What are the solutions to the equation (x-1)(x+1)=2(x^2-3) ?

The solutions to the equation (x-1)(x+1)=2(x^2-3) are x=sqrt(5),x=-sqrt(5)

Find the zeros of (x-1)(x+1)=2(x^2-3)

The zeros of (x-1)(x+1)=2(x^2-3) are x=sqrt(5),x=-sqrt(5)

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(x - 1) (x + 1) = 2 (x^{2} - 3)

x = 5, x = - 5

Solution steps

(x - 1) (x + 1) = 2 (x^{2} - 3)

Expand

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

Expand

2 (x^{2} - 3) : 2 x^{2} - 6

x^{2} - 1 = 2 x^{2} - 6

Move

1

to the right side

x^{2} = 2 x^{2} - 5

Move

2 x^{2}

to the left side

- x^{2} = - 5

Divide both sides by

- 1

x^{2} = 5

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = 5, x = - 5

(\frac{8}{125})^{- \frac{4}{3}}

2 x^{2} - 32

9 x - 1 < \frac{3}{4} (16 x - 2)

y^{4} + 10 y^{2} + 24

3 x - 3 = - 3 (1 - x)

What are the solutions to the equation (x-1)(x+1)=2(x^2-3) ?
The solutions to the equation (x-1)(x+1)=2(x^2-3) are x=sqrt(5),x=-sqrt(5)
Find the zeros of (x-1)(x+1)=2(x^2-3)
The zeros of (x-1)(x+1)=2(x^2-3) are x=sqrt(5),x=-sqrt(5)