Under this supposition, the value of x, and that of c-a, for the point C between A and B, both reduce to ; that is, when C 2 the lights are of equal intensity, the point of equal illumination is at the middle of the line AB. The value of x, and that of c x, for the points C and C", which lie on the prolongation of AB, both reduce to + c α or, to -c b that is, to infinity; which indicates, that the conditions of the question are absolutely impossible. It is evident, indeed, that they are so; for, when the intensity of the two lights is equal, no part lying on the prolongation of AB could be as much illuminated by the distant as by the nearer light: hence, the supposition of equal illumination, from which the equation of the problem is derived, is impossible; and this is shown in the analysis by the corresponding values of the unknown quantity becoming infinite. 4th. Let a = b, and c = 0. Under these suppositions, the value of x and of c-x, for the point of equal illumination between A and B, both reduce to 0, as indeed they ought to do, since the points A, B, and C, are then united in one. we see that it becomes, under the above suppositions, 0.x2-0.x = 0, which may be satisfied by giving to x any value whatever: hence, it is a case of indetermination. Indeed, since the two lights are of the same intensity, and are placed at the same point, they ought to illuminate equally every point of the straight line. C 5th. Let c = 0, and a and b be unequal. Under this supposition, both values of x, and both values of - x, will reduce to 0; and hence, there is but one point of the line that will be equally illuminated, and that is the point at which the two lights are placed. In this case, the equation of the problem reduces to which gives two values, (a - b) x2 = 0, x = 0, and x = 0. The preceding discussion presents a striking example of the precision with which the algebraic analysis responds to all the relations which exist between the quantities that enter into the enunciation of a problem. 151. Every equation which can be reduced to the form in which m and n are positive whole numbers, and 2p and q, known quantities, is called a trinomial equation. Hence, a trinomial equation contains three kinds of terms: viz., terms which contain the unknown quantity affected with two different exponents, and one or more known terms. If we suppose m = 2 and n = 1, the equation becomes a trinomial equation of the second degree. 152. The resolution of trinomial equations of the second degree, has already been explained, and the methods which were pursued are, with some slight modifications, applicable to all trinomial equations in which m = 2n, that is, to all equations of the form Let us take, as an example, the trinomial equation of the fourth degree, and by substituting 2p for the co-efficient of x2, and q for the absolute term, we have hence, = y, and consequently, x = ± √√ y, y2+2py = 9, and y = · p± √√ q + p2: We see that the unknown quantity has four values, since each of the signs and which affect the first radical can be combined in succession with each of the signs which affect the second; but these values taken two and two are numerically equal, and have contrary signs. 1. Take the equation EXAMPLES. x4-25x2 = 144. If we make x2=y, the equation becomes, which gives, y2 - 25y = 144, y = 16, and y = 9. Substituting these values, in succession, for y in the equation x2y, and there will result, 1st. x2 = 16, which gives x = + 4 and x= 4. Substituting these values, in succession, for y, and we have 1st. x2 = 2d. x2 = 8, which gives x = +2√2, and x =— 1, which gives x=+ √−1, and The last two values of x are imaginary. 3. Let us take the literal equation x-(2bc+4a2) x2 = b2c2. By making 2y, we have whence, and consequently, If we make √x=y, we have x = y2, and hence, 2y2-7y=99; it 153. Before resolving the general case of trinomial equations, may be well to remark that, the nth root of any quantity, is an expression which multiplied by itself n-1 times will produce the given quantity. |