Solving an Amazing System of Equations

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Solving an Amazing System of Equations | Math Olympiad Prep!

Рет қаралды 1,960

Ай бұрын

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 ...
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• Israeli Math Olympiad ...
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Пікірлер: 19

@tejpalsingh366 Ай бұрын

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 Ай бұрын

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)

@mohammedsaysrashid3587 Ай бұрын

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

@johnstanley5692 Ай бұрын

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 )

@user-ee7nw2rx9s Ай бұрын

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

@tejpalsingh366 Ай бұрын

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

@jaggisaram4914 Ай бұрын

❌ = 8 , -1. 🌱 = -1 , 8

@kassuskassus6263 Ай бұрын

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

@tunneloflight Ай бұрын

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

@ramunasstulga8264 Ай бұрын

1/5 marks for correct answer

@tunneloflight Ай бұрын

@@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 Ай бұрын

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

@tunneloflight Ай бұрын

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

@walterwen2975 Ай бұрын

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

@SidneiMV Ай бұрын

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) }*

@user-ny6jf9is3t Ай бұрын

(χ,y) =(8,-1) , (-1,8)

@paulortega5317 22 күн бұрын

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)