This equation is similar to the last, except in this, the second term of the first member has the sign

Here we must observe that the second power of the binomial 2 a xa2, the same as that of x+a with the

is x2 a,

exception of the sign of the second term.

In this equation, as before, we find two terms of the second power of a binomial; if we can find the other term we can easily solve the question.

Adding 4 to both members of the equation it becomes

Since

4x+4= 572 +4 576.

-2 in this corresponds to a, the root of the first menber is r-2. In fact, (x-2)2= x2-4x + 4. The root of 576 is 24.

22.

The former only an

The two values of x are 26 and·

swers the conditions of the question.

Proof. If the whole number, 26, had paid their shares, each would have paid 1435 shillings. But 22 only paid, consequently each paid 143 = 63 shillings.

3. There are two numbers, whose difference is 9, and whose sum multiplied by the greater produces 266. What are those numbers?

If we use the general formula as before, we have

Both values will answer the conditions of the question; for

In all the above examples, after the question was put into equation, the first thing done, was to reduce all the terms containing to one term, and those containing into another, and to place them in one member of the equation, and to collect all the terms consisting entirely of known quantities into the other. This must always be done. Moreover must have the sign + and its coefficient must be 1. The equation will then be in the following form.

p and q being any known quantities and either positive or negative.

Every equation, however complicated, consisting of terms which contain x2, and x, and known quantities may be reduced to this form.

We consider a2 and px as two terms of the second power of the binomial x + a in which

Hence the binomial x + a is equal to x+, and the third

Therefore the first member of the above equation may be

P is the

rendered a complete second power, of which x + 2

root, by adding to it. The same quantity must be added to

4

the second member, to preserve the equality.

The equation then becomes

x2 + px + P2=q+P2.,

9+

4

From the above observations we derive the following general rule for the solution of equations which contain the first and second powers of the unknown quantity.

ist. Prepare the equation, by collecting all the terms containing the first and second powers of the unknown quantity into the first member, and all the terms consisting entirely of known quantities into the other member. Unite all the terms containing the second power into one term, and all containing the first power into another. If the sign before the term containing the second power of the unknown quantity be not positive, make it so b". changing all the signs of both members. If the coefficient this term is not 1, make it so by dividing all the terms by its coefficient.

2d. Make the first member a complete second power. This is done by adding to both members the second power of half the coeffi cient of x (or of the first power of the unknown quantity.)

3d. Take the root of both members.

The root of the first me.nbe will be a binomial, the first term of which will be the unknown quantity, and the second will be half the coefficient of x as found above. The root of the second member must have the double sign ±.

4th. Transpose the term consisting of known quantities from the first to the second member, and the value of x will be found.

4. A and B sold 130 ells of silk (of which 40 ells were A's and 90 B's) for 42 crowns. of an ell more than B did. crown?

Let

Now A sold for a crown one third
How many ells did cach sell for a

the number of ells B sold for a crown; then x + the number A sold for a crown ;