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

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

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

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

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

x = 3, x = 1

Solution steps

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

Expand

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

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

Move

1

to the left side

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

Move

x^{2}

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 3, x = 1

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

\frac{7}{10} - \frac{1}{3} + 2

x^{2} + 12 x - 3 = 0

12 3 - 7108

\frac{3 x}{5} - \frac{x - 3}{2} = \frac{x + 2}{3}

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