Solve 4(a+1)^2-4a^2+20geq0

$$4 ( a + 1 ) ^ { 2 } - 4 a ^ { 2 } + 20 \geq 0$$

Solve for a

$$a\geq -3$$

Solution Steps

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

$$4\left(a^{2}+2a+1\right)-4a^{2}+20\geq 0$$

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

$$4a^{2}+8a+4-4a^{2}+20\geq 0$$

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

$$8a+4+20\geq 0$$

Add 4 and 20 to get 24.

$$8a+24\geq 0$$

Subtract 24 from both sides. Anything subtracted from zero gives its negation.

$$8a\geq -24$$

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

$$a\geq \frac{-24}{8}$$

Divide -24 by 8 to get -3.

$$a\geq -3$$