Solve 3/4+{left(3/2-2x/sqrt{3}right)}^2={left(sqrt{3}-xright)}^2+x^2/3

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

Solve for x (complex solution)

$$x\in \mathrm{C}$$

Short Solution Steps

Multiply both sides of the equation by 12, the least common multiple of 4,3.

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

Rationalize the denominator of \frac{2x}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

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

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

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

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

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

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

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

To raise \frac{9-4x\sqrt{3}}{6} to a power, raise both numerator and denominator to the power and then divide.

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

Express 12\times \frac{\left(9-4x\sqrt{3}\right)^{2}}{6^{2}} as a single fraction.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 9 times \frac{6^{2}}{6^{2}}.

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

Since \frac{9\times 6^{2}}{6^{2}} and \frac{12\left(9-4x\sqrt{3}\right)^{2}}{6^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 9\times 6^{2}+12\left(9-4x\sqrt{3}\right)^{2}.

$$\frac{324+972-864x\sqrt{3}+576x^{2}}{6^{2}}=12\left(\sqrt{3}-x\right)^{2}+4x^{2}$$

Combine like terms in 324+972-864x\sqrt{3}+576x^{2}.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{6^{2}}=12\left(\sqrt{3}-x\right)^{2}+4x^{2}$$

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

$$\frac{1296-864x\sqrt{3}+576x^{2}}{6^{2}}=12\left(\left(\sqrt{3}\right)^{2}-2\sqrt{3}x+x^{2}\right)+4x^{2}$$

The square of \sqrt{3} is 3.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{6^{2}}=12\left(3-2\sqrt{3}x+x^{2}\right)+4x^{2}$$

Use the distributive property to multiply 12 by 3-2\sqrt{3}x+x^{2}.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{6^{2}}=36-24\sqrt{3}x+12x^{2}+4x^{2}$$

Combine 12x^{2} and 4x^{2} to get 16x^{2}.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{6^{2}}=36-24\sqrt{3}x+16x^{2}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{36}=36-24\sqrt{3}x+16x^{2}$$

Subtract 36 from both sides.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{36}-36=-24\sqrt{3}x+16x^{2}$$

Add 24\sqrt{3}x to both sides.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{36}-36+24\sqrt{3}x=16x^{2}$$

Subtract 16x^{2} from both sides.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{36}-36+24\sqrt{3}x-16x^{2}=0$$

Multiply both sides of the equation by 36.

$$1296-864x\sqrt{3}+576x^{2}-1296+864\sqrt{3}x-576x^{2}=0$$

Reorder the terms.

$$-576x^{2}+576x^{2}+864\sqrt{3}x-864\sqrt{3}x+1296-1296=0$$

Combine -576x^{2} and 576x^{2} to get 0.

$$864\sqrt{3}x-864\sqrt{3}x+1296-1296=0$$

Combine 864\sqrt{3}x and -864\sqrt{3}x to get 0.

$$1296-1296=0$$

Subtract 1296 from 1296 to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Solve for x

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

Short Solution Steps

Multiply both sides of the equation by 12, the least common multiple of 4,3.

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

Rationalize the denominator of \frac{2x}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

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

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

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

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

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

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

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

To raise \frac{9-4x\sqrt{3}}{6} to a power, raise both numerator and denominator to the power and then divide.

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

Express 12\times \frac{\left(9-4x\sqrt{3}\right)^{2}}{6^{2}} as a single fraction.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 9 times \frac{6^{2}}{6^{2}}.

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

Since \frac{9\times 6^{2}}{6^{2}} and \frac{12\left(9-4x\sqrt{3}\right)^{2}}{6^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 9\times 6^{2}+12\left(9-4x\sqrt{3}\right)^{2}.

$$\frac{324+972-864x\sqrt{3}+576x^{2}}{6^{2}}=12\left(\sqrt{3}-x\right)^{2}+4x^{2}$$

Combine like terms in 324+972-864x\sqrt{3}+576x^{2}.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{6^{2}}=12\left(\sqrt{3}-x\right)^{2}+4x^{2}$$

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

$$\frac{1296-864x\sqrt{3}+576x^{2}}{6^{2}}=12\left(\left(\sqrt{3}\right)^{2}-2\sqrt{3}x+x^{2}\right)+4x^{2}$$

The square of \sqrt{3} is 3.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{6^{2}}=12\left(3-2\sqrt{3}x+x^{2}\right)+4x^{2}$$

Use the distributive property to multiply 12 by 3-2\sqrt{3}x+x^{2}.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{6^{2}}=36-24\sqrt{3}x+12x^{2}+4x^{2}$$

Combine 12x^{2} and 4x^{2} to get 16x^{2}.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{6^{2}}=36-24\sqrt{3}x+16x^{2}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{36}=36-24\sqrt{3}x+16x^{2}$$

Subtract 36 from both sides.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{36}-36=-24\sqrt{3}x+16x^{2}$$

Add 24\sqrt{3}x to both sides.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{36}-36+24\sqrt{3}x=16x^{2}$$

Subtract 16x^{2} from both sides.

$$\frac{1296-864x\sqrt{3}+576x^{2}}{36}-36+24\sqrt{3}x-16x^{2}=0$$

Multiply both sides of the equation by 36.

$$1296-864x\sqrt{3}+576x^{2}-1296+864\sqrt{3}x-576x^{2}=0$$

Reorder the terms.

$$-576x^{2}+576x^{2}+864\sqrt{3}x-864\sqrt{3}x+1296-1296=0$$

Combine -576x^{2} and 576x^{2} to get 0.

$$864\sqrt{3}x-864\sqrt{3}x+1296-1296=0$$

Combine 864\sqrt{3}x and -864\sqrt{3}x to get 0.

$$1296-1296=0$$

Subtract 1296 from 1296 to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

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