Solve (1/3)^2+1/3(2x+1)+x(x+1)+(1/2)^2geq0

$$( \frac { 1 } { 3 } ) ^ { 2 } + \frac { 1 } { 3 } ( 2 x + 1 ) + x ( x + 1 ) + ( \frac { 1 } { 2 } ) ^ { 2 } \geq 0$$

Solve for x

$$x\in \mathrm{R}$$

Steps Using the Quadratic Formula

Calculate \frac{1}{3} to the power of 2 and get \frac{1}{9}.

$$\frac{1}{9}+\frac{1}{3}\left(2x+1\right)+x\left(x+1\right)+\left(\frac{1}{2}\right)^{2}\geq 0$$

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

$$\frac{1}{9}+\frac{1}{3}\times 2x+\frac{1}{3}+x\left(x+1\right)+\left(\frac{1}{2}\right)^{2}\geq 0$$

Multiply \frac{1}{3} and 2 to get \frac{2}{3}.

$$\frac{1}{9}+\frac{2}{3}x+\frac{1}{3}+x\left(x+1\right)+\left(\frac{1}{2}\right)^{2}\geq 0$$

Least common multiple of 9 and 3 is 9. Convert \frac{1}{9} and \frac{1}{3} to fractions with denominator 9.

$$\frac{1}{9}+\frac{2}{3}x+\frac{3}{9}+x\left(x+1\right)+\left(\frac{1}{2}\right)^{2}\geq 0$$

Since \frac{1}{9} and \frac{3}{9} have the same denominator, add them by adding their numerators.

$$\frac{1+3}{9}+\frac{2}{3}x+x\left(x+1\right)+\left(\frac{1}{2}\right)^{2}\geq 0$$

Add 1 and 3 to get 4.

$$\frac{4}{9}+\frac{2}{3}x+x\left(x+1\right)+\left(\frac{1}{2}\right)^{2}\geq 0$$

Use the distributive property to multiply x by x+1.

$$\frac{4}{9}+\frac{2}{3}x+x^{2}+x+\left(\frac{1}{2}\right)^{2}\geq 0$$

Combine \frac{2}{3}x and x to get \frac{5}{3}x.

$$\frac{4}{9}+\frac{5}{3}x+x^{2}+\left(\frac{1}{2}\right)^{2}\geq 0$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{4}{9}+\frac{5}{3}x+x^{2}+\frac{1}{4}\geq 0$$

Least common multiple of 9 and 4 is 36. Convert \frac{4}{9} and \frac{1}{4} to fractions with denominator 36.

$$\frac{16}{36}+\frac{5}{3}x+x^{2}+\frac{9}{36}\geq 0$$

Since \frac{16}{36} and \frac{9}{36} have the same denominator, add them by adding their numerators.

$$\frac{16+9}{36}+\frac{5}{3}x+x^{2}\geq 0$$

Add 16 and 9 to get 25.

$$\frac{25}{36}+\frac{5}{3}x+x^{2}\geq 0$$

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.

$$\frac{25}{36}+\frac{5}{3}x+x^{2}=0$$

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 1 for a, \frac{5}{3} for b, and \frac{25}{36} for c in the quadratic formula.

$$x=\frac{-\frac{5}{3}±\sqrt{\left(\frac{5}{3}\right)^{2}-4\times 1\times \frac{25}{36}}}{2}$$

Do the calculations.

$$x=\frac{-\frac{5}{3}±0}{2}$$

Solutions are the same.

$$x=-\frac{5}{6}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x+\frac{5}{6}\right)^{2}\geq 0$$

Inequality holds for x\in \mathrm{R}.

$$x\in \mathrm{R}$$