Solve (4/3+2x)^2>1/3-(2x+1)(2x+1/3)

$$( \frac { 4 } { 3 } + 2 x ) ^ { 2 } > \frac { 1 } { 3 } - ( 2 x + 1 ) ( 2 x + \frac { 1 } { 3 } )$$

Solve for x

$$x\in \left(-\infty,-\frac{2}{3}\right)\cup \left(-\frac{1}{3},\infty\right)$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{4}{3}+2x\right)^{2}.

$$\frac{16}{9}+\frac{16}{3}x+4x^{2}>\frac{1}{3}-\left(2x+1\right)\left(2x+\frac{1}{3}\right)$$

Use the distributive property to multiply 2x+1 by 2x+\frac{1}{3} and combine like terms.

$$\frac{16}{9}+\frac{16}{3}x+4x^{2}>\frac{1}{3}-\left(4x^{2}+\frac{8}{3}x+\frac{1}{3}\right)$$

To find the opposite of 4x^{2}+\frac{8}{3}x+\frac{1}{3}, find the opposite of each term.

$$\frac{16}{9}+\frac{16}{3}x+4x^{2}>\frac{1}{3}-4x^{2}-\frac{8}{3}x-\frac{1}{3}$$

Subtract \frac{1}{3} from \frac{1}{3} to get 0.

$$\frac{16}{9}+\frac{16}{3}x+4x^{2}>-4x^{2}-\frac{8}{3}x$$

Add 4x^{2} to both sides.

$$\frac{16}{9}+\frac{16}{3}x+4x^{2}+4x^{2}>-\frac{8}{3}x$$

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$\frac{16}{9}+\frac{16}{3}x+8x^{2}>-\frac{8}{3}x$$

Add \frac{8}{3}x to both sides.

$$\frac{16}{9}+\frac{16}{3}x+8x^{2}+\frac{8}{3}x>0$$

Combine \frac{16}{3}x and \frac{8}{3}x to get 8x.

$$\frac{16}{9}+8x+8x^{2}>0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$\frac{16}{9}+8x+8x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 8 for a, 8 for b, and \frac{16}{9} for c in the quadratic formula.

$$x=\frac{-8±\sqrt{8^{2}-4\times 8\times \frac{16}{9}}}{2\times 8}$$

Do the calculations.

$$x=\frac{-8±\frac{8}{3}}{16}$$

Solve the equation x=\frac{-8±\frac{8}{3}}{16} when ± is plus and when ± is minus.

$$x=-\frac{1}{3} x=-\frac{2}{3}$$

Rewrite the inequality by using the obtained solutions.

$$8\left(x+\frac{1}{3}\right)\left(x+\frac{2}{3}\right)>0$$

For the product to be positive, x+\frac{1}{3} and x+\frac{2}{3} have to be both negative or both positive. Consider the case when x+\frac{1}{3} and x+\frac{2}{3} are both negative.

$$x+\frac{1}{3}<0 x+\frac{2}{3}<0$$

The solution satisfying both inequalities is x<-\frac{2}{3}.

$$x<-\frac{2}{3}$$

Consider the case when x+\frac{1}{3} and x+\frac{2}{3} are both positive.

$$x+\frac{2}{3}>0 x+\frac{1}{3}>0$$

The solution satisfying both inequalities is x>-\frac{1}{3}.

$$x>-\frac{1}{3}$$

The final solution is the union of the obtained solutions.

$$x<-\frac{2}{3}\text{; }x>-\frac{1}{3}$$