5 5 α Thus the roots are -a, -a±√10. 136. Reciprocal Equations. A reciprocal equation is one in which the coefficients are the same in order whether read backwards or forwards. Thus Let the two roots of the quadratic in y be a and ẞ; then the roots of the original equation will be the four roots of the two equations Ex. 3. To solve We have ax5+ bx + cx3 +cx2+bx+a=0. a (x5+1) + bx (x3 +1)+cx2 (x+1)=0, that is (x+1){α (x1 − x3 + x2 -- x+1)+ bx (x2 − x+1)+cx2} = 0. Hence x=-1, or else ax2 + (b − a) x3 + (a − b + c) x2 + (b − a) x+a=0. The last equation is a reciprocal equation of the fourth degree and is solved as in Ex. 2. 137. Roots found by inspection. When one root of an equation can be found by inspection, the degree of the equation can be lowered by means of the theorem of Art. 88. Ex. 1. Solve the equation x(x-1)(x-2)=a (a-1) (a-2). One root of the equation is clearly a. Hence xa is a factor of x (x − 1) (x − 2) – a (a − 1) (a − 2) = (x − a) { x2 - (3 − a) x + (a − 1) (a − 2)}. Hence one root of the equation is a, and the others are given by x2 - (3 − a) x + (a − 1) (a − 2) = 0. Ex. 2. Solve the equation x3+2x2 - 11x+6=0. Here we have to try to guess a root of the equation, and in order to do this we take advantage of the following principle: α If x = ± be a root of the equation ax + bxn−1 + ... + k=0, where β α k are integers and is in its lowest terms, then a will be a β factor of k and B a factor of a. As a particular case, if there are any rational roots of "+. +k=0, they will be of the form x=±a, where a is a factor of k. In the example before us the only possible rational roots are ±1, ±2, ±3 and ±6. It will be found that x=2 satisfies the equation, and we have (x − 2) (x2 + 4x-3)=x3 + 2x2 - 11x+6. Hence the other roots of the equation are given by Since xa and x=b both satisfy the equation, (x − a) (x − b) will divide (a− x) + (x − b)1 — (a - b)4, and as the quotient will be of the second degree, the equation formed by equating it to zero can be solved. We may however proceed as follows. The equation may be written (a − x)1 + (x − b)1 = { (a − x) + (x − b)}4 =(ax)+4(ax)3 (x-b)+6 (a− x)2 (x − b)2 .. 2(a-x) (x-b) {2 (a− x)2+3 (a− x) (x − b) + 2 (x − b)2}=0. Then the required roots are a, b and the roots of the quadratic x2-x (a+b)+2a2 - 3ab+2b2=0. a4 (x − a) (x − b) (x-b) (x-c) (x −c) (x − a) The equation is clearly satisfied by x=a, by x=b, and by x=c. Thus the roots are a, b, c, − (a+b+c). 138. Binomial Equations. The general form of a binomial equation is "+k=0. m The following are some of the cases of binomial equations which can be solved by methods already given― for the general case De Moivre's theorem in Trigonometry must be employed. Hence there are three roots of the equation x3=1; that is there are three cube roots of unity, which are Since x-1= (x − 1) (x + 1) (x + √ − 1) (x−√−1), the four fourth roots of unity are The latter equation is a reciprocal equation. Divide by x2, and we have Hence x2+1=0. x2+1=(x2+1)2 - 2x2= (x2+1−√√/2x) (x2+1+√2x). 139. Cube roots of unity. In the preceding article we found that the three cube roots of unity are 1, § ( − 1 + √ − 3), † ( − 1 − √ − 3). An imaginary cube root of unity is generally represented by w; or, when it is necessary to distinguish between the two imaginary roots, one is called w,, and the other so that 1, @1 and w are the three roots of the equation - 1 = 0. a3 () ရွာ Taking the above values, we have 1 + w ̧ + w2 = 1 + } ( − 1 + √− 3) + 1⁄2 ( − 1 − √ − 3) = 0, also @,w2 = {(−1 + √ − 3) (-1-√-3)=1. These relations follow at once from Art. 129; for the sum of the three roots of x3-1=0 is zero, and the product is 1. and 2 Again w2=(-1 + √ − 3)2 = } ( − 1 - √ −3) = w, 2 w,2 = 1 ( − 1 − √ — 3)2 = † ( − 1 + √ − 3) = w ̧, = 2 = so that w, w, and ww,. These relations follow at once from Thus if we square either of the imaginary cube roots of unity we obtain the other. Hence if o be either of the imaginary cube roots of unity, the three roots are 1, w and w2. We know that a3+b3+c3-3abc=(a+b+c) (a2+b2 + c2 - bc - ca - ab). Hence a+b+c is a factor of a3+b3 + c3 − 3abc, and this is the case for all values of a, b, c. Hence a+ (wb) + (w2c) is a factor of a3 + (wb)3 + (w2c)3 − 3a(wb) (w2c), that is of a3 +63 + c3 - 3abc; and a+w2b+wc can similarly be shewn to be a factor. Hence a3+b3+c3 − 3abc=(a+b+c) (a+wb+w2c) (a + w2b+wc). |