The Easy Way to Solve This Quartic Challenge | An Algebra Problem

[@bowsi9]

Master of substitutions

[@9허공]

Substitution t = x/2 is more easy to calculate.
(t - 2)^2 + (t - 1)^3 + t^4 - 2 = t^4 + t^3 - 2t^2 - t + 1 = (t^4 - 2t^2 + 1) + (t^3 - t)
= (t^2 -1)^2 + t(t^2 - 1) = (t^2 - 1)(t^2 + t - 1) = (t - 1)(t + 1)(t^2 + t - 1) = 0
=> t = { 1, -1, ( -1±√5 )/2 } => x = { 2, -2, -1±√5 }

[@Shobhamaths]

If u simplify the given expression
x^4+2x^3-8x^2-8x+16=0
x=2
x^3+4x^2-8=0
x=-2
x^2+2x-4=0
x=-1±√5👍

[@dorkmania]

Alternatively, for x/2 = t
(t - 2)² + (t - 1)³ + t⁴ = 2
=> (t - 2)² - 1 + (t - 1)³ + t⁴ - 1 = 2 - 1 - 1
=> (t - 3)(t - 1) + (t - 1)³ + (t² + 1)(t + 1)(t - 1) = 0
Factoring out (t - 1) or (x - 2)
t - 3 + (t - 1)² + (t² + 1)(t + 1) = 0
=> t + (t - 1)² - 4 + 1 + (t² + 1)(t + 1) = 0
=> (t + 1) + (t + 1)(t - 3) + (t² + 1)(t + 1) = 0
Factoring out (t + 1) or (x + 2)
1 + t - 3 + t² + 1 = 0
=> t² + t - 1 = 0
=> 2t = - 1 ± √5
Or x = ±2, - 1 ± √5

[@RyanLewis-Johnson-wq6xs]

(X-4)^2/8+(X-2)^3/16+(X^4)/32 X=2,-2,-1+Sqrt[5],-1-Sqrt[5]

[@CHIRANJITTTTTT]

you could use √5 instead of Sqrt[5]
🕉️🔱🇮🇳🚩🙏🚩🇮🇳🔱🕉️

[@Quest3669]

x= + -2; + - √5-1

[@RajeshKumar-wu7ox]

X=2 and x^3 +4x^2 -8 =0

[@gregevgeni1864]

After some algebra
x^4 +2x^3-8x^2 -8x +16 =0
or (x^2 +16/x^2) +2(x-4/x) -8 =0
Put x -4/x =t =>
t^2 +2t -10=0 etc ..

[@RealQinnMalloryu4]

(x^2 ➖ 16]/(8)^2+(x^3 ➖ 8)/(16)^3= {x^0+x^0 ➖}/64+{x^0+x^0+}/656 ={x^2/64+x^2/656}={x^4/720+x^4/32}=x^8/752 x^2^3/10^704^13 x^1^1^1/10^7^10^4^13^1 /2^5^7^1^2^5^2^2^1^1 1^1^1^1^1^1^1^2 1^2 (x ➖ 2x+1).

[@SidneiMV]

[(x - 4)/2]² + [(x - 2)/2]³ + (x/2)⁴ = 2
(x/2 - 2)² + (x/2 - 1)³ + (x/2)⁴ = 2
x/2 - 1 = u => x = 2(u + 1)
(u - 1)² + u³ + (u + 1)⁴ = 2
u⁴ + 4u³ + 6u² + 4u + 1 + u³ + u² - 2u + 1 - 2 = 0
u⁴ + 5u³ + 7u² + 2u = 0
u(u³ + 5u² + 7u + 2) = 0
u = 0 => *x = 2*
u³ + 5u² + 7u + 2 = 0
u³ + 8 + 5u² + 7u - 6 = 0
(u + 2)(u² - 2u + 4) + (u + 2)(5u - 3) = 0
(u + 2)(u² + 3u + 1) = 0
u = -2 => *x = -2*
u² + 3u + 1 = 0
u = (-3 ± √5)/2 => *x = -1 ± √5*

[@SidneiMV]

ANOTHER WAY
4(x - 4)² + 2(x - 2)³ + x⁴ - 32 = 0
x - 2 = u => x = u + 2
4(u - 2)² + 2u³ + (u + 2)⁴ - 32 = 0
u⁴ + 8u³ + 24u² + 32u + 16 + 2u³ + 4u² - 16u + 16 - 32 = 0
u⁴ + 10u³ + 28u² + 16u = 0
u(u³ + 10u² + 28u + 16) = 0
u = 0 => *x = 2*
u³ + 10u² + 28u + 16 = 0
(u³ + 64) + (10u² + 28u - 48) = 0
(u + 4)(u² - 4u + 16) + 2(u + 4)(5u - 6) = 0
(u + 4)(u² + 6u + 4) = 0
u = -4 => *x = -2*
u² + 6u + 4 = 0
u = (-6 ± 2√5)/2
u = -3 ± √5 => x = *-1 ± √5*