x2-24x-480x.

34x2-24x-2880.
4x2-6x=720.

5 x2-6x+9=729.

Extracting the root, 6x-3+27.

7x=327=30, or -24.

NOTE. In several of the preceding solutions, and in those that follow, it may be remarked, that of the two values found, only one is consistent with the conditions of the question, although both fulfil the conditions of the algebraic equation derived from it, thus showing that the algebraic equation is more general than the question from which it was derived. In the Algebra those values only are given which are consistent with the nature of the question.

6. Let the length (and consequently the price in shillings of a yard) of the shortest trench; then (x+4) will represent the length (and also the price in shillings) of a yard of the longest trench; .. by the question,

collecting (1)
(2)+2
Art. 82

Extracting root

1 x2+(x+4)2=400 shillings in L.20.
22x2+8x=384.
3x2+4x=192.
4x2+4x+4=196.
• +2=-14

6. x=14—2—12, or
7 and x+4=16, or -12.

-16.

7. Let x=the breadth of the frame, then the length of the glass and frame will be (24+-2x), and the width (16+2x), and the surface of the glass and frame will be (24+2x) (16+2a), while the surface of the glass is 24x16-384; and by the question, the first is double of the second; whence

1

(24+2x)(16+2x)=768. 2 4x2+80x+384-768.

(2)÷4, and transp. 3x2+20x=96.

4x+20x+100=196.
x+10=14.

Art. 82

Extracting root

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The other value is evidently inconsistent with the nature of the problem.

8. Let x=the first; then since the sum of the first and second is 10, the second will be (10-x); again, since they are in geometrical progression, x : (10—x) :: (10—x) : (10-x)2 the third; therefore by the question,

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