What are the solutions to the equation (x+1)^2-4(x+1)+2=0 ?

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

Find the zeros of (x+1)^2-4(x+1)+2=0

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

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(x + 1)^{2} - 4 (x + 1) + 2 = 0

x = 1 + 2, x = 1 - 2

Decimal Notation

x = 2.41421 \dots, x = - 0.41421 \dots

Solution steps

(x + 1)^{2} - 4 (x + 1) + 2 = 0

Expand

(x + 1)^{2} - 4 (x + 1) + 2 : x^{2} - 2 x - 1

x^{2} - 2 x - 1 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 1) = 22

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + 2, x = 1 - 2

\frac{4 x - 8}{x ^{2} - 2 x}

3 (5 n - 1) - 14 n + 9 = 10 (n - 4) - 6 n - 4 (n + 1)

2 x (x - 5)^{- 3} - 4 x^{2} (x - 5)^{- 4}

8 n - (2 n - 3) = 12

x (x - 1) (x + 4) > 0

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