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• Vieta's Formula
  #REDIRECT[[Vieta's formulas]]
  29 bytes (3 words) - 23:31, 3 June 2022

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• Factoring
  == Vieta's/Newton Factorizations == ...hat give a polynomial and ask a question about the roots. Combined with [[Vieta's formulas]], these are excellent factorizations that show up everywhere.
  3 KB (532 words) - 21:00, 13 January 2024
• Quadratic equation
  * [[Vieta's Formulas]]
  2 KB (264 words) - 11:04, 15 July 2021
• 2005 AIME I Problems/Problem 6
  ...= 1 - \sqrt{2006}</math>. Now transforming the original function and using Vieta's formula, <math>x^4-4x^3+6x^2-4x-2005=0</math> so <math>x_1 x_2 x_3 x_4 = \f
  4 KB (686 words) - 11:52, 13 June 2024
• 1983 AIME Problems/Problem 15
  ...interval <math>(-3,3)</math>, and another outside of it. We also see, by [[Vieta's Formulas]], that the average of the two roots is <math>\frac{5\cos \alpha}{
  20 KB (3,497 words) - 14:37, 27 May 2024
• 1986 AIME Problems/Problem 3
  ...atic need not be solved. The value of <math>ab</math> can be found through Vieta's.
  3 KB (545 words) - 22:44, 12 October 2023
• 1988 AIME Problems/Problem 13
  ...t <cmath>a= (u^8+v^8)(u^4+v^4)(u^2+v^2)(u+v).</cmath> A simple exercise of Vieta's gets us <math>a= \boxed{987}.</math>
  10 KB (1,595 words) - 15:30, 24 August 2024
• 1991 AIME Problems/Problem 8
  Let <math>x^2 + ax + 6a = (x - s)(x - r)</math>. Vieta's yields <math>s + r = - a, sr = 6a</math>. By [[Vieta's formulas]], <math>x_1 + x_2 = -a</math> where <math>x_1, x_2</math> are the
  2 KB (309 words) - 22:55, 25 July 2024
• 1993 AIME Problems/Problem 5
  ...th>r_1r_2+r_2r_3+r_3r_1+210\cdot 2(r_1+r_2+r_3)+3\cdot 210^2.</math> Using Vieta's (again) and plugging stuff in yields <math>-77+210\cdot 2\cdot -313+3\cdot
  5 KB (923 words) - 12:17, 16 September 2024
• 2003 AIME II Problems/Problem 9
  Now by [[Vieta's formulas|Vieta's]] we know that <math>-z_4-z_3-z_2-z_1=-1</math>, Let <math>S_k=z_1^k+z_2^k+z_3^k+z_4^k</math> then by [[Vieta's Formula]] we have
  3 KB (562 words) - 01:41, 11 September 2024
• 2002 AIME II Problems/Problem 15
  It follows from Vieta's Formulas that the product of the roots of this quadratic is <math>117m_1^2< ...2) r + 117 = 0.</cmath>We are told the product of the circles is 68, so by Vieta's formulas, <math>\frac{117}{k^2} = 68.</math> Hence, <math>k^2 = \frac{117}{
  7 KB (1,188 words) - 16:46, 23 June 2024
• 2005 AMC 10A Problems/Problem 10
  Note: you can also use Vieta's Formulae
  3 KB (450 words) - 19:33, 13 September 2024
• 1988 IMO Problems/Problem 6
  WLOG, let <math>r_1=a</math>. By Vieta's,
  4 KB (746 words) - 16:32, 16 January 2025
• 2007 AMC 10A Problems/Problem 20
  By Vieta's Formulas, note that the roots are reciprocals of each other. Therefore, bot
  4 KB (663 words) - 06:32, 4 November 2022
• 2004 AMC 12B Problems/Problem 23
  ...h>, and <math>t</math>, where <math>t = r+s</math>, we see that by Vieta's Formula's, <math>2004 = r+s+t = t + t = 2t</math>, and so <math>t = 1002</math>. Ther
  3 KB (533 words) - 13:52, 29 October 2023
• 2008 AIME II Problems/Problem 7
  ...math>(r + s)^3 - r^3 - s^3 = 3r^2s + 3rs^2 = 3rs(r + s).</cmath> Also by [[Vieta's formulas]] we have <cmath>rst = -rs(r + s) = -\dfrac{2008}{8} = -251</cmath By [[Vieta's formulas]], we have <math>r+s+t = 0</math>, and so the desired answer is <m
  7 KB (1,251 words) - 18:18, 2 January 2024
• 2008 AMC 10B Problems/Problem 9
  ...g both sides by <math>a</math>, we get <math>x^2 - 2x + b/a = 0</math>. By Vieta's formulas, the sum of the roots is <math>2</math>, therefore their average i
  1 KB (218 words) - 15:59, 13 April 2024
• 2008 AMC 10A Problems/Problem 18
  We can now apply Vieta's Formula which gives: <cmath>c = \sqrt{ \left( \frac{69}{4} \right)^2 - 80}
  8 KB (1,339 words) - 13:15, 1 August 2022
• 2009 AMC 12A Problems/Problem 17
  Using [[Vieta's formulas]] we get that the sum of these two roots is <math>\boxed{1}</math> ...um is: <math>\frac{b}{1-r} = r.</math> Thus, <math>b = r-r^2</math> and by Vieta's, the sum of the two possible values of <math>r</math> (<math>r_1</math> and
  3 KB (453 words) - 09:22, 6 October 2023
• 2001 AMC 12 Problems/Problem 23
  ...th>(a-k-n)^2=1</math>, and the imaginary part is positive. Furthermore, by Vieta's Formulas, we know that d must be a multiple of p and q, so <math>\frac{4d}{
  7 KB (1,304 words) - 08:53, 5 October 2024
• 2010 AIME II Problems/Problem 5
  ...h>a(b+c)+bc=ab+ac+bc=468</math>. Since these two equations look a lot like Vieta's for a cubic, create the polynomial <math>x^3-81x^2+468x=0</math> (leave the
  2 KB (392 words) - 22:49, 5 February 2022

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