What are the solutions to the equation (x+2)^4+x^4=82 ?

The solutions to the equation (x+2)^4+x^4=82 are x=1,x=-3,x=-1+sqrt(10)i,x=-1-sqrt(10)i

Find the zeros of (x+2)^4+x^4=82

The zeros of (x+2)^4+x^4=82 are x=1,x=-3,x=-1+sqrt(10)i,x=-1-sqrt(10)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x + 2)^{4} + x^{4} = 82

x = 1, x = - 3, x = - 1 + 10 i, x = - 1 - 10 i

Solution steps

(x + 2)^{4} + x^{4} = 82

Expand

(x + 2)^{4} + x^{4} : 2 x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16

2 x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16 = 82

Move

82

to the left side

2 x^{4} + 8 x^{3} + 24 x^{2} + 32 x - 66 = 0

Factor

2 x^{4} + 8 x^{3} + 24 x^{2} + 32 x - 66 : 2 (x - 1) (x + 3) (x^{2} + 2 x + 11)

2 (x - 1) (x + 3) (x^{2} + 2 x + 11) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x + 3 = 0 or x^{2} + 2 x + 11 = 0

Solve

x - 1 = 0 : x = 1

Solve

x + 3 = 0 : x = - 3

Solve

x^{2} + 2 x + 11 = 0 : x = - 1 + 10 i, x = - 1 - 10 i

The solutions are

x = 1, x = - 3, x = - 1 + 10 i, x = - 1 - 10 i

f^{4} (0.5 f - 0.25) = 6 + f

(x + 4) (x + 2) (x - 1) = 0

3 x^{3} - 12 x^{2} + 9 x - 4 = 0

x, x^{3} + y^{3} = a^{3}

- x^{3} - 1 = 0

What are the solutions to the equation (x+2)^4+x^4=82 ?
The solutions to the equation (x+2)^4+x^4=82 are x=1,x=-3,x=-1+sqrt(10)i,x=-1-sqrt(10)i
Find the zeros of (x+2)^4+x^4=82
The zeros of (x+2)^4+x^4=82 are x=1,x=-3,x=-1+sqrt(10)i,x=-1-sqrt(10)i