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

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

Evaluate

$$\frac{\sqrt{6}-\sqrt{2}}{4}\approx 0.258819045$$

Solution Steps

Rationalize the denominator of \frac{4-2\sqrt{3}}{2\sqrt{2}\left(\sqrt{3}-1\right)} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

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

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

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

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

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

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

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

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

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

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}.

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

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

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

Calculate 4 to the power of 2 and get 16.

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

The square of \sqrt{3} is 3.

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

Multiply 16 and 3 to get 48.

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

Calculate 4 to the power of 2 and get 16.

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

Subtract 16 from 48 to get 32.

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

Apply the distributive property by multiplying each term of 4\sqrt{2}-2\sqrt{6} by each term of 4\sqrt{3}+4.

$$\frac{16\sqrt{3}\sqrt{2}+16\sqrt{2}-8\sqrt{3}\sqrt{6}-8\sqrt{6}}{32}$$

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

$$\frac{16\sqrt{6}+16\sqrt{2}-8\sqrt{3}\sqrt{6}-8\sqrt{6}}{32}$$

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}.

$$\frac{16\sqrt{6}+16\sqrt{2}-8\sqrt{3}\sqrt{3}\sqrt{2}-8\sqrt{6}}{32}$$

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

$$\frac{16\sqrt{6}+16\sqrt{2}-8\times 3\sqrt{2}-8\sqrt{6}}{32}$$

Multiply -8 and 3 to get -24.

$$\frac{16\sqrt{6}+16\sqrt{2}-24\sqrt{2}-8\sqrt{6}}{32}$$

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

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

Combine 16\sqrt{6} and -8\sqrt{6} to get 8\sqrt{6}.

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

Factor

$$\frac{\sqrt{6} - \sqrt{2}}{4} = 0.2588190451025207$$