What are the solutions to the equation (x+1)^2+4=1 ?

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

Find the zeros of (x+1)^2+4=1

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

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(x + 1)^{2} + 4 = 1

x = 3 i - 1, x = - 3 i - 1

Solution steps

(x + 1)^{2} + 4 = 1

Move

4

to the right side

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

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

x + 1 = - 3 : x = 3 i - 1

Solve

x + 1 = - - 3 : x = - 3 i - 1

The solutions to the quadratic equation are:

x = 3 i - 1, x = - 3 i - 1

x, (x + y)^{2} + (x - y)^{2} = 0

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

6 x^{2} - 13 x - 15 = 0

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

- x^{2} + 9 = 9

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