Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Whence z=30=7 (900— 864) =305/36=30+6, by

the rule, And consequently x=30+6=36, or 30—6=24, the two

parts sought. 3. It is required to find two numbers such that their sum shall be 10(a), and the sum of their squares 586).

Let x=the greater of the two numbers,

Then will a-r=the less, And x2 +(a-x)=2x2 - 2ax+a=b, by the question, Or 2x2 - 2ax=b-a", by transposition,

b-a?
And x2

by division.
2
b-a?

1
Whence x=
*(

26-a?
4 2

by the rule,
And if 10 be put for a, and 58 for b, we shall have

10. 1
atov 116-100=7, the greater number,

+
2

101 And 10—3=

ē - 116–100=3, the less,
2

ax=

[ocr errors]

a

[ocr errors]

a

[ocr errors]
[graphic]

1

yalue of x.

16. Given vite–23–2(1+x=x2)=, to find the

. == v(x-7)+r(1-2=r.

1 1 Ans. x=-tav41

2

of x.

17. Given v

to find the value

Ans. x=1 +\v 5 18. Given xen — 2.03ni+x"=6, to find the value of x.

Ang. x=+iv 13

QUESTIONS PRODUCING QUADRATIC

EQUATIONS.

The methods of expressing the conditions of questions of this kind, and the consequent reduction of them, till they are brought to a quadratic equation, involving only one unknown quantity and its square, are the same as those already given for simple equations.

1. To find two numbers such that their difference shall be 8, and their product 240.

Let x equal the least number.

Then will x+= the greater, And x(x+8)=ra +8x=240, by the question, Whence r=-4+ 16+240=-4+256, by the com

mon rule, before given, Therefore x=16.4=12, the less number,

And x t-8=12+8=20, the greater. 2. It is required to divide the number 60 into two such parts, that their product sball be 864.

Let x=the greater part,

Then will 60 - x=the less, And x(60- x)=50x - x2=864, by the question, Or by changing the signs on both sides of the equation

x? - 60x= -864,

Whence x=30+7 (900-864)=305/36=30+6, by

the rule, And consequently x=30+6=36, or 30-6=24, the two

parts sought 3. It is required to find two numbers, such that their sum shall be 10(a), and the sum of their squares 58(b). Let x=the greater of the two numbers,

Then will a-(=the less, And x2 +(a – »)=2x2 - 2ax+ao=b, by the question, Or 2.2 - 2ax=b-a?, by transposition,

bwa? And x2

, by division. 2

[ocr errors]

ar =

[ocr errors]

b

1 Whence x=

a2 4

2

by the rule, And if 10 be put for a, and 58 for b, we shall have 10

1
+ v116-100=7, the greater runher,
2

[merged small][ocr errors][merged small]

4. Having sold a piece of cloth for 241., I gained as much per cent. as it cost me; what was the price of the cloth ?

Let x= pounds the cloth cost,

Then will 24-2=the whole gain,
But 100 : « :: 2 : 24-X, by the question.
Or x2=100(24-X)=2400 - 100x,

That is, xa +100x=2400,
Whence x=-50+2500+2400=-50+70=20

by the rule,
And consequently 202. =price of the cloth.

5. A person bought a number of sheep for 801., and if he had bought 4 more for the same money, he would have paid 1l. less for each ; how many did he buy? Let x represent the number of sheep,

80
Then will be the price of each,

[ocr errors]
[merged small][merged small][ocr errors][merged small]

80 80 But

+1, by the question,
2 x+4

80x
Or 80= +x, by multiplication,

x+4
And 80x+320=80x+x+4x, by the same,
Or, by leaving out 80x on each side, x2 + 4x=320,
Whence x=-2+4+320=-2+18, by the rule,

And consequently x=16, the number of sheep. 6. It is required to find two numbers, such that their sum, product, and difference of their squares, shall be all equal to each other.

Let x=the greater number, and y= the less.

Then { 2+y="_y:}by the question.

[ocr errors]

х2 – уг
Hence 13 =x-y, or x=y+1, by 2d equation.

x+y
And (y+1)+y=y(y+1) by 1st equation,
That is, 2y+1=ya +y; or y? - y=1,

1

1

Whence y=+v6+1)= + v5, by the rule,

Therefore y=+v5=1.6180 ...
And x=y+1=+*5=2.6180 :-

[ocr errors]

3 1

2 Where ... denotes that the decimal does not end.

7. It is required to find four numbers in arithmetical progression, such that the product of the two extremes shall be 45, and the product of the means 77.

Let x= least extreme, and y=common difference, Then x, x+y, x+2y, and x+3y, will be the four num

bers, Hence :

S x(x+3y)=+3xy=45 tion,

x+2y)=
And 2ya=77-45=32, by subtraction,

32

= 16 by division, and y=v16=4,

2 Therefore x3 + 3xy=x2 +12x=45, by the 1st equation, And consequently x=-6+ ✓ (36 + 45) = -6+9=3,

by the rule,
Whence the numbers are 3, 7, 11, and 15.

x+y(2429)=x2+3y+-242=77} by the ques

Or ya =

[ocr errors]

Rate TE64} by the question

,

8. It is required to find three numbers in geometrical progression, such that their sum shall be 14, and the sum of their squares 84.

Let x, y, and z be the three numbers,
Then xz=y?, by the nature of proportion,

5x+y+z=14
And

+ y2 +=84 Hence xtz=14-y, by the second equation, And 22 +223+2:=196 - 28y+yo, by squaring both

sides,
Or x2 +22+2ye=196 – 28y+ya by putting 2y2 for

its equal 2x2,
That is x2 + y2 +22=196 - 28y by subtraction,

Or 196 - 28y=84 by equality,

196--84 Hence y

=4, by transposition and division, 28

Again &z=y2=16, or r=- by the 1st equation,

[ocr errors]
[ocr errors]
[ocr errors]
« ΠροηγούμενηΣυνέχεια »