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Whence 2=30=(900— 864) =30536=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-r=the less, And a +(a -)=2x2 - 2axta =b, by the question, Or 2.02 - 2ax=b-a2, by transposition,

b-a And x2 - ar=

by division. 2

b-a2 Whence x=

avto)

6+2) V 26–43

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

10 1
= a +av 116–100=7, the greater cumber,

1

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yalue of x.

16. Given vito-x*–2(1+0=xa)=, to find the

Ans. ==*+/41 z.

of x.

17. Given

-X, to find the value

Ans. x=+5 18. Given xon – 2:03n itx"=6, to find the value of x.

Ang. x=VI+tv 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 ste==the greater. And x(x+8)=x2 +8x=240, by the question, Whence x=-47 716+240=-4+256, by the com

mon rule, before given, Therefore x= =16c-4=12, the less number,

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

Let x=the greater part,

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

x2 - 60x= -864,

Whence x=30+7 (900- 864) =30/36=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-x=the less, And x2 +(a - x)"=2x2 - 2axtar=b, by the question, Or 2.2 - 2ax=b-a2, by transposition,

b-a Andra

by division. 2

ax

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a

b-a?

1 Whence x=

+

✓ 26-a2 4 2

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

10 1

+ 116-100=7, the greater number, 2 2

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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 :: X : 24—%, by the question.
Or x2=100(24-=2400- 100x,

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

by the rule,
And consequently 201.=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 11. less for each ; how many did he buy? Let x represent the number of sheep,

80
Then will be the price of each,

80
And =price of each, if +4 cost 801.
x+4

80 80
But

+1, by the question,
* x+4

80x
Or 80= +x, by multiplication,

x+4
And 80x+320=80x+x3+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 x+y=xy

lity=_y} by the question.

2-ya
Hence 1= =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

, 2

+v(+1)=

y=+*+5=1.6180... And x=y+1= + 75=2.6180 ...

Therefore ya

2 Where .. i 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, x(x+3y) ***2 +3.y=45 Hence + y(x+2y)=x2 +3xy+2y2 =77} by the ques

x

Or ya=

And 2y2=77-45=32, by subtraction,
32

=16 by division, and y=16=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.

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 x2=y?, by the nature of proportion,
And

5x+y+z=14

x2 + y2 + Hence xtz=14-y, by the second equation, And x2 +223 +22=196 — 28y+yo, by squaring both

sides,
Or x2 +22+2y=196 - 28y+ya by putting 2ya 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

16
Again z=y=16, or x= by the 1st equation,

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