Solve (x-1)^2-(x+1)^2<2/3

$$( x - 1 ) ^ { 2 } - ( x + 1 ) ^ { 2 } < \frac { 2 } { 3 }$$

Solve for x

$$x>-\frac{1}{6}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{2}-2x+1-\left(x+1\right)^{2}<\frac{2}{3}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}-2x+1-\left(x^{2}+2x+1\right)<\frac{2}{3}$$

To find the opposite of x^{2}+2x+1, find the opposite of each term.

$$x^{2}-2x+1-x^{2}-2x-1<\frac{2}{3}$$

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

$$-2x+1-2x-1<\frac{2}{3}$$

Combine -2x and -2x to get -4x.

$$-4x+1-1<\frac{2}{3}$$

Subtract 1 from 1 to get 0.

$$-4x<\frac{2}{3}$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x>\frac{\frac{2}{3}}{-4}$$

Express \frac{\frac{2}{3}}{-4} as a single fraction.

$$x>\frac{2}{3\left(-4\right)}$$

Multiply 3 and -4 to get -12.

$$x>\frac{2}{-12}$$

Reduce the fraction \frac{2}{-12} to lowest terms by extracting and canceling out 2.

$$x>-\frac{1}{6}$$