Solve (x-1)(x+1)-2x>x^2+5

$$( x - 1 ) ( x + 1 ) - 2 x > x ^ { 2 } + 5$$

Solve for x

$$x<-3$$

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.

$$x^{2}-1-2x>x^{2}+5$$

Subtract x^{2} from both sides.

$$x^{2}-1-2x-x^{2}>5$$

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

$$-1-2x>5$$

Add 1 to both sides.

$$-2x>5+1$$

Add 5 and 1 to get 6.

$$-2x>6$$

Divide both sides by -2. Since -2 is negative, the inequality direction is changed.

$$x<\frac{6}{-2}$$

Divide 6 by -2 to get -3.

$$x<-3$$