Solve x={left(frac{sqrt{3}-1}{1+sqrt{3}*1}right)}^2

$$x= { \left( \frac{ \sqrt{ 3 } -1 }{ 1+ \sqrt{ 3 } \times 1 } \right) }^{ 2 }$$

Solve for x

$$x=7-4\sqrt{3}\approx 0.07179677$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{\sqrt{3}-1}{1+1\sqrt{3}} by multiplying numerator and denominator by 1-\sqrt{3}.

$$x=\left(\frac{\left(\sqrt{3}-1\right)\left(1-\sqrt{3}\right)}{\left(1+1\sqrt{3}\right)\left(1-\sqrt{3}\right)}\right)^{2}$$

Consider \left(1+1\sqrt{3}\right)\left(1-\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x=\left(\frac{\left(\sqrt{3}-1\right)\left(1-\sqrt{3}\right)}{1^{2}-\left(1\sqrt{3}\right)^{2}}\right)^{2}$$

Calculate 1 to the power of 2 and get 1.

$$x=\left(\frac{\left(\sqrt{3}-1\right)\left(1-\sqrt{3}\right)}{1-\left(1\sqrt{3}\right)^{2}}\right)^{2}$$

Expand \left(1\sqrt{3}\right)^{2}.

$$x=\left(\frac{\left(\sqrt{3}-1\right)\left(1-\sqrt{3}\right)}{1-1^{2}\left(\sqrt{3}\right)^{2}}\right)^{2}$$

Calculate 1 to the power of 2 and get 1.

$$x=\left(\frac{\left(\sqrt{3}-1\right)\left(1-\sqrt{3}\right)}{1-\left(\sqrt{3}\right)^{2}}\right)^{2}$$

The square of \sqrt{3} is 3.

$$x=\left(\frac{\left(\sqrt{3}-1\right)\left(1-\sqrt{3}\right)}{1-3}\right)^{2}$$

Subtract 3 from 1 to get -2.

$$x=\left(\frac{\left(\sqrt{3}-1\right)\left(1-\sqrt{3}\right)}{-2}\right)^{2}$$

To raise \frac{\left(\sqrt{3}-1\right)\left(1-\sqrt{3}\right)}{-2} to a power, raise both numerator and denominator to the power and then divide.

$$x=\frac{\left(\left(\sqrt{3}-1\right)\left(1-\sqrt{3}\right)\right)^{2}}{\left(-2\right)^{2}}$$

Expand \left(\left(\sqrt{3}-1\right)\left(1-\sqrt{3}\right)\right)^{2}.

$$x=\frac{\left(\sqrt{3}-1\right)^{2}\left(1-\sqrt{3}\right)^{2}}{\left(-2\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

$$x=\frac{\left(\left(\sqrt{3}\right)^{2}-2\sqrt{3}+1\right)\left(1-\sqrt{3}\right)^{2}}{\left(-2\right)^{2}}$$

The square of \sqrt{3} is 3.

$$x=\frac{\left(3-2\sqrt{3}+1\right)\left(1-\sqrt{3}\right)^{2}}{\left(-2\right)^{2}}$$

Add 3 and 1 to get 4.

$$x=\frac{\left(4-2\sqrt{3}\right)\left(1-\sqrt{3}\right)^{2}}{\left(-2\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt{3}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Add 1 and 3 to get 4.

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

Multiply 4-2\sqrt{3} and 4-2\sqrt{3} to get \left(4-2\sqrt{3}\right)^{2}.

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

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

$$x=\frac{16-16\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{\left(-2\right)^{2}}$$

The square of \sqrt{3} is 3.

$$x=\frac{16-16\sqrt{3}+4\times 3}{\left(-2\right)^{2}}$$

Multiply 4 and 3 to get 12.

$$x=\frac{16-16\sqrt{3}+12}{\left(-2\right)^{2}}$$

Add 16 and 12 to get 28.

$$x=\frac{28-16\sqrt{3}}{\left(-2\right)^{2}}$$

Calculate -2 to the power of 2 and get 4.

$$x=\frac{28-16\sqrt{3}}{4}$$

Divide each term of 28-16\sqrt{3} by 4 to get 7-4\sqrt{3}.

$$x=7-4\sqrt{3}$$

Assign x

$$x≔7-4\sqrt{3}$$