examples will show their application to a variety of such equa

Adding the square of p to both members, we have,

By Rule II, 4x2” +4px” + p2=4q +p2.

b √ x +

7. Given Vx+a+b√x+a⇒ 262, to find x.

(77.) Divide 34 into two such parts, that their product shall be 225.

Let x one of the parts; then will the other be 34 and taking their product, we have,

or,

x(34-x)=225, -
34x-x225;

or by changing the signs,

-

Ꮖ ;

therefore, x=17825, or 9.

In this case, the two values of x give the two parts into which the given number is to be divided, for, 25 x9= 225. 2. There are two numbers whose difference is 7, and half their product plus 30, is equal to the square of the lesser number. What are the numbers?

Let x be the less; then x+7 will be the greater; and by the conditions of the question.

x(x+7)

+30= x2.

2

Completing the square by Rule II.

4x2-28x+49=240 +49=289;

extracting the root, 2x-7=17;

12 is therefore the less, and 12+7=19, the greater; or, if we take the other value of x, which is -5, we shall have for the greater, −5+7=2. Either of these values will satisfy the conditions of the equation, for

19 × 12
2-

+30= 144, and

-5x2
2

+30=25.

3. Divide the number 30 into two such parts, that their product shall be equal to eight times their difference.

Let the lesser part; then 30-≈ will be the greater, and their difference will be

(30-x)-x=30—2x.

Then, by the conditions of the problem,

or,

x(30-x)=8(30 — 2x);

30x x2 240 16x.

Transposing and changing the signs,

x2 46x —— 240.

Completing the square by Rule I.

x2 46x+529- · 240 +529=

extracting the root,

x-23=17;

whence, x = 23 ± 17 = 6, or 40.

289;

Of these values of x, it is evident, that only one will satisfy the conditions of the question, for 40 cannot be the lesser part, nor any part of 30; 6 is therefore the value of x, and we have for the greater part, 30-6, or 24, which answers the conditions of the question, for

6 × 24=144

=

(30-6)-6)8, or 18 x8-144.

4. A person bought a number of sheep for $120. If there had been 8 more, each sheep would have cost him half a dollar less. What was the cost of a sheep?

120

the number of sheep, then is the price of one

Let x= sheep; but if he had bought 8 more for the same money, the

price of each sheep would have been

120

x+8

and by the con

ditions, the difference between these two prices is half a dollar;

x2+8x+161920+16=1936.

Extracting the root, x+4= ±44;

The value found for a, by making 44 positive, answers the required conditions; and the other value, — 48, would be the answer to another problem, which is thus enunciated :A person bought a number of sheep for $120. If there had been 8 less, each sheep would have cost him half a dollar more.

NOTE. It may sometimes occur, that after completing the square, the second member of the equation is negative, in which case, the root cannot be extracted (63). Such a result is generally an indication of some impossibility in the conditions from which the equation is derived, or an error in forming the equation. If we have the equation 2-2x —— 13, by completing the square, and reducing, we shall have

x= = 1 ± √ — 12,

in which the value of a contains an imaginary expression. For another example, let it be required to divide 20 into two such parts, that their product shall be 125.

Let a one part, then 20- x will be the other, and

x(20x)=123,