Solve x^2-1=(x+1)(x-1)

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

Solve for x (complex solution)

$$x\in \mathrm{C}$$

Solution Steps

Consider \left(x+1\right)\left(x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Subtract x^{2} from both sides.

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

Combine x^{2} and -x^{2} to get 0.

$$-1=-1$$

Compare -1 and -1.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

Consider \left(x+1\right)\left(x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Subtract x^{2} from both sides.

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

Combine x^{2} and -x^{2} to get 0.

$$-1=-1$$

Compare -1 and -1.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$