Dividing the given equation by a-1, we obtain the depressed equation -5x+6=0. The roots of this equation are 2 and 3. Therefore the three roots of the given equation are 1, 2, 3.

EXERCISES II.

Solve the following equations :

1. x3x2-17x+15=0, one root being 3.

2. x3

8x2+25x500, one root being 5.

3. 9x18x24x80, one root being 2.

4. 4 x 12 x2 11x+20=0, one root being – 4.

9x2+14x-3= 0, one root being 4.

6. 27 x8 63 x 340, one root being

3.

7. x + 3x3- 19 x2 - 27 x +90= 0, two roots being 2, — 5. 8. x4x3- 17 x2 + 5x + 60 = = 0, two roots being 4, - 3.

9. 12x 59 x3 + 33x2 + 99x+27= 0, two roots being 3, 3. 10. x 20 x2 - 21 x 20 = 0, two roots being 5, - 4.

Relation between the Roots and Coefficients.

9. We can, without loss of generality, divide both members of an equation by the coefficient of a". We thus obtain an equation of the form

wherein P1, P2, **, P, denote any numbers, real or imaginary

10. If r1, 72, 73 be the roots of the equation

Equating coefficients of like powers of x, we have

From these equations we obtain the following relations between the roots and the coefficients, when the coefficient of the highest power of x is unity:

The sum of the roots is equal to the coefficient of the second term with sign changed.

The sum of the products of the roots in pairs is equal to the coefficient of the third term.

The sum of the products of the roots taken three at a time is equal to the coefficient of the fourth term with sign changed; and so on. The product of the roots is equal to the last term if n be even, and to the last term with sign changed if n be odd.

Ex. Given x3 5 x2+3x+4 = 0.

The sum of the products of the roots, two at a time, is 3.
The product of the roots is 4.

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11. The following special cases may be noted:

(i.) If the sum of the roots be 0, the second term is wanting. (ii.) If one root, or more than one root, be 0, the last term is wanting.

12. It is important to notice that the relations given in the preceding articles cannot be used to obtain the solution of an equation.

4

Ex. From the equation 23-5x2+3x+4=0, we have r1+2+13=5 (1), r12+r1rs + rqr3=3 (2), ፲፬ = – 4 (3). Multiplying (1) by ri2, (2) by r1, (3) by 1, and adding the resulting products, we obtain 3-5+3 r1 + 4 = 0.

2

This is a mere statement that r1 satisfies the given equation, and in no way helps us to find its value.

13. The properties of Art. 10 can sometimes be used to solve an equation, if a relation between the roots be known. Ex. One root of the equation - 6x2 + 11 x − 6 = 0 is twice a second root. Solve the equation.

3

2r+3=11

From (1) and (2) we obtain r1 = 1, 4, and r3 = 3, 9.
Since these values of r1 and r, must also satisfy equation (3),

1, 3, is the only admissible solution.

Finally, r2, and the required roots are 1, 2, 3.

Symmetrical Functions.

14. As in Ch. XXI., Art. 15, the value of an expression. which is symmetrical in the roots can be found without solving the equation.

Ex. If 71, 72, 73, be the roots of the equation

find the value of

or

We have

x3 + P1x2 + PgxX + P3 = 0,

r12 + r22 + r32.

2

(r1 + r1⁄2 + r3)2 = r22 + r22 + r22 + 2 (~17‍2 + Tirs + r1⁄2o3),

whence

(− p1)2 = r2 + r22 + rz2 + 2 Pai

r22 + r22 + r32 = p12 - 2 P2.

Formation of an Equation from its Roots.

15. The relations of Art. 10 enable us to form an equation if its roots be known. We may assume that the coefficient of the highest power of the unknown number is 1.

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Ex. 1. Form the equation whose roots are 2, 3, 4.
We have

r1 + r2 + r3 = 5, the coefficient of x, with sign changed. rir2 + r1rg + Pos=-2, the coefficient of x; and rrr=-24, the term free from x, with sign changed. Therefore the required equation is 3-522-2x + 24 = 0.

Ex. 2. Form the equation whose roots are

1 + √−2, 1−√−2, 1+ √3, 1− √3.

The work will be simplified by first forming the quadratic equation whose roots are the conjugate imaginaries, and the quadratic equation whose roots are the conjugate surds, and then multiplying together these two equations. The first of these quadratic equations is -2x+3=0;

We could also have formed the equation in Ex. 1 by multiplying together the three linear factors which equated to 0 have as roots the given roots. We should thus have obtained (x+2)(x-3)(x-4)=0, or 3-5x-2x+24=0.

EXERCISES III.

Solve the following equations:

1. x3-3x2 - 4x+12= 0, the sum of two roots being 0.

2. 2x8 - 11 x2 + 12x+9= 0, two roots being equal.

3. x3- 6 x2 - x = 6, one root being 2 greater than a second. 4. 3 x3-25 x2 - 19x+9= 0, the product of two roots being 3. 5. x3-3x2-6x+8= 0, the roots being in A. P.

6. 2 x3-9 x2

27 x + 54 = 0, the roots being in G. P.

7. 8x3 30 x2 + 27 x 50, one root being twice the sum of the other two.

8. x4 — 7 x3 + 9 x2 + 7 x − 10 = 0, one root being equal and opposite to a second root.

9. 6 x 13 x3- 18 x2 + 52 x 240, one root being the reciprocal of a second root.

135 x2 – 135 x + 162 0, the roots being in G. P.

11. x4 14x3 + 51 x2

Given the equations

and

x2 + P1x2 + P2X + Ps = 0

x2 + P1x2 + P2x2 + P3x + P4 = 0,

express the following symmetrical functions in terms of the coefficients.

If r1, 72, 73, be the roots of the equation x3 + P1x2 + P2X + ps = 0, form

16. If, a rational fraction in its lowest terms, be a root of

wherein aŋ, α1, ..., a„ are now assumed to be integers, then a is divisible by s, and a, is divisible by r.

The second member of the last equation is integral. Consequently

aorn

8

must reduce to an integer; that is, aor" must be divisible by s. Since r, and therefore r", is not divisible by s, ao must be divisible by s.

Now, multiplying (1) by s, and dividing by r, we readily infer that an is divisible by r.

or

Ex. Forming the equation whose roots are,, 5, we obtain x3-37 x2+37x=0,

We see that 6 is divisible by 2 and 3, the denominators of the two fractional roots, and that 10 is divisible by 1, 2, and 5.