Harvard University Admission Question || Algebra Exam || 99% Failed Entrance Test

[@LloydCash-he1qv]

x + y = 2 y = 2 - x (1) Substitute into the quintic equation x^5 + y^5 = 152 x^5 + (2 - x)^5 = 152 √ x^5 + 32 - 80x + 80x^2 - 40x^3 + 10x^4 - x^5 = 152 x^4 - 4x^3 + 8x^2 - 8x - 12 = 0 (x^2 - 2x - 2)(x^2 - 2x + 6) = 0 x^2 - 2x - 2 = 0 or x^2 - 2x + 6 = 0 x = 1 ± √3 x = 1 ± i.√5 Resubstituting into (1) gives y = 1 ∓ √3 y = 1 ∓ i.√5 The solutions are x = 1 ± √3 , y = 1 ∓ √3 OR x = 1 ± i.√5 , y = 1 ∓ i.√5

[@Lemda_gtr]

Great job 👏 I’ve enjoyed

[@superacademy247]

Glad you enjoyed. Thanks 👍😊

[@edemafan4938]

Very powefull way if we are in front of a general case. But here as the factorisation is quite possible, it is the faster way.

[@superacademy247]

That is it. 😍🥰

[@Nikioko]

x⁵ + y⁵ = 152 x + y = 2 y = 2 − x x⁵ + (2 − x)⁵ = 152 x⁵ + 32 − 80x + 80x² − 40x³ + 10x⁴ − x⁵ = 152 10x⁴ − 40x³ + 80x² − 80x − 120 = 0 x⁴ − 4x³ + 8x² − 8x − 12 = 0 (x² − 2x + 6) (x² − 2x − 2) = 0 x₁,₂ = 1 ± √(1 − 6) = 1 ± √5 i x₃,₄ = 1 ± √(1 + 2) = 1 ± √3 x₁ = 1 − √5 i ∨ x₂ = 1 + √5 i ∨ x₃ = 1 − √3 ∨ x₄ = 1 + √3 𝕃ₓ = {1−√3, 1+√3, 1−√5i, 1+√5i} x and y are of course interchangeable, due to commutative properties: y₁ = x₂, y₂ = x₁, y₃ = x₄, y₄ = x₃

[@superacademy247]

Thanks for highlighting commutative addition property 🥰

[@walterwen2975]

Harvard University Admission Question: x⁵ + y⁵ = 152, x + y = 2; x, y =? (x + y)² = x² + y² + 2xy = 2² = 4, x² + y² = 2(2 - xy) (x + y)(x² + y²) = x³ + y³ + xy(x + y) = 4(2 - xy), x³ + y³ = 4(2 - xy) - 2xy = 2(4 - 3xy) (x² + y²)(x³ + y³) = x⁵ + y⁵ + (x²y²)(x + y) = 4(2 - xy)(4 - 3xy) 4(2 - xy)(4 - 3xy) - 2x²y² = x⁵ + y⁵ = 152, 2(8 - 10xy + 3x²y²) - x²y² = 76 5x²y² - 20xy + 16 - 76 = 5(x²y² - 4xy - 12) = 0, x²y² - 4xy - 12 = (xy + 2)(xy - 6) = 0 xy + 2 = 0; xy = - 2 or xy - 6 = 0; xy = 6 xy = - 2: (x - y)² = (x + y)² - 4xy = 2² - 4(- 2) = 12 = (2√3)², x - y = ± 2√3; x + y = 2 2x = 2 ± 2√3, x = 1 ± √3; y = 2 - x = 2 - (1 ± √3) = 1 -/+ √3 xy = 6: (x - y)² = (x + y)² - 4xy = 2² - 4(6) = - 20 = (2i√5)², x - y = ± 2i√5; x + y = 2 2x = 2 ± 2i√5, x = 1 ± i√5; y = 2 - x = 2 - (1 ± i√5) = 1 -/+ i√5 Answer check: x = 1 ± √3, y = 1 -/+ √3: (1 ± √3)⁵ = 1 ± 5√3 + 10(√3)² ± 10(√3)³ + 5(√3)⁴ ± (√3)⁵ x⁵ + y⁵ = (1 ± √3)⁵ + (1 -/+ √3)⁵ = 2(1 + 30 + 45) = 152; Confirmed x + y = (1 ± √3) + (1 -/+ √3) = 2; Confirmed x = 1 ± i√5, y = 1 -/+ i√5: (1 ± i√3)⁵ = 1 ± i√5 + 10(i√5)² ± 10(i√5)³ + 5(i√5)⁴ ± (i√5)⁵ x⁵ + y⁵ = (1 ± i√5)⁵ + (1 -/+ i√5)⁵ = 2(1 - 50 + 125) = 152; Confirmed x + y = (1 ± i√5) + (1 -/+ i√5) = 2; Confirmed Final answer: x = 1 + √3, y = 1 - √3; x = 1 - √3, y = 1 + √3; Two complex value roots, x = 1 + i√5, y = 1 - i√5 or x = 1 - i√5, y = 1 + i√5

[@superacademy247]

Thanks for the complete solution generation

[@smusic677]

Mepinna pakayata re wenakota kulu harakek wage kulappu wenawa.mul kale edalama ahemai.hodata danna deyak.

[@paulortega5317]

let f(n) = x^n + y^n f(0) = 2 f(1) = 2 since x + y = 2 f(n+2) = f(1)*f(n+1) - xy*f(n) let xy = a f(n+2) = 2*f(n+1) - a*f(n-2) f(0) = 2 f1) = 2 f(2) = 4 - 2a f(3) = 8 - 6a f4) = 16 - 16a + 2a^2 f(5) = 32 - 40a + 10a^2 = 152 10a^2 - 40a + 120 = 0 a^2 - 4a + 12 = 0 (a - 6)*(a + 2) = 0 a = 6 or -2 ❶xy = 6 x + y = 2 x + 6/x = 2 x^2 - 2x + 6 = 0 x = 1 ± sqrt(5) ❷xy = -2 x + y = 2 x - 2/x = 2 x^2 - 2x - 2 = 0 x = 1 ± sqrt(3)

[@RajaBabu-zh2fl]

22min 😂

[@key_board_x]

(x + y)² = x² + y² + 2xy → given: x + y = 2 4 = x² + y² + 2xy (x + y)³ = (x + y)².(x + y) (x + y)³ = (x² + 2xy + y²).(x + y) (x + y)³ = x³ + x²y + 2x²y + 2xy² + xy² + y³ (x + y)³ = x³ + y³ + 3x²y + 3xy² x³ + y³ = (x + y)³ - 3x²y - 3xy² x³ + y³ = (x + y)³ - 3xy.(x + y) → given: x + y = 2 x³ + y³ = 8 - 6xy (x + y)⁵ = (x + y)².(x + y)².(x + y) (x + y)⁵ = (x² + 2xy + y²).(x² + 2xy + y²).(x + y) (x + y)⁵ = (x⁴ + 2x³y + x²y² + 2x³y + 4x²y² + 2xy³ + x²y² + 2xy³ + y⁴).(x + y) (x + y)⁵ = (x⁴ + 6x²y² + 4x³y + 4xy³ + y⁴).(x + y) (x + y)⁵ = x⁵ + x⁴y + 6x³y² + 6x²y³ + 4x⁴y + 4x³y² + 4x²y³ + 4xy⁴ + xy⁴ + y⁵ (x + y)⁵ = x⁵ + y⁵ + 5x⁴y + 5xy⁴ + 10x³y² + 10x²y³ x⁵ + y⁵ = (x + y)⁵ - 5x⁴y - 5xy⁴ - 10x³y² - 10x²y³ x⁵ + y⁵ = (x + y)⁵ - 5xy.(x³ + y³) - 10x²y².(x + y) → given: x + y = 2 x⁵ + y⁵ = 32 - 5xy.(x³ + y³) - 20x²y² → recall: x³ + y³ = 8 - 6xy x⁵ + y⁵ = 32 - 5xy.(8 - 6xy) - 20x²y² x⁵ + y⁵ = 32 - 40xy + 30x²y² - 20x²y² x⁵ + y⁵ = 32 - 40xy + 10x²y² → given: x⁵ + y⁵ = 152 152 = 32 - 40xy + 10x²y² 10x²y² - 40xy - 120 = 0 x²y² - 4xy - 12 = 0 → let: z = xy z² - 4z - 12 = 0 Δ = (- 4)² - (4 * - 12) = 16 + 48 = 64 = 8² z = (4 ± 8)/2 z = 2 ± 4 First case: z = 6 → recall: z = xy → xy = 6 Second case: z = - 2 → recall: z = xy → xy = - 2 Resume: x + y = 4 ← this is the sum S xy = 6 or xy = - 2 ← this is the product P So x & y are the solution of the following equation: a² - Sa + P = 0 First case: xy = 6 and (x + y) = 2 a² - 2a + 6 = 0 Δ = (- 2)² - (4 * 6) = 4 - 24 = - 20 = 20i² a = (2 ± i√20)/2 a = (2 ± 2i√5)/2 a = 1 ± i√5 → x = 1 + i√5 Recall: y = 2 - x y = 2 - 1 - i√5 → y = 1 - i√5 → x = 1 - i√5 Recall: y = 2 - x y = 2 - 1 + i√5 → y = 1 + i√5 Second case: xy = - 2 and (x + y) = 2 a² - 2a - 2 = 0 Δ = (- 2)² - (4 * - 2) = 4 + 8 = 12 a = (2 ± √12)/2 a = (2 ± 2√3)/2 a = 1 ± √3 → x = 1 + √3 Recall: y = 2 - x y = 2 - 1 - √3 → y = 1 - √3 → x = 1 - √3 Recall: y = 2 - x y = 2 - 1 + √3 → y = 1 + √3

[@superacademy247]

Great 👌 solution. Thanks 😊

[@boguslawszostak1784]

Easy Peasy. x=u+v y=u-v x+y=2u=2 u=1 x^5+y^5=(1+v)^5+(1-v)^5=152 10 v^4 + 20 v^2 - 150 = 0 v^4 + 2v^2 - 15 = 0 (v^2+1)-16=0 (v^2+1-4)(v^2+1=4)=0 v^2=3 or v^2=-5 we have four solutions v=sqrt(3), u=1 , x=1+sqrt(3) v=-sqrt(3), u=1 , x=1-sqrt(3) v=i*sqrt(5), u=1 , x=1+i*sqrt(5) v=-i*sqrt(5), u=1 , x=1-i*sqrt(5)

[@andresfernandez985]

then try your hand at one of the problems of the clay mathematics institute. see if you can call any easy peasy too 🤠

[@boguslawszostak1784]

​@@andresfernandez985 What is your problem my friend? I simply pointed out that my favorite substitution turns a difficult problem into one that is easy to solve without having to guess what trick to apply.

[@annammathomas9368]

Too long.

[@asimkumerdas3497]

ONE STEP IS WRONG IN CASE. 1