Let two roots in the unary quadratic equation AX 2+BX+C = 0 (A ≠ 0 and △ = B 2-4ac ≥ 0) be X 1, and X 2 be X1+X2 =-B/AX1X2 = C. If b 2-4ac = 0, the equation has two equal real roots; If b 2-4ac

Edit this passage proved by Vieta's theorem.

The formula for finding the root of a quadratic equation with one variable is:

X = (-b √ b-4ac)/2a, then x 1=(-b+√b-4ac)/2a, x2 = (-b-√ b-4ac)/2ax1+x2 = (-b+√ b-4ac/2a. =(x 1+x2)？ Proof of the generalization of vieta theorem -2x 1x2 Let x 1, x2, ..., xn be n solutions of the unary n-degree equation ∑AiXi =0. Then: an (x-x1) (x-x2) ... (x-xn) = 0, so: an (x-x1) (x-x2) ... (x-xn) = ∑ aixi (open (x-xn) It is best to use the principle of multiplication). By comparing the coefficients, we can get: a (n-1) =-an (∑ xi) a (n-2) = an (∑ xixj) ... A0 = [(-1) ]× an×∏ xiso: ∑ xi = ∑.

Arithmetic progression formula: the sum of the first n terms is Sn=na 1+n(n- 1)d/2 or Sn=(a 1+an)n/ 2 if m+n=p+q, then: am+an=ap+aq exists; If m+n=2p, then: am+an=2ap and above are all positive integer values of the nth item in text translation. An = first term+(number of terms-1)× sum of n terms before tolerance. Sn= (first period+last period) × number of periods. ⊙(n- 1) Number of terms = (last term-first term) ⊙ Tolerance+1 When the series is an odd term, the sum of the first n terms = middle term × series is an even term. Find the sum of the first term and the last term and divide it by the formula 2an+ 1 =.