Math, asked by kalpit1731, 1 year ago

How to find sum of roots of equatons of different degree?

Answers

Answered by Sweetbuddy
17

p(x)=xn+p1xn−1+⋯+pn−1x+p0=(x−r1)(x−r2)⋯(x−rn).

When you multiply out (x−r1)⋯(x−rn), and then put together the powers of x, you get a polynomial expression in which the coefficients are written in terms of the roots. For example, to get the coefficient of x, note that you will get an x when you expand and multiply the x in the first binomial by each of the constants in the rest, giving you (−1)n−1(r2⋯rn)x. Another when you multiply the x in the second binomial by each of the constants in the rest; another when you multiply the x in the third binomial by the constants in the rest; etc.

If you work this out, you'll find that when you expand (x−r1)⋯(x−rn) and then group together the powers of x, you will have:

The coefficient of xn is 1.

The coefficient of xn−1 is −(r1+⋯+rn).

The coefficient of xn−2 is

r1r2+r1r3+⋯+r1rn+r2r3+⋯+rn−1rn.

The coefficient of xn−3 is

−(r1r2r3+r1r2r4+⋯+r1r2rn+r1r3r4+⋯+rn−2rn−1rn).

The coefficient of x is

(−1)n−1(r1r2⋯rn−1+r1r2⋯rn−2rn+⋯+r2⋯rn).

The constant coefficient is

(−1)n(r1⋯rn).

But for two polynomials to be equal they have to be equal coefficient by coefficient. So that means that:

p1 =(−1)1(r1+⋯+rn) p2 =(−1)2(r1r2+r1r3+⋯+r1rn+r2r3+⋯+rn−1rn) p3 =(−1)3(r1r2r3+r1r2r4+⋯+r1r2rn+r1r3r4+⋯+rn−2rn−1rn) ⋮ pn =(−1)n(r1⋯rn).

That is:

The sum of the roots is −p1;

The sum of all products of two roots is (−1)2p2;

The sum of all products of three roots is (−1)3p3;

The sum of all products of four roots is (−1)4p4;

The sum of all products of n−1 roots is (−1)n−1pn−1;

The product of all roots is (−1)npn.

Which is what the text you cite says.

How about a polynomial that is not monic,

p(x)=p0xn+p1xn−1+⋯+pn−1x+pn,p0≠0 ?

Note that r is a root of p(x) if and only if it is a root of

P(x)=

1

p0

p(x)=xn+

p1

p0

xn−1+⋯+

pn−1

p0

x+

pn

p0

.

So the argument above applies to P(x).

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