What are the solutions to the equation (x-1)^4+(x-5)^4=82 ?

The solutions to the equation (x-1)^4+(x-5)^4=82 are x=2,x=4,x=3+5i,x=3-5i

Find the zeros of (x-1)^4+(x-5)^4=82

The zeros of (x-1)^4+(x-5)^4=82 are x=2,x=4,x=3+5i,x=3-5i

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(x - 1)^{4} + (x - 5)^{4} = 82

x = 2, x = 4, x = 3 + 5 i, x = 3 - 5 i

Solution steps

(x - 1)^{4} + (x - 5)^{4} = 82

Expand

(x - 1)^{4} + (x - 5)^{4} : 2 x^{4} - 24 x^{3} + 156 x^{2} - 504 x + 626

2 x^{4} - 24 x^{3} + 156 x^{2} - 504 x + 626 = 82

Move

82

to the left side

2 x^{4} - 24 x^{3} + 156 x^{2} - 504 x + 544 = 0

Factor

2 x^{4} - 24 x^{3} + 156 x^{2} - 504 x + 544 : 2 (x - 2) (x - 4) (x^{2} - 6 x + 34)

2 (x - 2) (x - 4) (x^{2} - 6 x + 34) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or x - 4 = 0 or x^{2} - 6 x + 34 = 0

Solve

x - 2 = 0 : x = 2

Solve

x - 4 = 0 : x = 4

Solve

x^{2} - 6 x + 34 = 0 : x = 3 + 5 i, x = 3 - 5 i

The solutions are

x = 2, x = 4, x = 3 + 5 i, x = 3 - 5 i

d^{4} + 4 d^{3} - 5 d^{2} - 36 d - 36 = 0

x^{4} + i = 0

x^{4} - 4 x^{3} + x^{2} + 6 x + 2 = 0

x^{5} - 5 x + 3 = 0

(z^{2} + 25)^{2} = 0

What are the solutions to the equation (x-1)^4+(x-5)^4=82 ?
The solutions to the equation (x-1)^4+(x-5)^4=82 are x=2,x=4,x=3+5i,x=3-5i
Find the zeros of (x-1)^4+(x-5)^4=82
The zeros of (x-1)^4+(x-5)^4=82 are x=2,x=4,x=3+5i,x=3-5i