Math Olympiad | How to solve for X in this problem ??
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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
asnwer=2x
Решается в уме
2
X=2
x = 2, 2±2*i
Should it not be 4 roots in this equation?
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.
Could you explain It on detall please?
@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
一看就知道有2
x = 2, 2±2*i