Solve frac{(1-sqrt{3})^2}{(2+sqrt{3})^2}

$$\frac { ( 1 - \sqrt { 3 } ) ^ { 2 } } { ( 2 + \sqrt { 3 } ) ^ { 2 } }$$

Evaluate

$$52-30\sqrt{3}\approx 0.038475773$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 1 and 3 to get 4.

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

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

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

The square of \sqrt{3} is 3.

$$\frac{4-2\sqrt{3}}{4+4\sqrt{3}+3}$$

Add 4 and 3 to get 7.

$$\frac{4-2\sqrt{3}}{7+4\sqrt{3}}$$

Rationalize the denominator of \frac{4-2\sqrt{3}}{7+4\sqrt{3}} by multiplying numerator and denominator by 7-4\sqrt{3}.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}$$

Consider \left(7+4\sqrt{3}\right)\left(7-4\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}.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{7^{2}-\left(4\sqrt{3}\right)^{2}}$$

Calculate 7 to the power of 2 and get 49.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-\left(4\sqrt{3}\right)^{2}}$$

Expand \left(4\sqrt{3}\right)^{2}.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-4^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-16\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-16\times 3}$$

Multiply 16 and 3 to get 48.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-48}$$

Subtract 48 from 49 to get 1.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{1}$$

Anything divided by one gives itself.

$$\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)$$

Use the distributive property to multiply 4-2\sqrt{3} by 7-4\sqrt{3} and combine like terms.

$$28-30\sqrt{3}+8\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$28-30\sqrt{3}+8\times 3$$

Multiply 8 and 3 to get 24.

$$28-30\sqrt{3}+24$$

Add 28 and 24 to get 52.

$$52-30\sqrt{3}$$

Expand

$$52-30\sqrt{3}$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 1 and 3 to get 4.

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

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

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

The square of \sqrt{3} is 3.

$$\frac{4-2\sqrt{3}}{4+4\sqrt{3}+3}$$

Add 4 and 3 to get 7.

$$\frac{4-2\sqrt{3}}{7+4\sqrt{3}}$$

Rationalize the denominator of \frac{4-2\sqrt{3}}{7+4\sqrt{3}} by multiplying numerator and denominator by 7-4\sqrt{3}.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}$$

Consider \left(7+4\sqrt{3}\right)\left(7-4\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}.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{7^{2}-\left(4\sqrt{3}\right)^{2}}$$

Calculate 7 to the power of 2 and get 49.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-\left(4\sqrt{3}\right)^{2}}$$

Expand \left(4\sqrt{3}\right)^{2}.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-4^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-16\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-16\times 3}$$

Multiply 16 and 3 to get 48.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-48}$$

Subtract 48 from 49 to get 1.

$$\frac{\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{1}$$

Anything divided by one gives itself.

$$\left(4-2\sqrt{3}\right)\left(7-4\sqrt{3}\right)$$

Use the distributive property to multiply 4-2\sqrt{3} by 7-4\sqrt{3} and combine like terms.

$$28-30\sqrt{3}+8\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$28-30\sqrt{3}+8\times 3$$

Multiply 8 and 3 to get 24.

$$28-30\sqrt{3}+24$$

Add 28 and 24 to get 52.

$$52-30\sqrt{3}$$