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

The solutions to the equation 2(x^2-1)=x are x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4

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

The zeros of 2(x^2-1)=x are x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4

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

x = \frac{1 + 17}{4}, x = \frac{1 - 17}{4}

Decimal Notation

x = 1.28077 \dots, x = - 0.78077 \dots

Solution steps

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

Expand

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

2 x^{2} - 2 = x

Move

x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 2 (- 2) = 17

x_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 17}{2 \cdot 2}, x_{2} = \frac{- ( - 1 ) - 17}{2 \cdot 2}

x = \frac{- ( - 1 ) + 17}{2 \cdot 2} : \frac{1 + 17}{4}

x = \frac{- ( - 1 ) - 17}{2 \cdot 2} : \frac{1 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 17}{4}, x = \frac{1 - 17}{4}

w^{2} + 13 w + 19 = 0

(- x - 3)^{2} = 8

4 m^{2} + 17 m + 12 = 0

(m - 2) x^{2} + 2 x = 8 x^{2} + 26 - m

- 64 + k^{2} = 0

What are the solutions to the equation 2(x^2-1)=x ?
The solutions to the equation 2(x^2-1)=x are x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4
Find the zeros of 2(x^2-1)=x
The zeros of 2(x^2-1)=x are x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4