So I stumbled upon this problem in my recent exam,
Let \(f(x) = x^8 - 4x^7 + 7x^6 + a_6x^5 + a_5x^4 + ... + a_2x + a_1\) where \(a_i \isin R\). Let \(x_1, x_2, ..., x_8 \isin R\) be zeros of \(f(x)\). Then find the possible values of \(a_1\)

The solution goes as follows:
By Veita's theorem, we have:
\(\sum_{i=1}^{8}x_i = 4\)
\(\sum_{i≠j}x_ix_j\) = 7
\(\therefore\) we have \(\sum_{}x_i^2 = 2\)
Applying \(RMS \ge AM\) on \(x_1, x_2, ..., x_8\)
we get
\(\sqrt{\tfrac{x_1^2 + x_2^2 + ... + x_8^2}{8}} \ge \tfrac{x_1 + x_2 + ... + x_8}{8}\)
\(\implies \tfrac{1}{2} = \tfrac{1}{2}\)
And the equality holds, iff the terms are equal.
\(\therefore x_1 = x_2 = x_3 = ... = x_8 = K\) And we have \(K = \tfrac{1}{2}\), and \(a_1 = \tfrac{1}{256}\)

Let us begin with the same approach for generalization for equal zeros.
Let us assume \(f(x) = a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + ... + a_1x + a_0\), with \(a_i, x_i \isin R\) and \(x_1, x_2, ..., x_n\) are zeros of \(f(x)\)
Let us say \(a_n = 1\) just so it makes the generalisation easier
\(\sum_{}x_i = -a_{n-1}\)
\(\sum_{i≠j}x_ix_j = a_{n-2}\)
\(\therefore\) we have \(\sum_{}x_i^2 = a_{n-1}^2 - 2a_{n-2}\)
Applying \(RMS \ge AM\) or Cauchy Schwarz on \(x_1, x_2, ..., x_n\)
We have,
\(\sqrt{\tfrac{x_1^2 + x_2^2 + ... x_n^2}{n}} \ge \thickspace\mid\tfrac{x_1 + x_2 + ... + x_n}{n}\mid\)
Substituting values, we get:
\(n(a_{n-1}^2 - 2a_{n-2}) \ge a_{n-1}^2\)

Since all the zeros are equal, we get,
\(n(a_{n-1}^2 - 2a_{n-2}) = a_{n-1}^2\)

And so this equality holds iff all the zeros are equal! Therefore, we have an condition for equal zeros i.e. if all the zeros are equal, the equality holds, and if the equality holds, then all the zeros must be equal (if they are real, ofcourse).

Interestingly, when we put \(n = 2\), and let \(a_1 = b, a_0 = c\)
We get:
\(2(b^2 - 2c) = b^2\)
\(\implies b^2 - 4c = 0\)
And ofcourse in the starting we assumed, \(a_n = 1\), accounting for that we have
\(b^2 - 4ac = 0\)
Which indeed as we learn in Highschool is the condition for equal zeros for a quadratic polynomial.

Now, let's put \(n = 3\) for more fun, (letting \(a_2 = a, a_1 = b\))
\(3(a^2 - 2b) = a^2\)
\(\implies\) \(a^2 = 3b\)
Therefore in a cubic polynomial \(x^3 + ax^2 + bx + c\), where \(a,b,c \isin R\)
If all the zeros of the polynomial are real, and \(a^2 = 3b\)
Then all the zeros must be equal!!

I worked on one more condition, which involved \(AM \ge GM\) inequality. That condition is:

\((\tfrac{-a_{n-1}}{n})^n = (-1)^na_0\)