lating coefficients of like powers of x in this identity,

-P1 = S1 = sum of the roots;

P2 = S2 = sum of the products of the roots taken two at a time;

-PS = sum of the products of the roots taken three at a 3 time;

- 1)"p=S= product of the roots.

If the coefficient of x" is p., then on dividing each term by the equation becomes

and, with the notation of Art. 521, we have

Za=-2, Zab=22, Zabc=

Σα

Σαβ

P2

Po

Po

Example 1. Solve the equations

P3

Po

abc...k(1) P

Po

x+ay+a2z=a3, x+by+b2z=b3, x+cy + c2z=c3.

From these equations we see that a, b, c are the values of t which satisfy the cubic equation

hence

t3 - zt2 - yt - x=0;

z=a+b+c, y = − (be+ca+ab), x=abc.

Example 2. If a, b, c are the roots of the equation 23 +P1x2+P2x+P3=0, form the equation whose roots are a2, b2, c2.

The required equation is (y-a2) (y-b2) (y - c2)=0,

or

that is, But hence

or

or

Thus the required equation is

(x2+P1x2+Pqx+P3) (1⁄23 −Р1x2+ − P3)=0,

540. The student might suppose that the relations established in the preceding article would enable him to solve any proposed equation; for the number of the relations is equal to the number of the roots. A little reflection will shew that is this not the case; for suppose we eliminate any n-1 of the quantities a, b, c,... k and so obtain an equation to determine the remaining one; then since these quantities are involved symmetrically in each of the equations, it is clear that we shall always obtain an equation having the same coefficients; this equation is therefore the original equation with some one of the roots a, b, c,...k substituted for x.

Let us take for example the equation

x2 + P ̧x2 +P2x+P2 = 0;

and let a, b, c be the roots; then

Multiply these equations by a3, -a, 1 respectively and add; thus

which is the original equation with a in the place of x.

The above process of elimination is quite general, and is applicable to equations of any degree,

541. If two or more of the roots of an equation are connected by an assigned relation, the properties proved in Art. 539 will sometimes enable us to obtain the complete solution.

Example 1. Solve the equation 4x3- 24x2+23x+18=0, having given that the roots are in arithmetical progression.

Denote the roots by a-b, a, a+b; then the sum of the roots is 3a; the sum of the products of the roots two at a time is 3a2-62; and the product of the roots is a (a2 – b2); hence we have the equations

from the first equation we find a=2, and from the second b= ±

since these values satisfy the third, the three equations are consistent.

Example 2. Solve the equation 24x3 – 14x2 - 63x+45=0, one root being double another.

542. Although we may not be able to find the roots of an equation, we can make use of the relations proved in Art. 539 to determine the values of symmetrical functions of the roots.

Example 1. Find the sum of the squares and of the cubes of the roots of the equation x3-px2+qx − r=0.

Denote the roots by a, b, c; then

Now

a+b+c=p, bc+ca+ab=q.

a2+b2+c2 = (a+b+c)2 − 2 (bc+ca+ab)

= p2 - 2q.

Again, substitute a, b, c for x in the given equation and add ; thus a3 +b3 + c3 − p (a2+b2+c2)+q (a+b+c) −3r=0;

.. a3+b3+c3=p (p2 − 2q) − pq+3r

5. x4-16x3+86x2 – 176x+105=0, two roots being 1 and 7.

6. 4x3+16x2- 9x-36=0, the sum of two of the roots being zero. 7. 4x3+20x2 - 23x+6=0, two of the roots being equal.

8. 3x3-26x2+52x-24-0, the roots being in geometrical progression.

9. 2x3-x2-22x-24-0, two of the roots being in the ratio of 3: 4.

10. 24x3+46x2+9x-9=0, one root being double another of the roots.

11. 8x4 - 2x3-27x2+6x+9=0, two of the roots being equal but opposite in sign.

12. 54x3-39x2 - 26x+16=0, the roots being in geometrical progression.

13. 32x3 – 48x2+22x-3=0, the roots being in arithmetical progression.

14. 6x4-29x3+40x2-7x-12=0, the product of two of the roots being 2.

15. x1- 2x3-21x2+22x+40=0, the roots being in arithmetical progression.

16. 27x-195x3 +494x2 – 520x+192=0, the roots being in geometrical progression.

17. 18x3 +81x2+121x+60=0, one root being half the sum of the other two.

18. If a, b, c are the roots of the equation x3 − px2+qx−r=0, find the value of

19. If a, b, c are the roots of 3+qx+r=0, find the value of

(1) (b-c)2+(c-a)2+(a - b)2.

(2) (b+c)-1+(c+a)−1+(a+b)−1.

20. Find the sum of the squares and of the cubes of the roots of

x2+qx2+rx+8=0.

21. Find the sum of the fourth powers of the roots of

x3+qx+r=0.

543. In an equation with real coefficients imaginary roots occur in pairs.

Suppose that f(x) = 0 is an equation with real coefficients, and suppose that it has an imaginary root a + ib; we shall shew that aib is also a root.

The factor of f(x) corresponding to these two roots is

(x — a―ib) (x — a + ib), or (x − a)2 + b2.

Let f(x) be divided by (x-a)2+b2; denote the quotient by Q, and the remainder, if any, by Rx+R'; then

f(x) = Q {(x − a)2 + b2 } + Rx + R'.

In this identity put x = a + ib, then f(x)=0 by hypothesis; also (x − a)2 + b2 = 0; hence R (a + ib) + R' = 0. .

Equating to zero the real and imaginary parts,

Ra + R'=0, Rb=0;

and b by hypothesis is not zero,

.. R=0 and R' = 0.

Hence f(x) is exactly divisible by (x-a)2 + b2, that is, by

(x − a − ib) (x − a + ib);

hence xα-ib is also a root.

=

544. In the preceding article we have seen that if the equation f(x) = 0 has a pair of imaginary roots a± ib, then (x -- a)2 + b2 is a factor of the expression f(x).