If, then, any problem furnishes an equation in which q is negative, and numerically greater than 2., we infer that the 4 conditions of the question are incompatible with each other. Suppose it is required to divide 6 into two parts such that their product shall be 10. Let a represent one of the parts, and 6—x the other part. Then, by the conditions, 2(6—2)=10; whence æ? - 6x=-10, and x=30V-1. The imaginary value of x indicates that it is impossible to find two numbers whose sum is 6 and product 10. From the preceding proposition, it appears that 9 is the greatest product which can be obtained by dividing 6 into two parts and multiplying them together. Discussion of Particular Problems. 282. In discussing particular problems which involve equations of the second degree, we meet with all the different cases which are presented by equations of the first degree, and some peculiarities besides. All the different cases enumerated in Chapter X. are presented by Prob. 19, page 195, when we make different suppositions upon the values of a, m, and n; but we need not dwell upon them. The peculiarities exhibited by equations of the second degree are double values of x, and imaginary values. 283. Double Values of the Unknown Quantity. We have seen that every equation of the second degree has two roots. Sometimes both of these values are applicable to the problem which furnishes the equation. Thus, in Prob. 20, page 195, we obtain either 100 or 180 miles for the distance between the places C and D. Let E represent the position of A when B sets out on his journey. Then, if we suppose CD equals 100 miles, ED will equal 55 miles, of which A will travel 30 miles (being 6 miles an hour for 5 hours), and B will travel 25 miles (being 5 miles an hour for 5 hours). If we suppose CD equals 180 miles, ED will equal 135 miles, of which A will travel 54 miles (being 6 miles an hour for 9 hours), and B will travel 81 miles (being 9 miles an hour for 9 hours). This problem, therefore, admits of two positive answers, both equally applicable to the question. Problems 22 and 23, page 196, are of the same kind. In Problem 18, page 195, one of the values of is positive and the other negative. А C с B Let the weaker magnet be placed at A, and the stronger aç B; then C will represent the position of a needle equally attracted by both magnets. According to the first value, the distance AC=8 inches, and CB=12 inches. Now, at the distance of 8 inches, the attraction of the weaker magnet will be 4 represented by gz; and at the distance of 12 inches, the at 9 traction of the other magnet will be represented by 122 , and these two powers are equal; for 4 9 82122 But there is another point, C', which equally satisfies the conditions of the question; and this point is 40 inches to the left of A, and therefore 60 inches to the left of B; for 4 9 4020602 284. Imaginary Values of the Unknown Quantity. We have seen that an imaginary root indicates impossible conditions in the proposed question which furnished the equation. In several of the preceding problems the values of x become imaginary in particular cases. When will the values of in Prob. 6, page 192, be imaginary? Ans. When l>a. What is the impossibility involved in this supposition ? Ans. It is impossible that the product of two numbers can be greater than the square of half their sum. When will the values of x in Prob. 11, page 194, be imaginary? Ans. When a?>b; or (2a)?> 46. What is the impossibility involved in this supposition ? Ans. The square of the sum of two numbers can not be greater than twice the sum of their squares. When will the values of x in Prob. 17, page 195, be imaginary? Ans. When a’>b; or (2a): > 86. What is the impossibility of this supposition ? Ans. The cube of the sum of two numbers can not be greater than four times the sum of their cubes. When will the values of x in Prob. 4, page 180, be imaginary, and what is the impossibility of this supposition ? 285. Geometrical Construction of Equations of the Second Degree.—The roots of an equation of the second degree may be represented by a simple geometrical figure. This may be done for each of the four forms: First form.—The first form gives for æ the two values р po р 4 Draw the line AB, and make it equal to VI. From B draw BC perpendicular to AB, and make it equal P to Join A and C; then AC will repre2' p A= = &+9, and 2-V&+ D sent the value of +9. For AC°= 4 А B ABP + BC? (Geom., Prop. 11, Bk. IV.). With C as a centre, and CB as a radius, describe a circle cutting AC in D, and AC produced in E. For the first value of x the radical is positive, and is set off from A toward C; then is set off from C to D; and AD, estimated from A P ; , to D, represents the first value of x. 2 For the second value of xwe begin at E, and set off EC P equal to - ; we then set off the minus radical from C to A; 2 and EA, estimated from E to A, represents the second value of x. + V +9, and <=?-V pa X= +9. -V X= (p2 -VA Second form. The second form gives for x the two values p р , x= 2 4 The first value of x is represented by AE estimated from A to E. The second value is +DC-CA, the latter being estimated from C to A. Hence the second value is represented by DA estimated from D to A. Third form.— The third form gives for x the two values р P 2 4 -9. Draw an indefinite line FA, and from any point, as A, set P a off a distance AB=-. We set off its 2 value to the left, because is negative. P 2 F D A At B draw BC perpendicular to FA, and make it equal to Vq. With C as a centre, and a radius equal P to 9, describe an are of a circle cutting FA in D and E. Now 2' pa I will be BD or BE. The first value of 4 x will be represented by – AB+BE, which is equal to – AE. The second will be represented by - AB-BD, which is equal to - AD; so that both of the roots are negative, and are estimated in the same direction, from A toward the left. Fourth form.—The fourth form gives for x the two values p p2 -2 V 2 4 2. Construct the radical part of the value of x as in the last Then, since is positive, we set off its value AB from p а P =R-V case. A toward the right. To AB we add BD, which gives AD for the first value of x; and from AB we subtract BE, which leaves AE for the second value of x. Α Ε Both val. ues are positive, and are estimated in the same direction, from A toward the right. Equal Roots.—If the radius CE be taken equal to CB, that P is, if Vq is equal to ., the arc described with the centre C will 2' be tangent to AF, the two points D and E will unite, and the two values of x become equal to each other. In this case the radical part of the value of x becomes zero. Imaginary Roots.—If the radius of the circle described with the centre C be taken less than CB, it will not meet the line AF. In this case q is numerical p2 4 ly greater than is , and the radical part of the |