What are the solutions to the equation (x+2)^4-81=0 ?

The solutions to the equation (x+2)^4-81=0 are x=1,x=-5

Find the zeros of (x+2)^4-81=0

The zeros of (x+2)^4-81=0 are x=1,x=-5

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(x + 2)^{4} - 81 = 0

x = 1, x = - 5

Solution steps

(x + 2)^{4} - 81 = 0

Move

81

to the right side

(x + 2)^{4} = 81

For

(g (x))^{n} = f (a)

, n is even, the solutions are

Solve

The solutions are

x = 1, x = - 5

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

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

5 y^{3} + 2 y^{2} - 45 y - 18 = 0

(5 x - 2)^{3} + 10 = 18

10 x^{4} - 17 x^{2} + 3 = 0

What are the solutions to the equation (x+2)^4-81=0 ?
The solutions to the equation (x+2)^4-81=0 are x=1,x=-5
Find the zeros of (x+2)^4-81=0
The zeros of (x+2)^4-81=0 are x=1,x=-5