Solve m^2+9<(1-m)^2

$$m ^ { 2 } + 9 < ( 1 - m ) ^ { 2 }$$

Solve for m

$$m<-4$$

Solution Steps

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

$$m^{2}+9<1-2m+m^{2}$$

Add 2m to both sides.

$$m^{2}+9+2m<1+m^{2}$$

Subtract m^{2} from both sides.

$$m^{2}+9+2m-m^{2}<1$$

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

$$9+2m<1$$

Subtract 9 from both sides.

$$2m<1-9$$

Subtract 9 from 1 to get -8.

$$2m<-8$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$m<\frac{-8}{2}$$

Divide -8 by 2 to get -4.

$$m<-4$$