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3. Let 57-90.-270=945, to find in this equation the values of I. Solu. By transposition,

510-90r—270=945=579—90x=945+270= 1215. Then divide by 5, and we get to -18.x = 243. Complete the square, and it becomes

-187 +81=243 +81 = 324. Extract now the square root, and 2--9=+324=+18.

Therefore =+18+9=27, or -9;

And x=27, ora 9=3, or 0-9. Nole.-Without the formality of substituting ye 'for 36, and y for 308, we proceeded at once to the solution of the equation, 26. 18x3

243, as exemplified in the second step of the operation.

4. Find, in the equation 3ax2–26=4x-5c, the values of x. Solu. First, by transposition, we get 3 ax®-4x=26-50.

4' 26-50 Then divide by 3a, and it is re

3a Complete the square, and

4 4 26-50 4 6ab-15ac+4
-1 +

+
30

3a

3a

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Examples for Practice 3. Let z? — 78+1=9, to find the values of x.

Ans. x=8, or 1. 6. Given 4r—7x=492, required the values of x.

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7. Let 2-6r+19=13, to find r's values.

Ans. I=4.732, or 1.268. 8. Given 5x3 + 4x=25, required the values of r.

Ans. x=1.871, or 2.671.

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9. Let - 1=5+11, to find the values of x.

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=

x+1 13 14. Given

+

to find the values of r. x+1

Ans. x=2, or – 3. 15. Given 2r*—7x2=9, to find the values of x.

Ans. x=+3, or tv

16. Let 2.r*-=496, required the values of x.

Ans. x=14, or EVO

.

17. Given x-x=a to find the values of x.

Ans. x= 3

1+ 40 +1

2

18. Let 22% +15=3x required the values of x.

3+N-11 ans. x

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: 19. Given 3x -- =10, to find the values of r. 4

Ans. x=6+V4. 20. Let Hare-201=c, required the values of x.

b+1 1,2 + 4ac Ans. x1

4a 21. Given

+
to find the values of x.

Ans. x=l+VI-a. $75) PROBLEMS, or questions, involving quadratic equations, are solved by the same process as that for the solution of those which contain simple equations. But as every quadratic

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equation contains two values of the unknown quantity ; ques. tions of this class may admit of a double solution. And as the particular nature of the question must determine whether one or both of these values apply to it, no general rule can be laid down ; yet may the processes of solution of Problems involving quadratic equations be sufficiently understood by observing the conclusions deduced from the following examples.

Exam. I. To find that number to which if you add 12, and multiply the sum by the number required, the product shall be 589.

: Solu. Let I = the number sought ;

Then (r+12)x=589, or x2 + 127 = 589 by the question. Complete the square and we have x4 + 127+36 = 625; hence, Extract the root, we get 5+6 = V025 = 25

Therefore, r = 25—6= 19, the number sought.

- x2

+ v144

Exam. II. To divide the number 56 into two such parts, that their product shall be 640.

Solu. Let I = one part ; then 56 x = the other part ; And I x (56 — x) = the product of the two parts.

Hence, 2 x (56 - x) = 640, or 56x = 640 by the question, or ro

56r =

- 640; hence By completing the square, r? 56r + 784 = 784 640 = 144 ; Therefore, x 28

+ 12 ; 3nd, consequently, x= 28 + 12 = 40 or 16.

Note. The two values of the unknown quantity are, in this instance, th two parts into which the given number was required to be divided.

Exam. III. Let the difference of two numbers be 7, and half their product plus 30 be equal to the square of the less number; it is required to find those two numbers: Solu. Let I = the less, then x + 7 = the greater number, I

+ 30 = half their product plus 30.
2
7

x2 + 7x
+ 30 = x2 (square of the less) or

+ 30 1 Il by the question.

Multiply now by 2, and r? + 77 + 60 2x2. Or, by transposition, r? - 78= 60; hence

49

49 289 By completing the square ro - 7r + 60 +

And 1(1 +7)

Hence, 1(r + )

2

2

ΕΣ

7 Consequently, r

1289 17

=+ Therefore I =

+ 17 + 7

= 12, or 5 = less part ;

2 And x + 7 = 12 + 7 or — 5 + 7 = 19 or 2 = the greater part.

Note. This problem, we see, admits of a double solution; for whether 12 and 19 or -5 and 2 be taken as the corresponding less and greater parts, the conditions of the questiou are answered by each pair of pumbers. Illus. Thus the difference of 12 and 19.......

is 7. That of 5 and 2..

is 7.

(12 x 19) Half product of 12 and 19 + 30 =

+ 30=:44 =12'

2 = square of the less,

(- 5 X 2) That of 5 and 2 + 30 =

+ 30 = 25 x

2 - 512

= square of the less. Exam. IV. To divide the number 30 into two such parts that their product may be equal to eight times their difference. Solu. Let r = the less ; then 30 r= the greater part, And 30 - I - I, or 30

2x = their difference. Hence, x x (30 — 3) =8 (30 2x), by the question,

or, which is the same, 30r - 20 *240 16r. Or, by transposition, 22 , 46.r 240.

Complete the square, and x? 46r + 529 = 529-240=289,
Therefore, I – 23 = + ✓ 289 = + 17,
Whence, r = 23+ 17 = 40 or 6 =less part.

And 30-I = 30–40 or 30 – 6=-10 or 24 = - greater part. Here the equation gives for the less part 40 and 6; but as 40 cannot possibly be a part of 30, we take 6 for the less part which allows 24 for the greater; and the two munibers 24 and 6 answer the conditions of the question.

Exam. V. A merchant lought cloth for 33l. 158. which he sold again at 21. 8s. per piece, and gained by the bargain as much as one piece cost him : what did he gain by the bargain? Solu. Let I= the number of pieces.

675
Then = the number of shillings which each piece cost;
and 48% = the number of shillings which he sold the whole for.
Therefore 482 675 what he gained by the bargain.

675
But 481 675 -, by the question,

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225

and 2

225
16

by transposition and division.

16

+

32+

225 50625 225 50025 65025 Then comp. the sqr. and i' -2 +

i 16 1024 16 1024 102+' 225

05025

+ 255 Therefore I

65025

; ✓ 1024 32

1024 +255 + 225

30 and r =

15 or 32

32 Bat it is obvious the number required is 15, because the conditions of the question are such that there cannot be a negative or fractional number of pieces.

Exam. V. A and B set off at the same time to a place at the distance of 150 miles from that which they left; À travels 3 miles an hour faster than B, and arrives at his journey's end 8 hours 20 minutes before him : it is required to find what rate each person travelled per hour. Solu. Let I the rate per hour at which В travels. Then I + 3 = the rate per hour at which A travels.

150
And = the number of hours for which В travels.

150
Also = the number of hours for which A travels.

1+3 But A arrives 8 hours 20 minutes (8$ hours) sooner at his journey's end than B; 150 130 150

25

150 Hence + 8 =

+ x +3

2 + 3

3 But, by reduction, we get 3% + 38 = 54.

9

9 225 · And completing the square x® + 3.x + = 54 +

4
3
1225

+ 15
Therefore, it

;
3
And I

= 6 or 9 miles an hour for B.

2 Whence r + 3 = 9 or — 6 miles an hour for A. Since this is a question of motion, it is evident that if motion to a place be reckoned po«i'ive, motion from the same place must be reckoned negative : But A and B are moving to a given place, therefore the positive numbers 6 and 9 are taken for the respective rates per hour of and B.

Exam. VI. It is required to divide the number 14 into two such parts that their product shall be 50. Solu. Let x = one part; then 14

-= the other part; Hence, by the question, I X (14 - 3) = 50, or 141 - = 50;

Therefore, x? 147 = 50, And x2 - 14r + 49=-50+ 49= -1, or r-7=+T;

Therefore, r=7+ - 1. But both the values of x are impossible, because the question itself is impracticable.

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+ 15

+ 1.5

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