Solve (1-t)^2-t^2>3

$$( 1 - t ) ^ { 2 } - t ^ { 2 } > 3$$

Solve for t

$$t<-1$$

Solution Steps

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

$$1-2t+t^{2}-t^{2}>3$$

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

$$1-2t>3$$

Subtract 1 from both sides.

$$-2t>3-1$$

Subtract 1 from 3 to get 2.

$$-2t>2$$

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

$$t<\frac{2}{-2}$$

Divide 2 by -2 to get -1.

$$t<-1$$