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

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

Evaluate

$$\frac{\sqrt{2}}{2}\approx 0.707106781$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

Add 2 and 4 to get 6.

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

Subtract 4 from 6 to get 2.

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

Rationalize the denominator of \frac{2+2\sqrt{3}}{2\sqrt{2}\left(\sqrt{3}+1\right)} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

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

Use the distributive property to multiply 2+2\sqrt{3} by \sqrt{2}.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

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

$$\frac{2\sqrt{2}+2\sqrt{6}}{4\sqrt{3}+4}$$

Rationalize the denominator of \frac{2\sqrt{2}+2\sqrt{6}}{4\sqrt{3}+4} by multiplying numerator and denominator by 4\sqrt{3}-4.

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

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}.

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

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

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

Calculate 4 to the power of 2 and get 16.

$$\frac{\left(2\sqrt{2}+2\sqrt{6}\right)\left(4\sqrt{3}-4\right)}{16\left(\sqrt{3}\right)^{2}-4^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(2\sqrt{2}+2\sqrt{6}\right)\left(4\sqrt{3}-4\right)}{16\times 3-4^{2}}$$

Multiply 16 and 3 to get 48.

$$\frac{\left(2\sqrt{2}+2\sqrt{6}\right)\left(4\sqrt{3}-4\right)}{48-4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{\left(2\sqrt{2}+2\sqrt{6}\right)\left(4\sqrt{3}-4\right)}{48-16}$$

Subtract 16 from 48 to get 32.

$$\frac{\left(2\sqrt{2}+2\sqrt{6}\right)\left(4\sqrt{3}-4\right)}{32}$$

Use the distributive property to multiply 2\sqrt{2}+2\sqrt{6} by 4\sqrt{3}-4.

$$\frac{8\sqrt{3}\sqrt{2}-8\sqrt{2}+8\sqrt{3}\sqrt{6}-8\sqrt{6}}{32}$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$\frac{8\sqrt{6}-8\sqrt{2}+8\sqrt{3}\sqrt{6}-8\sqrt{6}}{32}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$\frac{8\sqrt{6}-8\sqrt{2}+8\sqrt{3}\sqrt{3}\sqrt{2}-8\sqrt{6}}{32}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{8\sqrt{6}-8\sqrt{2}+8\times 3\sqrt{2}-8\sqrt{6}}{32}$$

Multiply 8 and 3 to get 24.

$$\frac{8\sqrt{6}-8\sqrt{2}+24\sqrt{2}-8\sqrt{6}}{32}$$

Combine -8\sqrt{2} and 24\sqrt{2} to get 16\sqrt{2}.

$$\frac{8\sqrt{6}+16\sqrt{2}-8\sqrt{6}}{32}$$

Combine 8\sqrt{6} and -8\sqrt{6} to get 0.

$$\frac{16\sqrt{2}}{32}$$

Divide 16\sqrt{2} by 32 to get \frac{1}{2}\sqrt{2}.

$$\frac{1}{2}\sqrt{2}$$