Question


Answer

This equation is a special form of quartic equation. Noticed 82 is the sum of 81 and 1. 81 is the fourth power of 3 and 1 is the fourth power of 1.

(x+2)^4+x^4 = 82 = 81+1

(x+2)^4-1 = 9^2-x^4

[(x+2)^2+1][(x+2)^2-1]=(9-x^2)(9+x^2)

(x^2+4x+5)(x+3)(x+1) = (3+x)(3-x)(9+x^2)
(1)

Since both sides contain a common factor x+3, we get one of the solutions of the function

x = -3

Factor out the common factor, (1) is simplified to the equation

(x^2+4x+5)(x+1) = (3-x)(9+x^2)

x^3+4x^2+5x+x^2+4x+5 = 27+3x^2-9x-x^3

Combine the like terms and simplify, we get

x^3+x^2+9x-11=0
(2)

Noticed if we substitute 1 to x, the left hand side is

1^3+1^2+9\times 1-11=0

Therefore x=1 is another solution of the equation

Let's factor the equation (2) considering the factor of x-1

x^3-1+x^2-1+9x-9=0

(x-1)(x^2+x+1+x+1+9)=0

(x-1)(x^2+2x+11)=0

It's found that the equation has one real number and 2 imaginary numbers

x=1

or

x = \dfrac{-2\pm i\sqrt{40} }{2} =-1\pm i\sqrt{10}

In summary, the equation has 4 solutions - 2 real numbers and 2 imaginary numbers



Steven Zheng posted 1 hour ago

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