(3.) Given x1+x3+x2+3x¬100=0, to find the number and situation of the real roots. hence two roots are real, and two imaginary; and the real roots must have contrary signs, for the last term of the equation is negative. To find the situation of the roots. In this example, the function X1 vanishes for æ=-1, and for the same value of x, the functions X and X, have contrary signs, agreeably to Lemma 2, and writing + or for 0 gives the same number of variations. The initial figures of the root are, therefore, 2 and —3. The same otherwise, by Budan's method. If the roots of this equation be all real, the permanencies and variation indicate three negative roots and one positive root. In the transformation by 2, one variation is left, and in transforming by 3, there is no variation left; therefore, the positive root is between 2 and 3. Here two variations are lost in the direct transformation, and no variations are left in the reciprocal transformation; therefore, the two roots in the interval 0 and 1 are imaginary. Hence the negative root is obviously situated between 3 and -4. U To find the negative root, we have the following operation: but this we shall leave for the student to perform, and the two roots will be found to be (4.) Find the roots of the equation x3+2x1+3x3+4x2+5x—20=0. Here we have X = x+2x1+3x3+4x2+5x-20 X1=5x+8x3+9x2+8x+5 X=-13x+14 Hence the difference of variations of sign indicates the existence of one real and four imaginary roots, the real root being situated between 1 and 2. hence the real root is nearly 1∙125790; and by using another period of ciphers we should have the root correct to ten places of decimals, with very little additional labour. ADDITIONAL EXAMPLES FOR PRACTICE. (1.) Find all the roots of the equation x3-3x-1=0. (2.) Find all the roots of the equation x3-22x-24=0. (3.) Find the roots of the equation a3+x2-500=0. (4.) Find the roots of the equation a3+x2+x-100=0. (5.) Find the roots of the equation 2x3-3x2—4x—10—0. (6.) Find the roots of the equation x1—12x2+12x-3=0. (7.) Find the roots of the equation x1-8x3+14x2+4x-8=0. (8.) Find the roots of the equation x1—x3+2x2+x-4=0. (9.) Find the roots of the equation x5—10x3+6x+1=0. (10.) Find the roots of the equation x3+3x1+2x3-3x2-2x-2=0. (11.) Find all the roots of the equation x+4x3-3x1—16x3+11x2+12x-9=0. (12.) Given the equations 4x2-5xy+6y2 2x-10y+20=0 to find the values of x and y. 480y1-125y3+3923y2-7176y+25740=0, contains the four values of y, and then x is found from the relation 177. In discussing algebraical problems, it is frequently necessary to introduce inequations, that is, expressions connected by the sign 7. Generally speaking, the principles already detailed for the transformation of equations, are applicable to inequations also. There are, however, some important exceptions which it is necessary to notice, in order that the student may guard against falling into error in employing the sign of inequality. These exceptions will be readily understood by considering the different transformations in succession.* I. If we add the same quantity to, or subtract it from, the two members of any inequation, the resulting inequation will always hold good, in the same sense as the original inequation. That is, if a7b, then a+a' 7b+a', and a—a' 7 b—a'• * Example in Inequations.-The double of a number diminished by 6 is greater than 24; and triple the number diminished by 6 is less than double the number increased by 10. Required a number which will fulfil the conditions. Let a represent a number fulfilling the conditions of the question; then, in the language of inequations, we have 2x-67 24, and 3x-642x+10. From the former of these inequations we have 2x 7 30, or x 715; and from the latter we get 3x-2x 10+6, or x ≤ 16; therefore 15 and 16 are the limits, and any number between these limits will satisfy the conditions of the question. Thus, if we take the number 15.9 we have 15.9 × 2-67 24 by 1.8, whilst 15.9 × 3-6 Z 15.9 × 2+ 10 by 1. Other examples may easily be formed The truth of this proposition is evident from what has been said with reference to equations. This principle enables us, as in equations, to transpose any term from one member of an inequation to the other, by changing its sign. II. If we add together the corresponding members of two or more inequations which hold good in the same sense, the resulting inequation will always hold good in the same sense as the original individual inequations. III. But if we subtract the corresponding members of two or more inequations which hold good in the same sense, the resulting inequation WILL NOT always hold good in the same sense as the original inequations. Take the inequations 47, 23, we have still 4 227 3, or 24. But take 910 and 68, the result is 9- 67 (not) 10—8, or 372. We must therefore avoid as much as possible making use of a transformation of this nature, unless we can assure ourselves of the sense in which the resulting inequality will subsist. IV. If we multiply or divide the two members of an inequation by a positive quantity, the resulting inequation will hold good in the same sense as the original inequation. This principle will enable us to clear an inequation of fractions. |