Solve (2a+2)^2-4(a^2-1)<0

$$( 2 a + 2 ) ^ { 2 } - 4 ( a ^ { 2 } - 1 ) < 0$$

Solve for a

$$a<-1$$

Solution Steps

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

$$4a^{2}+8a+4-4\left(a^{2}-1\right)<0$$

Use the distributive property to multiply -4 by a^{2}-1.

$$4a^{2}+8a+4-4a^{2}+4<0$$

Combine 4a^{2} and -4a^{2} to get 0.

$$8a+4+4<0$$

Add 4 and 4 to get 8.

$$8a+8<0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$8a<-8$$

Divide both sides by 8. Since 8 is positive, the inequality direction remains the same.

$$a<\frac{-8}{8}$$

Divide -8 by 8 to get -1.

$$a<-1$$