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

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

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

The zeros of (2x+1)^2=1+(x+1)(x-1) are x=-1/3 ,x=-1

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

x = - \frac{1}{3}, x = - 1

Decimal Notation

x = - 0.33333 \dots, x = - 1

Solution steps

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

Expand

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

Expand

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

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

Move

x^{2}

to the left side

3 x^{2} + 4 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

4^{2} - 4 \cdot 3 \cdot 1 = 2

x_{1, 2} = \frac{- 4 \pm 2}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 4 + 2}{2 \cdot 3}, x_{2} = \frac{- 4 - 2}{2 \cdot 3}

x = \frac{- 4 + 2}{2 \cdot 3} : - \frac{1}{3}

x = \frac{- 4 - 2}{2 \cdot 3} : - 1

The solutions to the quadratic equation are:

x = - \frac{1}{3}, x = - 1

x^{2} - 2 x + 6

4 x^{2} - 13 = 9 x

(x + 5)^{2} - 3 (x + 5) - 10 = 0

w^{2} = 12 w

8 x^{2} + 20 x = 48

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