Solve x^2+(sqrt{3}-frac{sqrt{3}}{3}-x)^2-(2-frac{2sqrt{3}}{3})^2

$$x ^ { 2 } + ( \sqrt { 3 } - \frac { \sqrt { 3 } } { 3 } - x ) ^ { 2 } - ( 2 - \frac { 2 \sqrt { 3 } } { 3 } ) ^ { 2 }$$

Evaluate

$$2x^{2}-\frac{4\sqrt{3}x}{3}+\frac{8\sqrt{3}}{3}-4$$

Short Solution Steps

Combine \sqrt{3} and -\frac{\sqrt{3}}{3} to get \frac{2}{3}\sqrt{3}.

$$x^{2}+\left(\frac{2}{3}\sqrt{3}-x\right)^{2}-\left(2-\frac{2\sqrt{3}}{3}\right)^{2}$$

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

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

The square of \sqrt{3} is 3.

$$x^{2}+\frac{4}{9}\times 3-\frac{4}{3}\sqrt{3}x+x^{2}-\left(2-\frac{2\sqrt{3}}{3}\right)^{2}$$

Multiply \frac{4}{9} and 3 to get \frac{4}{3}.

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

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

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{3}{3}.

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

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

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

Do the multiplications in 2\times 3-2\sqrt{3}.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-2\sqrt{3}\right)^{2}.

$$2x^{2}+\frac{4}{3}-\frac{4}{3}\sqrt{3}x-\frac{36-24\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{3^{2}}$$

The square of \sqrt{3} is 3.

$$2x^{2}+\frac{4}{3}-\frac{4}{3}\sqrt{3}x-\frac{36-24\sqrt{3}+4\times 3}{3^{2}}$$

Multiply 4 and 3 to get 12.

$$2x^{2}+\frac{4}{3}-\frac{4}{3}\sqrt{3}x-\frac{36-24\sqrt{3}+12}{3^{2}}$$

Add 36 and 12 to get 48.

$$2x^{2}+\frac{4}{3}-\frac{4}{3}\sqrt{3}x-\frac{48-24\sqrt{3}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$2x^{2}+\frac{4}{3}-\frac{4}{3}\sqrt{3}x-\frac{48-24\sqrt{3}}{9}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 9 is 9. Multiply \frac{4}{3} times \frac{3}{3}.

$$2x^{2}+\frac{4\times 3}{9}-\frac{4}{3}\sqrt{3}x-\frac{48-24\sqrt{3}}{9}$$

Since \frac{4\times 3}{9} and \frac{48-24\sqrt{3}}{9} have the same denominator, subtract them by subtracting their numerators.

$$2x^{2}+\frac{4\times 3-\left(48-24\sqrt{3}\right)}{9}-\frac{4}{3}\sqrt{3}x$$

Do the multiplications in 4\times 3-\left(48-24\sqrt{3}\right).

$$2x^{2}+\frac{12-48+24\sqrt{3}}{9}-\frac{4}{3}\sqrt{3}x$$

Do the calculations in 12-48+24\sqrt{3}.

$$2x^{2}+\frac{-36+24\sqrt{3}}{9}-\frac{4}{3}\sqrt{3}x$$

Expand

$$2x^{2}-\frac{4\sqrt{3}x}{3}+\frac{8\sqrt{3}}{3}-4$$

Short Solution Steps

Combine \sqrt{3} and -\frac{\sqrt{3}}{3} to get \frac{2}{3}\sqrt{3}.

$$x^{2}+\left(\frac{2}{3}\sqrt{3}-x\right)^{2}-\left(2-\frac{2\sqrt{3}}{3}\right)^{2}$$

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

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

The square of \sqrt{3} is 3.

$$x^{2}+\frac{4}{9}\times 3-\frac{4}{3}\sqrt{3}x+x^{2}-\left(2-\frac{2\sqrt{3}}{3}\right)^{2}$$

Multiply \frac{4}{9} and 3 to get \frac{4}{3}.

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

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

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{3}{3}.

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

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

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

Do the multiplications in 2\times 3-2\sqrt{3}.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-2\sqrt{3}\right)^{2}.

$$2x^{2}+\frac{4}{3}-\frac{4}{3}\sqrt{3}x-\frac{36-24\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{3^{2}}$$

The square of \sqrt{3} is 3.

$$2x^{2}+\frac{4}{3}-\frac{4}{3}\sqrt{3}x-\frac{36-24\sqrt{3}+4\times 3}{3^{2}}$$

Multiply 4 and 3 to get 12.

$$2x^{2}+\frac{4}{3}-\frac{4}{3}\sqrt{3}x-\frac{36-24\sqrt{3}+12}{3^{2}}$$

Add 36 and 12 to get 48.

$$2x^{2}+\frac{4}{3}-\frac{4}{3}\sqrt{3}x-\frac{48-24\sqrt{3}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$2x^{2}+\frac{4}{3}-\frac{4}{3}\sqrt{3}x-\frac{48-24\sqrt{3}}{9}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 9 is 9. Multiply \frac{4}{3} times \frac{3}{3}.

$$2x^{2}+\frac{4\times 3}{9}-\frac{4}{3}\sqrt{3}x-\frac{48-24\sqrt{3}}{9}$$

Since \frac{4\times 3}{9} and \frac{48-24\sqrt{3}}{9} have the same denominator, subtract them by subtracting their numerators.

$$2x^{2}+\frac{4\times 3-\left(48-24\sqrt{3}\right)}{9}-\frac{4}{3}\sqrt{3}x$$

Do the multiplications in 4\times 3-\left(48-24\sqrt{3}\right).

$$2x^{2}+\frac{12-48+24\sqrt{3}}{9}-\frac{4}{3}\sqrt{3}x$$

Do the calculations in 12-48+24\sqrt{3}.

$$2x^{2}+\frac{-36+24\sqrt{3}}{9}-\frac{4}{3}\sqrt{3}x$$