Solve left(2x+1right)^2-4*1(x^2+1)>0

$${ \left(2x+1 \right) }^{ 2 } -4 \times 1( { x }^{ 2 } +1) > 0$$

Solve for x

$$x>\frac{3}{4}$$

Solution Steps

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

$$4x^{2}+4x+1-4\times 1\left(x^{2}+1\right)>0$$

Multiply 4 and 1 to get 4.

$$4x^{2}+4x+1-4\left(x^{2}+1\right)>0$$

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

$$4x^{2}+4x+1-4x^{2}-4>0$$

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

$$4x+1-4>0$$

Subtract 4 from 1 to get -3.

$$4x-3>0$$

Add 3 to both sides. Anything plus zero gives itself.

$$4x>3$$

Divide both sides by 4. Since 4 is positive, the inequality direction remains the same.

$$x>\frac{3}{4}$$