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

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

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

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

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

x = 0, x = - 1

Solution steps

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

Expand

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

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

Move

1

to the left side

x^{2} + 2 x = x

Move

x

to the left side

x^{2} + x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot 0 = 1

x_{1, 2} = \frac{- 1 \pm 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 1}{2 \cdot 1}, x_{2} = \frac{- 1 - 1}{2 \cdot 1}

x = \frac{- 1 + 1}{2 \cdot 1} : 0

x = \frac{- 1 - 1}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 0, x = - 1

0.2 x^{2} - 1.5 x + 2.9 = 0

x^{2} - 10 x + 19 = 3

x, I n (- x) = I n (x^{2} - 6)

(0.3 - x) (1.1 - x) - (- 0.3) (0.4) = 0

- 6 x^{2} + 48 x = 0

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