255. Equations are sometimes proposed which are intended to be solved, partly by inspection, and partly by ordinary methods; we will give two examples. Bring the fractions on each side of the equation to a common denominator; thus (x+4)-(x-4) (9 + x)2 - (9− x)2 = x2-16 81-22 Here it is obvious that x=( 0 is a root. To find the other roots we begin by dividing both sides of the equation by 4x; thus Thus there are three roots of the proposed equation, namely, 0, 6, -6. Hence it is obvious that x=a is a root. We may write the equation a3-a3=7a2(x-a); and to find the other roots we begin by dividing by x-a. x2+ax+a2=7a2. Thus By solving this quadratic we shall obtain x=2a or -3a. Thus there are three roots of the proposed equation, namely, a, 2a, —~ 3a. * EXAMPLES. XXVII. 14. 16. 17. 13. x=7(2-x). (x+9)=2x-3. 15. √(x+8)— √(x+3)= √x. 6 /(1-3)+52=7. (3x-3)+ √(5x-19)=√(2x+8). 18. (2x+1)- √(7x−27) = √(3x+4). XXVIII. Problems which lead to Quadratic Equations. 258. Find two numbers such that their sum is 15, and their product is 54. Let x denote one of the numbers, then 15-x will denote the other number; and by supposition If we take x=9 we have 15-x=6, and if we take x=6 we have 15-x=9. Thus the two numbers are 6 and 9. Here although the quadratic equation gives two values of x, yet there is really only one solution of the problem. 259. A person laid out a certain sum of money in goods, which he sold again for £24, and lost as much per cent. as he laid out: find how much he laid out. Let x denote the number of pounds which he laid out; then x-24 will denote the number of pounds which he lost. Now by supposition he lost at the rate of x per cent., that is the loss was the fraction X of the cost; therefore 100 From this quadratic equation we shall obtain x=40 or 60. Thus all we can infer is that the sum of money laid out was either £40 or £60; for each of these numbers satisfies all the conditions of the problem. 260. The sum of £7. 4s. was divided equally among a certain number of persons; if there had been two fewer persons, each would have received one shilling more find the number of persons. Let x denote the number of persons; then each person 144 received shillings. If there had been x-2 persons x 144x=144(x − 2) + x(x − 2) ; x2-2x=288. From this quadratic equation we shall obtain x=18 or 16. Thus the number of persons must be 18, for that is the only number which satisfies the conditions of the problem. The student will naturally ask whether any meaning can be given to the other result, namely -16, and in order to answer this question we shall take another problem closely connected with that which we have here solved. 261. The sum of £7. 4s. was divided equally among a certain number of persons; if there had been two more persons, each would have received one shilling less: find the number of persons. Let x denote the number of persons. Then proceeding as before we shall obtain the equation Thus in the former problem we obtained an applicable result, namely 18, and an inapplicable result, namely -16; and in the present problem we obtain an applicable result, namely 16, and an inapplicable result, namely 18. 262. In solving problems it is often found, as in Art. 260, that results are obtained which do not apply to the problem actually proposed. The reason appears to be, that the algebraical mode of expression is more general than ordinary language, and thus the equation which is a proper representation of the conditions of the problem will also apply to other conditions. Experience will convince the student that he will always be able to select the result which belongs to the problem he is solving. And it will be often possible, by suitable changes in the enunciation of the original problem, to form a new problem corresponding to any result which was inapplicable to the original problem; this is illustrated in Article 261, and we will now give another example. 263. Find the price of eggs per score, when ten more in half a crown's worth lowers the price threepence per score. Let x denote the number of pence in the price of a score of eggs, then each egg costs pence; and therefore the number of eggs which can be bought for half a crown 20 per score less, each egg would cost pence, and the number of eggs which could be bought for half a crown If the price were threepence From this quadratic equation we shall obtain x=15 or 12. Hence the price required is 15d. per score. It will be found that 12d. is the result of the following problem; find the price of eggs per score when ten fewer in half a crown's worth raises the price threepence per score. |