Math Olympiad | A Tricky Quartic Equation with Two Algebraic Methods

[@user-kp2rd5qv8g]

Let x-3=t. Then, (t+3)^2 (t-3)^2 + t^2 = 51 > (t^2-9)^2 + t^2 = 51. Let t^2 = a. So, (a-9)^2 + a = 51 > a^2 -17a + 30 = 0 > a= 2,15 > t= +/-√2, +/-√15 > x = 3+/-√2, 3+/-√15.

[@user-kt1dm9jz5t]

X=3+(15)^(1/2), 3-(15)^(1/2), 3+(2)^(1/2), 3-(2)^(1/2).

[@dorkmania]

x²(x - 6)² + (x - 3)² = 51
=> (x² - 6x)² + (x - 3)² = 51
=> (x² - 6x + 9 - 9)² + (x - 3)² = 51
=> ((x - 3)² - 9)² + (x - 3)² = 51
=> ((x - 3)² - 9)² + (x - 3)² - 9 = 42
Taking (x - 3)² - 9 = t and solving for t² + t - 42 = 0

[@tejpalsingh366]

X= + - √15+3; + - √2+3

[@Shobhamaths]

x=3±√15;x=3±√2 👍
let t =x-3;t^4-17t^2+30=0;t^2=15, t^2=2

[@user-ny6jf9is3t]

χ=3+ -(ριζα15), χ=3+ -(ριζα2)
Εβαλα χ-3=y κλπ

[@SidneiMV]

x - 3 = u => x = u + 3
(u + 3)²(u - 3)² + u² = 51
(u² - 9)² + u² - 51 = 0
u⁴ - 18u² + 81 + u² - 51 = 0
u⁴ - 17u² + 30 = 0
(u² - 15)(u² - 2) = 0
u² = 15 => u = ± √15 => *x = 3 ± √15*
u² = 2 => u = ± √2 => *x = 3 ± √2*
(3 ± √2)²(-3 ± √2)² + (±√2)² = (2 - 9)² + 2 =
= (-7)² + 2 = 51
(3 ± √15)²(-3 ± √15)² + (±√15)² = (15 - 9)² + 15 =
= 6² + 15 = 51

[@RealQinnMalloryu4]

(x^2)^2={x^4 ➖ 36x^2}= 36x^2+(x^2 ➖ )= 7 {36x2+7}= 43x^2 20^23x^2 10^10^10^13x^2 2^52^52^56^7x^2 1^1^1^1^1^16^7^1x^2 6^1^1x^2 3^21^1x^2 3^1^1^1x^2 3x^2 (x ➖ 3x+2).