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 -a, is x? — 2 a x + a, the same as that of x + a with the 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. It may be found as follows, No and 2 ax = -42 2 a=-4 which gives al = 4 . Adding 4 to both members of the equation it becomes ght - 4x + 4 = 572 + 4 = 576. Since — 2 in this corresponds to a, the root of the first mentber is x — 2. In fact, (x - 2) = x - 4x +4. The root of 576 is 24. Hence IC -2= + 24 x = 2 + 24. The two values of x are 26 and — 22. The former only an swers the conditions of the question. Proof. If the whole number, 26, had paid their shares, each would have paid 42 = 54 shillings. But 22 only paid, consequently each paid 43 = 61 shillings. 3. There are two numbers, whose difference is 9, and whose sum multiplied by the greater produces 266. What are those numbers ? Let x = the greater ; then x — 9 = the less, 2x –9 = their sum.' By the conditions X (2 x — 9) = 266 If we use the general formula as before, we have x = x Both values will answer the conditions of the question ; for 14 + 5 = 19 and 19 x 14 = 266 also - 94+ (-181) = -28 and - 28 X - 94 = 266. In all the above examples, after the question was put into equation, the first thing done, was to reduce all the terms containing me to one term, and those containing it 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 xmust have the sign + and its coefficient must be 1. The equation will then be in the following form. goo + px=a. p and q being any known quantities and either positive or negative. Every equation, however complicated, consisting of terms which contain x», and x, and known quantities may be reduced to this form.. Let the equation be 3 x _ 15? 5 4x = ? Clearing of fractions it becomes 140 0 — 12.212 - 70 +6x=75_ 5r.. Transposing and uniting terms 146.- 7 x = 145 Changing all the signs in both members 7 x— 146 x =— 145 Dividing by 7 (the coefficient of x) To solve the equation i tpx=9. We consider x and p x as two terms of the second power of the binomial x + a in which 2 a x = px term ce Hence the binomial x + a is equal to x + ?, and the third term of the second power is mee. In fact (+) (0+2) = x+pI+ Therefore the first member of the above equation may be rendered a complete second power, of which « + is the root, by adding to it . The same quantity must be added to the second member, to preserve the equality. The equation then becomes pro + px + = + Taking the root of both members *+=+(2+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 membei. Unite ali the terms containing the second power into one term, and all containing the first power into another. If the sign brjore the term containing the second power of the unknown quantity be not positive, make it so be. 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 coeffcient of x (or of the first power of the unknown quantity.) 3d. Take the root of both members. The root of the first me.nba- 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. Now A sold for a crown one third of an ell more than B did. How many ells did cach sell for a crown? Let x = the number of ells B sold for a crown; then x + = the number A sold for a crown; |