What are the solutions to the equation (x+1)^2=2(x-3) ?

The solutions to the equation (x+1)^2=2(x-3) are x=sqrt(7)i,x=-sqrt(7)i

Find the zeros of (x+1)^2=2(x-3)

The zeros of (x+1)^2=2(x-3) are x=sqrt(7)i,x=-sqrt(7)i

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

x = 7 i, x = - 7 i

Solution steps

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

Expand

(x + 1)^{2} : x^{2} + 2 x + 1

Expand

2 (x - 3) : 2 x - 6

x^{2} + 2 x + 1 = 2 x - 6

Move

1

to the right side

x^{2} + 2 x = 2 x - 7

Move

2 x

to the left side

x^{2} = - 7

For

x^{2} = f (a)

the solutions are

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

x = - 7, x = - - 7

- 7 = 7 i

- - 7 = - 7 i

x = 7 i, x = - 7 i

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

x^{2} - 4 x + 3 x - 2 = 28

6 a^{2} + 7 a - 3 = 0

12 x^{2} + 40 x - 32 = 0

2 x^{2} = x + 4

What are the solutions to the equation (x+1)^2=2(x-3) ?
The solutions to the equation (x+1)^2=2(x-3) are x=sqrt(7)i,x=-sqrt(7)i
Find the zeros of (x+1)^2=2(x-3)
The zeros of (x+1)^2=2(x-3) are x=sqrt(7)i,x=-sqrt(7)i