What are the solutions to the equation (x+2)^4=x^4+16 ?

The solutions to the equation (x+2)^4=x^4+16 are x=0,x=-3/2+i(sqrt(7))/2 ,x=-3/2-i(sqrt(7))/2

Find the zeros of (x+2)^4=x^4+16

The zeros of (x+2)^4=x^4+16 are x=0,x=-3/2+i(sqrt(7))/2 ,x=-3/2-i(sqrt(7))/2

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} + 16

x = 0, x = - \frac{3}{2} + i \frac{7}{2}, x = - \frac{3}{2} - i \frac{7}{2}

Solution steps

(x + 2)^{4} = x^{4} + 16

Expand

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

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

Move

16

to the right side

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

Move

x^{4}

to the left side

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

Factor

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

8 x (x^{2} + 3 x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x^{2} + 3 x + 4 = 0

Solve

x^{2} + 3 x + 4 = 0 : x = - \frac{3}{2} + i \frac{7}{2}, x = - \frac{3}{2} - i \frac{7}{2}

The solutions are

x = 0, x = - \frac{3}{2} + i \frac{7}{2}, x = - \frac{3}{2} - i \frac{7}{2}

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

x^{3} - 5 x^{2} + 80 - 16 x = 0

x^{3} - 7 x^{2} + 14 x - 8 = 0

x^{3} - 2 x^{2} - 15 x = 0

x^{3} + 9 x^{2} + 23 x + 15 = 0

What are the solutions to the equation (x+2)^4=x^4+16 ?
The solutions to the equation (x+2)^4=x^4+16 are x=0,x=-3/2+i(sqrt(7))/2 ,x=-3/2-i(sqrt(7))/2
Find the zeros of (x+2)^4=x^4+16
The zeros of (x+2)^4=x^4+16 are x=0,x=-3/2+i(sqrt(7))/2 ,x=-3/2-i(sqrt(7))/2