Solve frac{4+(sqrt{3}-1)^2-6}{2*2*(sqrt{3}+1)}

$$\frac { 4 + ( \sqrt { 3 } - 1 ) ^ { 2 } - 6 } { 2 \times 2 \times ( \sqrt { 3 } + 1 ) }$$

Evaluate

$$\frac{\sqrt{3}}{2}-1\approx -0.133974596$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

Add 4 and 4 to get 8.

$$\frac{8-2\sqrt{3}-6}{2\times 2\left(\sqrt{3}+1\right)}$$

Subtract 6 from 8 to get 2.

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

Multiply 2 and 2 to get 4.

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

Use the distributive property to multiply 4 by \sqrt{3}+1.

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

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

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

Consider \left(4\sqrt{3}+4\right)\left(4\sqrt{3}-4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

Calculate 4 to the power of 2 and get 16.

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

The square of \sqrt{3} is 3.

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

Multiply 16 and 3 to get 48.

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

Calculate 4 to the power of 2 and get 16.

$$\frac{\left(2-2\sqrt{3}\right)\left(4\sqrt{3}-4\right)}{48-16}$$

Subtract 16 from 48 to get 32.

$$\frac{\left(2-2\sqrt{3}\right)\left(4\sqrt{3}-4\right)}{32}$$

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

$$\frac{16\sqrt{3}-8-8\left(\sqrt{3}\right)^{2}}{32}$$

The square of \sqrt{3} is 3.

$$\frac{16\sqrt{3}-8-8\times 3}{32}$$

Multiply -8 and 3 to get -24.

$$\frac{16\sqrt{3}-8-24}{32}$$

Subtract 24 from -8 to get -32.

$$\frac{16\sqrt{3}-32}{32}$$