What are the solutions to the equation (x-3)^4+(x-5)^4=2 ?

The solutions to the equation (x-3)^4+(x-5)^4=2 are x=4,x=4+sqrt(6)i,x=4-sqrt(6)i

Find the zeros of (x-3)^4+(x-5)^4=2

The zeros of (x-3)^4+(x-5)^4=2 are x=4,x=4+sqrt(6)i,x=4-sqrt(6)i

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(x - 3)^{4} + (x - 5)^{4} = 2

x = 4, x = 4 + 6 i, x = 4 - 6 i

Solution steps

(x - 3)^{4} + (x - 5)^{4} = 2

Expand

(x - 3)^{4} + (x - 5)^{4} : 2 x^{4} - 32 x^{3} + 204 x^{2} - 608 x + 706

2 x^{4} - 32 x^{3} + 204 x^{2} - 608 x + 706 = 2

Move

2

to the left side

2 x^{4} - 32 x^{3} + 204 x^{2} - 608 x + 704 = 0

Factor

2 x^{4} - 32 x^{3} + 204 x^{2} - 608 x + 704 : 2 (x - 4)^{2} (x^{2} - 8 x + 22)

2 (x - 4)^{2} (x^{2} - 8 x + 22) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 4 = 0 or x^{2} - 8 x + 22 = 0

Solve

x - 4 = 0 : x = 4

Solve

x^{2} - 8 x + 22 = 0 : x = 4 + 6 i, x = 4 - 6 i

The solutions are

x = 4, x = 4 + 6 i, x = 4 - 6 i

z^{4} = 128 + 1283 i

x^{6} - 16 x^{2} = 0

2 x^{6} + 7 x^{3} = 15

x + 1 = x^{3} - 2 x - 5

8 x^{4} + 10 x^{2} - 3 = 0

What are the solutions to the equation (x-3)^4+(x-5)^4=2 ?
The solutions to the equation (x-3)^4+(x-5)^4=2 are x=4,x=4+sqrt(6)i,x=4-sqrt(6)i
Find the zeros of (x-3)^4+(x-5)^4=2
The zeros of (x-3)^4+(x-5)^4=2 are x=4,x=4+sqrt(6)i,x=4-sqrt(6)i