Solve (2x+1)^2-4(x+1)^2leq-2

$$( 2 x + 1 ) ^ { 2 } - 4 ( x + 1 ) ^ { 2 } \leq - 2$$

Solve for x

$$x\geq -\frac{1}{4}$$

Solution Steps

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

$$4x^{2}+4x+1-4\left(x+1\right)^{2}\leq -2$$

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

$$4x^{2}+4x+1-4\left(x^{2}+2x+1\right)\leq -2$$

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

$$4x^{2}+4x+1-4x^{2}-8x-4\leq -2$$

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

$$4x+1-8x-4\leq -2$$

Combine 4x and -8x to get -4x.

$$-4x+1-4\leq -2$$

Subtract 4 from 1 to get -3.

$$-4x-3\leq -2$$

Add 3 to both sides.

$$-4x\leq -2+3$$

Add -2 and 3 to get 1.

$$-4x\leq 1$$

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

$$x\geq -\frac{1}{4}$$