Solving an Amazing System of Equations | Math Olympiad Prep!

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Solving an Amazing System of Equations | Math Olympiad Prep!
Welcome to our Math Olympiad Prep series! In this video, we'll tackle an amazing system of equations that's sure to challenge and sharpen your problem-solving skills. Perfect for those preparing for the Math Olympiad or anyone looking to enhance their math prowess, this tutorial will guide you step-by-step through the solution process. Don't miss out on this exciting opportunity to master a complex system of equations and boost your confidence for the competition!
Stay tuned, and let’s solve this together! If you enjoy the video, please like, subscribe, and hit the bell icon for more Math Olympiad tips and tricks.
Topics covered:
System of equations
Algebra
Problem solving
Algebraic identities
Algebraic manipulations
Solving systems of equations
Factorization
Cubic equations
Synthetic division
Rational root theorem
Math enthusiast
Math tutorial
Math Olympiad
Math Olympiad Preparation
#mathtutorial #systemofequations #problemsolving #mathhelp #algebra #learnmath #mathtricks #studytips #education #solvingequations #mathskills #mathenthusiast #stem #math
Additional Resources:
• Easier Than You Think ...
• Two Victorious Ways to...
• Israeli Math Olympiad ...
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Опубликовано:

3 окт 2024

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Комментарии : 19

@mohammedsaysrashid3587 3 месяца назад

A wonderful introduction thanks for sharing (x,y)= (8,-1)(-1,8)

@Quest3669 3 месяца назад

Shorts... Let x= a^3; y=b^3 then eqn turns to a^6+b^6=65= 2^6+ (-1)^6 And; a+b= 1 Hence ,(a;b)= (2; -1); (-1; 2 ) X; y=(8; -1);( -1; 8) Another questions can be made by Putting a+b= -1 or -3

@woobjun2582 3 месяца назад

The given can be a +b = 1 ...(eq1) a^6 + b^6 = 65 ...(eq2) with 3rd-root(x, y) = a, b. Then, squaring (eq1) a^2 +b^2 = 1 -2ab ...(eq3) and cubing (eq3) (a^2 +b^2)^3 = (1 -2ab)^3; a^6 +b^6 +3(a^2+b^2)(ab)^2 = 1 - 6ab +12(ab)^2 -8(ab)^3 and with (eq2) 65 +3(a^2+b^2)(ab)^2 = 1 - 6ab +12(ab)^2 -8(ab)^3. Rearranging with t = ab 2t^3 -9t^2 +6t +64 =0, which can be by SDM, (t +2)(2t^2 -13t +32) = 0 and thus t +2 = 0; t = ab = -2; while 2t^2 -13t +32 = 0 yields t = ab = cmplx. Therefore, a +b = 1; ab = -2 that is a^2 -a -2 = 0; a^2 -2a +a -2 = 0; a(a -2) +(a -2) = 0; (a+1)(a -2) = 0 a = -1, 2 and thus by (eq1) b = 2, -1 That is, (a,b) = (-1, 2), (2, -1) since 3rd-root(x, y) = a, b (x, y) = (-1, 8), (8, -1)

@johnstanley5692 2 месяца назад

let g1 =a+b-1 (=0), g2=a^6+b^6-65 (=0). Since g1 is 1st order in 'a' and 'b' we can use synthetic division to eliminate one of the components for example using 'a' as reference then remainder g2/g1 = p(b)=2*b^6 - 6*b^5 + 15*b^4 - 20*b^3 + 15*b^2 - 6*b - 64; p(b) = (b - 2)*( b + 1)*( 2*b^4 - 4*b^3 + 15*b^2 - 13*b + 32) so (b=2, a=-1), ( b=-1, a=-2 )

@jaggisaram4914 3 месяца назад

❌ = 8 , -1. 🌱 = -1 , 8

@АндрейПергаев-з4н 3 месяца назад

Возведем в куб первое уравнение х+у+3(ху)^(1/3)=1 Второе уравнение (х+у)^2=65+2ху А дальше ху=k^3 И далее как на видео Только на видео к этому моменту пришли на 6 минуте, вместо того чтобы потратить 2 минуты

@Quest3669 3 месяца назад

(X, Y)=( 8; -1); (-1; 8)

@tunneloflight 3 месяца назад

By observation. x,y = (8,-1), (-1,8)

@ramunasstulga8264 3 месяца назад

1/5 marks for correct answer

@tunneloflight 3 месяца назад

@@ramunasstulga8264 Just look at the problem. Sum of squares = 65. Sum of cube roots =1. Smooth curves, so clearly there are at most two roots (one positive, one negative). Sum of perfect squares -> text 8 and 1 = 65. Try cube roots -> 2 and 1. Sum must be 1 so +8, and -1. Fits. X and Y are interchangeable. So two solutions. It takes longer to write this than to see the answer.

@ramunasstulga8264 3 месяца назад

@@tunneloflight yes, but I mean you would only get one mark for this

@tunneloflight 3 месяца назад

@@ramunasstulga8264 Why would I care about marks? Marks are meaningless twaddle.

@kassuskassus6263 3 месяца назад

(x,y)=(-1,8); (8,-1)

@SidneiMV 3 месяца назад

x⅓ + y⅓ = 1 x² + y² = 65 (x⅓ + y⅓)³ = 1 x + y + 3(xy)⅓(x⅓ + y⅓) = 1 x + y + 3(xy)⅓ = 1 x + y = a 3(xy)⅓ = 1 - a => xy = (1 - a)³/27 x² + y² + 2xy = a² xy = (a² - 65)/2 (1 - a)³/27 = (a² - 65)/2 2(a - 1)³ + 27(a² - 65) = 0 a = 7 (by inspection) x + y = 7 xy = (7² - 65)/2 => xy = -8 t² - 7t - 8 = 0 (t + 1)(t - 8) = 0 *(x,y) = { (-1, 8); (8, -1) }*

@Fjfurufjdfjd 3 месяца назад

(χ,y) =(8,-1) , (-1,8)

@walterwen2975 3 месяца назад

Math Olympiad Prep: ³√x + ³√y = 1, x² + y² = 65, x, y ϵR; x, y = ? (³√x + ³√y)³ = (³√x)³ + (³√y)³ + 3(³√xy)(³√x + ³√y) = x + y + 3(³√xy) = 1 (x + y)² = [1 - 3(³√xy)]², x² + y² + 2xy = 65 + 2xy = 1 + 9(³√x²y²) - 6(³√xy) 2xy - 9(³√x²y²) + 6(³√xy) + 64 = 0, Let: u = ³√xy, xy = u³, ³√x²y² = u²; u ϵR 2u³ - 9u² + 6u + 64 = 0, (2u³ + 4u²) - (13u² - 6u - 64) = 0 2u²(u + 2) - (13u - 32)(u + 2) = (u + 2)(2u² - 13u + 32) = 0 2u² - 13u + 32 = 0, Δ = 13² - 8(32) = - 87 < 0, u ϵR; Rejected u + 2 = 0, u = - 2 = ³√xy; xy = (- 2)³ = - 8, x + y = 1 - 3(³√xy) = 1 + 6 = 7 y = 7 - x, xy = x(7 - x) = 7x - x² = - 8, x² - 7x - 8 = 0, (x + 1)(x - 8) = 0 x + 1 = 0, x = - 1, y = 7 - x = 8 or x - 8 = 0, x = 8, y = 7 - x = - 1 Answer check: x = - 1, y = 8 or x = 8, y = - 1 ³√x + ³√y = - 1 + 2 = 2 - 1 = 1, x² + y² = 1 + 8² = 8² + 1 = 65; Confirmed Final answer: x = - 1, y = 8 or x = 8, y = - 1

@RajeshKumar-wu7ox 3 месяца назад

(x,y)=(-1,8),(1,-8),(-8,1),(8,-1)

@paulortega5317 2 месяца назад

similar idea let u = x^(1/3), x=u^3 let v = y^(1/3), y=v^3 u + v = 1 u^6 + v^6 = 65 goal: find u*v let u*v=a let f(n) = u^n + v^n f(0) = 2 f(1) = 1 f(n+1) = f(1)f(n) - u*v*f(n-1) = f(n) - a f(2) = 1 - 2*a f(3) = 1 - 3*a f(4) = 1 - 4*a + 2*a^2 f(5) = 1 - 5*a + 5*a^2 f(6) = 1 - 6*a + 9*a^2 - 2*a^3 = 65 Solve for a: 2*a^3 - 9*a^2 + 6*a -1 = 65 (a + 2)*(2*a^2 - 13*a + 32) = 0 only real solution is a = -2 u*v = -2 u + (-2/u) = 1 u^2 - u - 2 = 0 (u - 2)*(u + 1) = 0 (u,v) = (2,-1), (-1,2) (x,y) = (8,-1), (-1,8)