A Nice System of Equations | Math Olympiad Algebra

A Nice System of Equations | Math Olympiad Algebra
Welcome to infyGyan! In this video, we'll explore a captivating system of equations that's perfect for sharpening your problem-solving skills. Join us as we break down each step of the solution, providing clear explanations and helpful tips along the way. Whether you're gearing up for a Math Olympiad or just enjoy tackling challenging algebra problems, this video is for you. Can you solve it? Give it a try and let us know in the comments!
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Topics covered:
System of equations
Algebra
Math Olympiad
Math Olympiad Training
Algebraic identities
Algebraic manipulations
Solving systems of equations
Substitution
Problem Solving
Math tutorial
Math Olympiad Preparation
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Thanks for watching!

[@user-kp2rd5qv8g]

Let √x=a, √y=b and ab=t. The given equations are: a+b=3, a^4+b^4 = 17/2ab. These, upon manipulations, yield 4t^2-89t+162=0 > t =81/4, 2. The first does not yield real x,y. If t=2, a+b=3 and ab=2 > a=2,1 and b = 1,2 > (x,y) = (4,1), (1,4).

[@user-ny6jf9is3t]

(χ,y)=(1,4) ή (4,1)

[@tejpalsingh366]

(X;Y)=(4;1); (1;4)

[@paulortega5317]

A suggestion. At 2:21 you are assigning ab = y. But y is already in use. That can be confusing. Better to use z or other.

[@infyGyan]

Mistake is all mine.
Thanks for the suggestion.🙏

[@RealQinnMalloryu4]

(4)+(4)/(1)+(6)+(3)/(1) =17/2 = 8.1 2^3.1^1 1^2.1^1 2.1 (y ➖ 2x+1). (1)+(2)= 3(y ➖ 2x+1) .

[@walterwen2975]

Math Olympiad Algebra: (x√x)/√y + y√y)/√x = 17/2, √x + √y = 3, x, y ϵR⁺
(√x + √y)² = 3², x + y + 2√(xy) = 9; √(xy) = [9 - (x + y)]/2
(x√x)/√y + y√y)/√x = [(x√x)√x + (y√y)√y)/√(xy) = (x² + y²)/√(xy) = 17/2
[2(x² + y²)]/[9 - (x + y)] = 17/2, 4(x² + y²) = 17[9 - (x + y)] = 153 - 17(x + y)
4(x² + y²) = 4(x + y)² - 8xy = 4(x + y)² - 8[9 - (x + y)]²/4
= 4(x + y)² - 2[9 - (x + y)]² = 4(x + y)² - 2[81 - 18(x + y) + (x + y)²]
= 2(x + y)² + 36(x + y) - 162 = 153 - 17(x + y)
2(x + y)² + 53(x + y) - 315 = 0, [(x + y) - 5][2(x + y) + 63] = 0, x, y ϵR⁺
2(x + y) + 63 > 0; (x + y) - 5 = 0, x + y = 5
√(xy) = [9 - (x + y)]/2 = (9 - 5)/2 = 2, xy = 4; y = 5 - x, x(5 - x) = 4
x² - 5x + 4 = 0, (x - 1)(x - 4) = 0; x - 1 = 0 or x - 4 = 0
x = 1, y = 5 - x = 5 - 1 = 4 or x = 4, y = 5 - 4 = 1
Answer check:
(x√x)/√y + y√y)/√x = (x² + y²)/√(xy) = 17/2, √x + √y = 3
x = 1, y = 4 or x = 4, y = 1
(x² + y²)/√(xy) = (1 + 4²)/√4 = (4² + 1)/√4 = 17/2; Confirmed
√x + √y = √1 + √4 = √4 + √1 = 3; Confirmed
Final answer:
x = 1, y = 4 or x = 4, y = 1

[@SidneiMV]

√x = a => x = a²
√y = b => y = b²
a³/b + b³/a = 17/2
a + b = 3
(a⁴ + b⁴)/ab = 17/2
a⁴ + b⁴ = 17ab/2
(a + b)⁴ = 81
a⁴ + b⁴ + 4ab(a² + b²) + 6a²b² = 81
17ab/2 + 4ab[(a + b)² - 2ab] + 6(ab)² = 81
17ab/2 + 4ab(9 - 2ab) + 6(ab)² = 81
17ab/2 + 36ab - 8(ab)² + 6(ab)² = 81
2(ab)² - 89ab/2 + 81 = 0
4(ab)² - 89ab + 162 = 0
ab = (89 ± 73)/8
ab = 81/4 or ab = 2
a + b = 3
ab = 81/4
a + b = 3
ab = 2
t² - 3t + 2 = 0
(t - 1)(t - 2) = 0
a = 1 => b = 2 => *(x, y) = (1, 4)*
a = 2 => b = 1 => *(x, y) = (4, 1)*
a + b = 3
ab = 81/4
t² - 3t + 81/4 = 0 => no real solutions