I just can't solve the problem or find the solution online:
1.(1977.)Let a and b be 2 solutions of .Prove that is the solution of
I tried Vietes formulas and I tried to divide the second equation with
Thanks
I just can't solve the problem or find the solution online:
1.(1977.)Let a and b be 2 solutions of .Prove that is the solution of
I tried Vietes formulas and I tried to divide the second equation with
Thanks
You can certainly do it using Vieta's formulas, but it will be laborious. I don't see any simpler way, though.
Let be the roots of . The Vieta formulas tell you that , and . You want to find the equation whose roots are and .
The coefficients of that equation are again given by the Vieta equations. For example, and . Those two are easy, but the others are harder. For example, . To evaluate that, notice that , and it follows that . I haven't even tried to do or , but it should be possible, using the standard relations among symmetric polynomials.
You are told that a and b are two of the solutions of the fourth-degree equation , but you don't know which two. So your solution must be an equation whose roots contain all possible products of pairs of roots. That means that if the four roots of the original equation are a, b, c, d, then the solution equation must have all six products ab, ac, ad, bc, bd, cd as its roots.
I can't do the last part of the problem for you, because I haven't done it myself (and I don't have the energy to try). Maybe someone else can step in here? The link that I gave to the Wikipedia page on symmetric polynomials should have all the background information needed.
There is one additional coefficient of the sixth-degree polynomial that I verified: the coefficient of x should be (the negative of) the sum of the product of the roots taken five at a time. You can check that this is equal to . As I said before, it's the remaining two coefficients (those of and ) that are going to be more difficult.