EXERCISES On Affected Cubic and Biquadratic Equations. 1. Find the values of x in the equation х x3-6x2+11x=6, (268). 6. Find an approximate value of x in the equation x2+10x2+5x=260, (269). 7. Find an approximate value of x in the equation x3-15x2+63x=50. 8. Find an approximate value of x in the equation x3-17x2+54x=350. 9. Find an approximate value of x in the equation x4-3x2-75x=10000. 10. Find an approximate value of x in the equation 2x4-16x3+40x2-30x+1=0. х 11. Find an approximate value of x in the equation (}x2-15)2+x√x=90. Ans. x 4.117'. Ans. x 1.028.' Ans. x 14.95'. Ans. x 10.23'. Ans. x 1.284'. Ans. x 10.52'. ELIMINATION BY THE METHOD OF COMMON DIVISOR. (269.) The following is a general method of Elimination, and, for Equations of the higher degrees, it will sometimes be found preferable to any other. Transpose all the terms of the two Equations to one side; then divide one into the other, and the remainder into the divisor, and so on, as in finding the Greatest Common Measure, (66,) until one of the two unknown quantities is eliminated from the remainder; and put this remainder =0. y2-24 x(y2—24)+10y) x2 (y2 −24) +x ( y3 −24y)—10y2+240(x+y x2(y2-24)+10xy xy+2y2-24x(y3-24y)+10y2 (y2-24 x(y3-24y)+2y4-48y2-24y2+576 In the remainder -10xy-20y2+240, we cancel the factor 10, and change the signs, for the next divisor. In dividing into this divisor, we take the binomial y2-24 for the quotient, and multiply the divisor by this binomial. The first remainder is equal to 0, because the divisor and dividend are each equal to 0; and it follows hence that each subsequent remainder is equal to 0. The operation will be much more simple if we divide the first equation by the second: the result will be the same. (270.) A Determinate Problem is one in which the given condi tions determine the values of the unknown or required quantities. A Determinate Problem is represented by as many independent equations as there are different conditions to be expressed, or unknown quantities to be determined, (120.) All the Problems which have hitherto been proposed in this work, are determinate; and no example of this kind need be here given. 2. Indeterminate Problems. (271) An Indeterminate Problem is one in which the given conditions do not determine the values of the required quantities,-admitting either of an unlimited number of values to those quantities, or else of a variety of values, within certain limits. An Indeterminate Problem is represented either by a less number of independent Equations than there are unknown quantities to be determined, or by an identical equation. We give an example of each of these forms of indeterminateness. EXAMPLE I. To find three numbers such that the first shall be 5 less than the second, and the sum of the second and third shall be 12. This Problem contains but two conditions; and if we represent the three required numbers by x, y, and z, we shall have only the two Equations y-x=5; y+z=12. By subtracting the first equation from the second, we have x+2=7. This equation will admit of an unlimited number of values of x and z; for we may assume any value whatever for one of the letters, as x, and determine thence the corresponding value of z. Thus if x=1, z=63; if x=1, z=6}; if x=1, z=61, &c.; and from the values of x or z, we might obtain the corresponding values of y from one of the given equations. If, however, the required numbers were limited to integral values. the third equation would be satisfied only by x=1, 2, 3, 4, 5, or 6, and z=6, 5, 4, 3, 2, or 1. In the first equation y=5+x, which would give EXAMPLE II. To find a number such, that of it, diminished by of it, and by 5. shall be equal to of the excess of 5 times the number above 60 The equation of this problem will be This last is an identical Equation, which will be satisfied by attributing to x any numerical value whatever. The problem is therefore entirely indeterminate. We may obtain an expression for the value of x from the last equation. Thus, by transposition, 52–5=60–60. By adding similar terms, and retaining x as a symbol in the first member, we have 0x=0; which gives x=8. Hence is a symbol of an indeterminate quantity. The same thing will appear from considering that the quotient of 0-0 is any quantity whatever; inasmuch as the divisor 0x any quantity will produce the dividend 0, (43). 3. Impossible Problems. (272.) An Impossible Problem is one in which there is some condition, expressed or implied, which cannot be fulfilled. An Impossible Problem is represented by a greater number of independent equations than there are unknown quantities to be determined; or by an equation in which the value of the unknown quantity is negative-zero-infinite-or imaginary. We subjoin an example of each of these forms of impossibility. EXAMPLE I. To find two numbers whose sum shall be 10, difference 2, and product 20. Representing the two numbers by x and y, we shall have x+y=10; x-y=2; xy=20. From the first and second equations the values of x and y will be found to be x=6, and y=4. The third equation cannot, therefore, be fulfilled; that is, the problem is impossible. If the third equation were xy=24, the problem would be possible, but this would not be an independent equation, since it may be derived from the other two. find Thus, squaring the first and second equations, and subtracting, we 4xy=96, or xy=24. EXAMPLE II. To find a number which, added to 17 and to 53, will make the first sum equal to of the second. If x represent the number, the equation will be 17+x= 53+x From this equation we shall find x=-5. This number, added to 17 and 53, gives 12 and 48, and 12=1 of 48. The problem is impossible in an arithmetical sense, according to which addition always implies augmentation; and it is in this sense only that the problem would be considered. To make it arithmetically consistent, it should be stated thus : To find a number which, subtracted from 17 and from 53, will make the first remainder equal to of the second. |