Solve (x/2-2/3)^2-(x/2+1/3)^2geqfrac{4}{3}

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

Solve for x

$$x\leq -1$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 3 is 6. Multiply \frac{x}{2} times \frac{3}{3}. Multiply \frac{2}{3} times \frac{2}{2}.

$$\left(\frac{3x}{6}-\frac{2\times 2}{6}\right)^{2}-\left(\frac{x}{2}+\frac{1}{3}\right)^{2}\geq \frac{4}{3}$$

Since \frac{3x}{6} and \frac{2\times 2}{6} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{3x-2\times 2}{6}\right)^{2}-\left(\frac{x}{2}+\frac{1}{3}\right)^{2}\geq \frac{4}{3}$$

Do the multiplications in 3x-2\times 2.

$$\left(\frac{3x-4}{6}\right)^{2}-\left(\frac{x}{2}+\frac{1}{3}\right)^{2}\geq \frac{4}{3}$$

To raise \frac{3x-4}{6} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3x-4\right)^{2}}{6^{2}}-\left(\frac{x}{2}+\frac{1}{3}\right)^{2}\geq \frac{4}{3}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 3 is 6. Multiply \frac{x}{2} times \frac{3}{3}. Multiply \frac{1}{3} times \frac{2}{2}.

$$\frac{\left(3x-4\right)^{2}}{6^{2}}-\left(\frac{3x}{6}+\frac{2}{6}\right)^{2}\geq \frac{4}{3}$$

Since \frac{3x}{6} and \frac{2}{6} have the same denominator, add them by adding their numerators.

$$\frac{\left(3x-4\right)^{2}}{6^{2}}-\left(\frac{3x+2}{6}\right)^{2}\geq \frac{4}{3}$$

To raise \frac{3x+2}{6} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3x-4\right)^{2}}{6^{2}}-\frac{\left(3x+2\right)^{2}}{6^{2}}\geq \frac{4}{3}$$

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

$$\frac{\left(3x-4\right)^{2}}{6^{2}}-\frac{9x^{2}+12x+4}{6^{2}}\geq \frac{4}{3}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{\left(3x-4\right)^{2}}{6^{2}}-\frac{9x^{2}+12x+4}{36}\geq \frac{4}{3}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 6^{2}.

$$\frac{\left(3x-4\right)^{2}}{36}-\frac{9x^{2}+12x+4}{36}\geq \frac{4}{3}$$

Since \frac{\left(3x-4\right)^{2}}{36} and \frac{9x^{2}+12x+4}{36} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(3x-4\right)^{2}-\left(9x^{2}+12x+4\right)}{36}\geq \frac{4}{3}$$

Do the multiplications in \left(3x-4\right)^{2}-\left(9x^{2}+12x+4\right).

$$\frac{9x^{2}-24x+16-9x^{2}-12x-4}{36}\geq \frac{4}{3}$$

Combine like terms in 9x^{2}-24x+16-9x^{2}-12x-4.

$$\frac{-36x+12}{36}\geq \frac{4}{3}$$

Divide each term of -36x+12 by 36 to get -x+\frac{1}{3}.

$$-x+\frac{1}{3}\geq \frac{4}{3}$$

Subtract \frac{1}{3} from both sides.

$$-x\geq \frac{4}{3}-\frac{1}{3}$$

Subtract \frac{1}{3} from \frac{4}{3} to get 1.

$$-x\geq 1$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x\leq -1$$