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Equating the two values found for the trinomial (z+y+x), we

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· z―y=y—x; hence

2y=x+2.

Substituting 2y for x+z in Equation (A), and reducing,

y2-yx=4;

y2-4

from which we shall find x=

y

Substituting this value of x in the first Equation,

1

y4 — 8y2+16

+2y2-4-37.

y2

Reducing this Equation, we find

3y4-49y2——16. :

From this Equation, we shall find y=4.

y2-4

Since x=

y

we find =x3; and z is now easily found equal to 5.

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PROBLEMS

In Quadratic Equations of one or more Unknown
Quantities.

1. The area of a rectangular lot of ground is 384 square rods, and its length is to its breadth as 3 is to 2. Required the length and breadth of the lot.

The area, in square measure, of a rectangle, is expressed by the product of the number of linear units in its length × the number of linear units in its breadth.

The length and breadth must be taken in the same denomination in multiplying the area will be found in the corresponding denomination of square measure.

Let a represent the length, and y the breadth of that lot.
By the conditions of the problem, we shall then have

xy=384; xy::3:2.

Converting this proportion into an Equation (149), we find

2x=3y;

from which and the first Equation, x and y will be found equal to 24 and 16 rods, respectively.

2. The length of a rectangular garden exceeds its breadth by 6 rods, and its area is 216 square rods. What are the length and breadth of the garden?

Let x represent the length, and y the breadth; then

x=y+6;

xy=216;

which will give x=18, and y=12 rods.

3. Find two numbers whose sum shall be 24, and whose product shall be equal to 35 times their difference.

Let x represent the greater, and y the less number; then

x+y=24; and
xy=35(x-y).

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4. Divide a line 20 inches in length into two such parts that the rectangle or product of the whole line and one of the parts shall be equal to the square of the other part.

Let x represent the shorter, and y the longer part.

The Equations will then be

x+y=20;

20x=y2.

These Equations will give x=30-10/5, and y=10√/5-10.

5. Find the dimensions of a rectangular field, so that its length shall be equal to twice its breadth, and its area 800 square rods.

Let x represent its length, and y its breadth; then

x=2y; and

xy=800;

from which we shall find x=40, and y=20 rods.

6. The sum of the two digits of a certain number is 10, and if their product be increased by 40, the digits will be reversed. What is the number?

Let x represent the tens' figure, and y the units' figure of the nur ber; then

10y+x represents the number with its digits reversed. We shall, therefore, have the following Equations:

x+y=10;

xy+40=10y+x.

From the first Equation,

x=10—y.

Substituting this value of x in the second Equation, and reducing, y2-y=30; hence

y=6; and x is easily found 4.

=

Therefore, the number is 46.

7. The sum of two fractions is 13, and the sum of their reciprocals is 3. What are the two fractions?

Let x and y represent the fractions; then

x+y=1}; and

1 1

+ =31. x y

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(3)

1

6

210

3

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8. The perimeter, or sum of the four sides of a rectangle, is 112 rods, and its area is 720 square rods. What are the length and breadth of the rectangle?

Let x represent its length, and y its breadth; then

2x+2y=112;

xy=720.

These two Equations will give x=36, and y=20 rods.

9. Divide the number 60 into two such parts that their product shall be to the sum of their squares as 2 to 5.

Let x represent the less, and y the greater part; then

x+y=60; and xy: x2+y2::2:5.

Converting this proportion into an Equation,

5xy=2x2+2y2; or

2x2+2y2-5xy=0.

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Squaring the first Equation, and multiplying the result by 2, we have (4) ·2x2+2y2+4xy=7200.

Subtracting the third Equation from the fourth,

800

9xy=7200; hence x=

y

Substituting this value of x in the first Equation, we shall find y= 40; then x=60-40=20.

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10. A merchant bought a piece of cloth for $120, and after cutting off 4 yards, sold the remainder for what the whole cost him-by which he made $1 a yard on what he sold. How many yards did the piece contain?

Let x represent the number of yards; then

120

X

120

represents what the merchant paid per yard;

represents what he received per yard.

X

-4

By the conditions of the problem, we then have

120

X- 4

120

=— +1; hence

X

=24 yards.

11. Divide the number 100 into two such parts that the difference of their square roots shall be 2.

Let x represent the greater, and y the less part; then

x+y=100; and
√x-√y=2.

Transposing Vy to the second member, and squaring both sides,

x=4+4√y+y.

Substituting this value of x in the first Equation,

4+4y+2y=100.

This Equation may be reduced to

2√y+y=48.

Transposing y, and squaring both sides,

4y=2304-96y+y2.

This Equation will give y=36; hence 100-36=64, is the value

of x.

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