As a particular example, let n = 2, a = 4, b = 3, and the given equation becomes x4-8 x2=9 and from this may be found in the usual way, by assuming %, the values of ; but these values are more readily found from the preceding general formula (1), which be comes *=√{4±√(16+9) } = √(4 ± 5) = √9 or√1 = ± 3 or±√ -1 1. Given x4. EXERCISES. 6x2+10=2, to find a 2. Given 26. 4x3 284, to find x - 3. Given an. ·2x2n = a, to find x 4. Given 52-2x=√(5 x2-2x) +12 3x 6. Given 2 (2—3x+11) = x2-3x+8 QUESTIONS PRODUCING QUADRATIC EQUATIONS CONTAINING ONLY ONE UNKNOWN QUANTITY. EXAMPLES. 1. Find a number such that, if 8 times the number be added to its square, the sum shall be 65. Let then the number, x2+8x=65 and the values of x are found by the usual process of solution to be Either of these values satisfies the equation. If, however, the negative solution - 13 be considered as the positive number 13 to be subtracted, then the enunciation of the question would require to be so far modified as was done in simple equations, by changing the condition of addition into subtraction (252); the cause of which change is evident, by substituting for x in the equation. But the enunciation without any change will be adapted to both values, by observing that the algebraical addition of a negative quantity is the same as the arithmetical subtraction of a positive one (56.) 2. Find a number such that, if 4 times its square be diminished by 6 times the number itself, the remainder is 70. the number, Let 3. Divide the number 24 into two parts, such that their product shall be 95. Let x = one of the parts, then 24 x= the other, 4. Find a number such that, if 15 be added to its square, the sum shall be 8 times the number. Let then the number, x2+15=8x from which is found a 5 or 3 (286.) It appears from this example that a question producing a quadratic equation sometimes admits of two positive solutions, which belong literally to the enunciation without any modification. Positive solutions are also called direct, and negative indirect, solutions. 5. To divide a number a into two parts, so that their product shall be equal to b2. Ꮳ The expression for the value of x shows the limitation of the conditions; for its value is possible, provided 462 does α not exceed a2. When 4 62 a2, or b= the value of x is a and that of a 2' a a 2' = so that in this case the 2 2' two parts are equal, each being a half of the given number. This is also the greatest value that can have; and hence the product is greatest when the number is divided into two equal parts. This result is analogous to the 27th proposition of the 6th book of Euclid. 6. There are three numbers in continued proportion; the sum of the first and second is 10, and the third exceeds the second by 24: required the numbers. also (321) x:10 x=10-x:34-x x2 - 20x+100 = 34 x and from this equation is found When x= = 2, 10—x = 8, 34-x=32, and the numbers are 2, 8, and 32. When x 25, 10 numbers are 25, X= - 15, and 9. ·15, 34- -9, and the The meaning of the negative solution 15 will be understood by considering that the addition of a negative quantity is the same as the subtraction of the same taken positively, and the first condition of the question then becomes (56) 25 + (− 15) = 25 − (+ 15) — 25 — 15 = 10; and the second is 9-(- 15) = 9 + (+ 15) = 9 +15=24. This circumstance indicates that - 15 may be changed to +15, provided that, instead of the condition of the sum of the first and second numbers being 10, their difference be 10; and the second condition may, for a similar reason, be changed into this, that the sum of the second and third is 24. The numbers, therefore, in this last case, after this modification in the enunciation of the question, are 25, 15, and 9, and are its direct solution; which is just the same as the solution 25, -15, and 9 of the given question, with the conventional meaning assigned to a negative number. If lines be taken to represent these numbers, then reckoning a certain point in an assumed straight line to be the beginning or origin, an inch to be the unit of measure, and the portion of the line towards the right of the origin to be positive, and towards the left negative, the numbers 2, 8, 32, are to be reckoned towards the right; and of the numbers 25, -15, 9, the first and last are to be reckoned towards the right, and the second 15 towards the left. Then considering the principle in art. (57), the sum of 25 and — 15 is 10, and (61) the excess of 9 above 15 is 24; so that these numbers satisfy the question. Though it follows from the conventional rule that the sum of the lines 25 and 15 is equal to 10, this merely means that the difference between 25 and +15 is 10, and the only use of this arbitrary rule is merely to generalise the solution, so as to comprehend in one formula the opposite conditions of sum and difference. If the question referred to weights instead of numbers or lines, the weights could also be represented by lines (44), and the preceding remarks apply therefore to the weights. For example, the negative weight 15 (an expression which, understood literally, is absurd) merely means a weight of 15 to be subtracted, wherever addition is indicated in the conditions of the question, and conversely. 7. A person bought a number of oxen for £80; and if he had bought 4 more for the same money, the price of each would have been £1 less: how many did he purchase? the number of oxen, Let the price paid for each, the price that would have been paid for each, and from this equation is found by the rule x=18-2 = 16 or 20 The number of oxen bought is therefore 16. The negative value 20 to this question suggests as usual the necessity of modifying the conditions of the question, in order that the value 20 may be literally or directly applicable. The necessary modification to be made will be perceived by substituting the negative value in the preceding equation, when it gives which shows that the condition of buying 4 more must be changed into that of buying 4 less; and hence, also, the price, instead of being £1 less, will now be £1 more. The question thus modified has for its direct solution the value 20, and its enunciation is, 8. A person bought a number of oxen for £80, and if he had bought 4 less for the same money, the price would have been £1 more for each: how many did he purchase? |