A tricky Oxford University Entrance Interview | Exponent Aptitude Test | 99% Failed Admission Exam

[@ez4214]

let x=m+2.5, y=2.5-m then (m+2.5)^4 + (2.5-m)^4=97 then 2m^4+75m^2-18.875=0 then m^2=1/4 or -151/4 then get m then get x, y

[@PeterChapman-rg6gr]

X = 2 or 3 and Y = 3 or 2. in both cases X + Y = 5 and X to the power of 4 + Y to the power of 4 = 97. 30 seconds to solve in my head.

[@TheNizzer]

Trivial. Basic problem when introducing exponents. +/-2, +/-3. No need for any algebra. Absolutely not an interview question for anything. Solution: 16+81=97.

[@georgesbv1]

negative don't work. Getting complex solutions need work.

[@stevenjohnson1143]

Simple x can equal positive or negative 3,2 same thing for y

[@saschavogt5083]

Since x+y=5 it cant be negative 2 or 3

[@YAWTon]

No, x and y cannot be negative. And there are two complex solutions.

[@prollysine]

we get , y^4 - 10y^3 + 75y^2 - 250y + 264 = 0 , (y-2)(y^3-8y^2+59y-132)=0 , y=2 ,
*(-2) , +1 -2 *(-3) , y^3 - 8y^2 + 59y - 132 = 0 , (y-3)(y^2-5y+44)=0 , y=3 ,
-8 +16 +1 -3 y^2-5y+44=0 , y=(5+/-V(25-176))/2 ,
+59 -118 -5 +15 y=(5+/-V(-151))/2 , y=(5+/-i*V151)/2 ,
-132 +264 = 0 , +44 - 132 = 0 , y= 2 , 3 , (5+i*V151)/2 , (5-i*V151)/2 ,
x=5-y , sulu , ( x , y ) = ( 3 , 2) , ( 2 , 3 ) , ( 5 - (5+i*V151)/2 , (5+i*V151)/2 ) , ( 5 - (5-i*V151)/2 , (5-i*V151)/2 ) ,

[@key_board_x]

(x + y)² = x² + y² + 2xy → given: x + y = 5
25 = x² + y² + 2xy
x² + y² = 25 - 2xy
(x + y)⁴ = [(x + y)²]²
(x + y)⁴ = [x² + y² + 2xy]²
(x + y)⁴ = x⁴ + x²y² + 2x³y + x²y² + y⁴ + 2xy³ + 2x³y + 2xy³ + 4x²y²
(x + y)⁴ = x⁴ + y⁴ + 6x²y² + 4x³y + 4xy³
(x + y)⁴ = x⁴ + y⁴ + 6x²y² + 4xy.(x² + y²) → recall: x² + y² = 25 - 2xy
(x + y)⁴ = x⁴ + y⁴ + 6x²y² + 4xy.(25 - 2xy)
(x + y)⁴ = x⁴ + y⁴ + 6x²y² + 100xy - 8x²y²
(x + y)⁴ = x⁴ + y⁴ - 2x²y² + 100xy
2x²y² - 100xy = x⁴ + y⁴ - (x + y)⁴ → given: x + y = 5
2x²y² - 100xy = x⁴ + y⁴ - 625 → given: x⁴ + y⁴ = 97
2x²y² - 100xy = 97 - 625
2x²y² - 100xy = - 528
2x²y² - 100xy + 528 = 0
x²y² - 50xy + 264 = 0 → let: a = xy
a² - 50a + 264 = 0
Δ = (- 50)² - (4 * 264) = 2500 - 1056 = 1444 = 38²
a = (50 ± 38)/2
a = 44 or a = 6 → recall: a = xy
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First case: a = 44 → recall: a = xy
xy = 44 ← this is the product P
x + y = 5 ← this is the sum S
x & y are the solution of the equation:
z² - Sz + P = 0
z² - 5z + 44 = 0
Δ = (- 5)² - (4 * 44) = 25 - 176 = - 151 = 151i²
z = (5 ± i√151)/2
First solution: z = (5 + i√151)/2
x = (5 + i√151)/2 → given: x + y = 5 → y = 5 - x
y = 5 - [(5 + i√151)/2]
y = (10 - 5 - i√151)/2
y = (5 - i√151)/2
Second solution: z = (5 - i√151)/2
x = (5 - i√151)/2 → given: x + y = 5 → y = 5 - x
y = 5 - [(5 - i√151)/2]
y = (10 - 5 + i√151)/2
y = (5 + i√151)/2
------------------------------------------------------------------------------
Second case: a = 6 → recall: a = xy
xy = 6 ← this is the product P
x + y = 5 ← this is the sum S
x & y are the solution of the equation:
z² - Sz + P = 0
z² - 5z + 6 = 0
Δ = (- 5)² - (4 * 6) = 25 - 24 = 1
z = (5 ± 1)/2
z = 3 or z = 2
Third solution: z = 3
x = 3 → given: x + y = 5 → y = 5 - x
y = 2
Fourth solution: z = 2
x = 2 → given: x + y = 5 → y = 5 - x
y = 3
------------------------------------------------------------------------------
Resume: (x ; y)
[(5 + i√151)/2 ; (5 - i√151)/2]
[(5 - i√151)/2 ; (5 + i√151)/2]
(3 ; 2)
(2 ; 3)