Math Olympiad | How to solve for X in this problem ??

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Learncommunolizer

Learncommunolizer

Місяць тому

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КОМЕНТАРІ: 15
@comdo777
@comdo777 13 днів тому
asnwer=2x
@oahuhawaii2141
@oahuhawaii2141 Місяць тому
The long way to solve this is to expand the left hand side (LHS) using the binomial expansion of (x+y)⁴ that has coefficients of 1:4:6:4:1. The x⁴ term will appear on both sides and cancel out to leave a cubic equation with lots of powers of 4, which are factored out. In the original form, it's easy to see that 2 is a solution, so (x-2) can be factored out from the cubic equation, leaving a quadratic equation, which can be solved by the quadratic formula. Instead of all that work, or factoring the difference of squares, I came up with a shortcut. Since both sides are raised to the 4th power, I took the 4th root, and made sure to put in the 4 roots of 1: (x-4)⁴ = x⁴ = 1⁴ * x⁴ x-4 = ±x, ±i*x x±x = 4; x*(1±i) = 4 2*x = 4; 0 = 4; x = 4/(1±i) x = 2; ∅; x = 2*(1±i) x = 2, 2±2*i
@yuusufliibaan1380
@yuusufliibaan1380 Місяць тому
❤❤❤ thanks 💯💯💯
@learncommunolizer
@learncommunolizer Місяць тому
Thank you very much!!👍👍👍
@user-do9mh4mh6z
@user-do9mh4mh6z Місяць тому
2
@user-uq5ou6ue5d
@user-uq5ou6ue5d Місяць тому
Решается в уме
@salahelmaslouhy2277
@salahelmaslouhy2277 Місяць тому
X=2
@oahuhawaii2141
@oahuhawaii2141 Місяць тому
x = 2, 2±2*i
@antonemelianov1369
@antonemelianov1369 Місяць тому
Should it not be 4 roots in this equation?
@oahuhawaii2141
@oahuhawaii2141 Місяць тому
Nope. The x⁴ term shows up on both sides and cancel out to leave a cubic equation, so 3 roots. There's 1 real root, and 2 complex conjugate roots.
@alexmata7325
@alexmata7325 Місяць тому
Could you explain It on detall please?
@oahuhawaii2141
@oahuhawaii2141 Місяць тому
@alexmata7325: The long way to solve this is to expand the left hand side (LHS) using the binomial expansion of (x+y)⁴ that has coefficients of 1:4:6:4:1. The x⁴ term will appear on both sides and cancel out to leave a cubic equation with lots of powers of 4, which are factored out. In the original form, it's easy to see that 2 is a solution, so (x-2) can be factored out from the cubic equation, leaving a quadratic equation, which can be solved by the quadratic formula. 1*x⁴*(-4)⁰ + 4*x³*(-4)¹ + 6*x²*(-4)² + 4*x¹(-4)³ + 1*x⁰*(-4)⁴ = x⁴ x⁴ - 16*x³ + 96*x² - 256*x + 256 = x⁴ - 16*x³ + 96*x² - 256*x + 256 = 0 x³ - 6*x² + 16*x - 16 = 0 (x - 2)*(x² - 4*x + 8) = 0 x = 2, 2 ± 2*i
@yojaychang
@yojaychang Місяць тому
一看就知道有2
@oahuhawaii2141
@oahuhawaii2141 Місяць тому
x = 2, 2±2*i
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