Given that a, B and C are the roots of the equation x ^ 3 + PX + q = 0, find a + B + C =?, and give the process and theoretical basis, It seems to have something to do with Veda's theorem, but I don't know how to use it But the answer is 0,

Given that a, B and C are the roots of the equation x ^ 3 + PX + q = 0, find a + B + C =?, and give the process and theoretical basis, It seems to have something to do with Veda's theorem, but I don't know how to use it But the answer is 0,

In the quadratic equation AX ^ 2 + BX + C (a is not 0), let two roots be x and y, then x + y = - B / A, xy = C / A. WIDA's theorem can also be used in higher order equations. Generally, for an equation of degree n ∑ AIX ^ I = 0, its roots are denoted as x1, X2 We have ∑ xi = (- 1) ^ 1 * a (n-1) / a (n) ∑ xixj = (- 1) ^ 2 * a (n-2)

The equation x ^ 2 + PX + q = 0 about X is changed into the form of (x + a) ^ 2 = B. when Q and P satisfy what relation, the equation has real roots? Find the roots of the equation

X ^ 2 + PX + q = 0 to (x + P / 2) ^ 2 = P ^ 2 / 4-q
When p ^ 2 / 4-q > = 0, the equation has real roots
In this case, P ^ 2 > = 4q
So the root of the equation is (P ^ 2 / 4-q) - P / 2

Proof of n-th equation having n roots at most
It seems to use Rolle's theorem

It can be proved by Rolle's theorem that the equation of degree n has n + 1 roots, which are x1, x2. X (n + 1)
According to Rolle's theorem, (x1, x2) (X2, x3); (xn, X (n + 1)) has Y1, Y2... YN, such that f '(Y1) = f' (Y2) =. F '(yn) = 0
Because the equation of degree n becomes an equation of degree n-1 after deriving once, that is to say, the equation of degree (n-1) has n roots
By analogy, using Rolle's theorem constantly, it is found that the equation of degree 1 has two roots or the equation of degree 0 (the equation of degree 0 is a constant) has one root, which is obviously contradictory

Let the equation x ^ n = a (a > 0, n is a rational number), and prove that the equation has and has only one root

Wrong
Counter example: x ^ 2 = 1, x = 1 or - 1, with 2 roots