From (i) and (ii) we have by subtraction

Similarly, since ẞ-y±0, we have from (ii) and (iii)

a (B+ y) +b=0..............

From (iv) and (v) we have by subtraction

..(v).

.(vi).

Y 0.

Now (vi) cannot be true unless a = 0, for a Also when a = 0, it follows from (iv) that b=0, and then from (i) that c= = 0.

Thus the quadratic equation ax2 + bx+c=0 cannot have more than two different roots, unless a = b = C= = 0; and when a, b, c are all zero it is clear that the equation ax2 + bx + c = O will be satisfied for all values of x, that is to say the equation is an identity.

The equation is clearly satisfied by x=a, and also by x-b; hence a, b are roots of the equation, and these are the only roots of the quadratic equation. [The equation is not an identity, for it is not satisfied by x=c.]

The equation is satisfied by x=a, by x=b, or by x=c. Hence, as it is only of the second degree in x, it must be an identity.

The equation is satisfied by x=a, by x=b, and by x=c; and the equation is not an identity, since the coefficient of x3 is not zero. Hence the roots of the cubic are a, b, c.

Ex. 4. Shew that, if

(a-a)x+(a-ẞ)2 y + (a− y)2 z=(a-8),

(b-a)x+(b-ẞ) y + (b − y)2 z = (b − 5)2,

(c-a)x+(c-B)2 y + (c− y)2 z = (c — d)3,

then will

(da)x+(d-ẞ)3 y + (d− y)2 z = (d – 8)2,

where d has any value whatever.

The equation

(X-a)x+(X-ẞ)2y+(X-y)2 z=(X-8)2

is a quadratic equation in X, and it has the three roots a, b, c. therefore satisfied when any other quantity d is put for X.

It is

128. Relations between the roots and coefficients of a quadratic equation.

If we put a and ẞ for the roots of the equation ax2 + bx + c = 0, we have

The formulae (i) and (ii) giving the sum and the product of the roots of a quadratic equation in terms of the coefficients are very important,

129. Relations between the roots and the coefficients of any equation. By the following method relations between the roots and the coefficients of an equation of any degree may be obtained.

in x

We have seen that if the expression of the nth degree

n-2

n-3

ax" + bx"1+cx"2 + dx"3 + ....

vanish for the n values x =α, x= B, x=y, &c., then will

n-1

2-2

ax" + bx”-1 + cx"-2 + dx"3 + ... = a (x − a) (x − B) (x − y)...

ax2

We have therefore only to find the continued product (x-a) (x-B) (xy)...... and equate the coefficients of the corresponding powers of a on the two sides of the last equation.

For example, if a, B, y be the roots of the cubic equation ax3 + bx2 + cx + d = 0, we have

ax3 + bx2 + cx+da (x − a) (x − B) (x − y)

= a {x3- (a + B+ y) x2 + (By +ya+aß) x — aẞy}.

Hence, equating coefficients, we have

a

It should be remarked that the sum of the roots of the above cubic equation will be zero provided that b = 0, that is provided that the term one degree lower than the highest is absent.

We may make use of the above to prove certain identical relations between three quantities whose sum is zero. For a, b, c will be the roots of the cubic x3 +px+q=0, provided that a+b+c=0, and that p and q satisfy the relations

then

.(i),

.(ii).

(iii),

...(iv).

(v).

(vi).

Multiply the equations (v) in order by an-3, bn-3, cn-3, and add;

an+bn+c2+p (an−2+bn−2+ cn−2) + q (an−3+bn−3 + cn−3)=0. Hence we have in succession

130. Equations with given roots. Although we cannot in all cases find the roots of a given equation, it is very easy to solve the converse problem, namely the problem of finding an equation which has given roots.

For example, to find the equation whose roots are 4 and 5.

We want to find an equation which is satisfied when x=4, or when x=5; that is when x-4=0, or when x-5=0; and in no other The equation required must be

cases.

for this is an equation which is a true statement when x-4=0, or when x-5=0, and in no other case

*

Again, to find the equation whose roots are 2, 3 and – 4.

We have to find an equation which is satisfied when x-2=0, or when x-3=0, or when x+4=0, and in no other case. The equation must therefore be (x − 2) (x − 3) (x+4)=0,

Ex. 1. If a, ẞ are the roots of the equation ax2+ bx+c=0, find the

Ex. 2. If a, ẞ, y be the roots of the equation ax3 + bx2+cx+d=0, find the equation whose roots are ẞy, ya, aß.

The required equation is

(x – By) (x − ya) (x − aß)=0,

that is x3- x2 (By+ya+aß) + xaßy (a +ẞ+ y) − a22y2=0.

* The equation x2 9x+200 is certainly an equation with the proposed and with no other roots; but to prove that is the only equation with the proposed and with no other roots, it must be assumed that every equation has a root.

If, for example, the equation x5+7x2 -2=0 had no roots, then (x −4) (x −5) (x5+7x2-2)=0 would also be an equation with the proposed roots and with no others.

The proof of the proposition that every equation has a root is given in works on the Theory of Equations.