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

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

Evaluate

$$\frac{\sqrt{2}\left(\sqrt{3}+1\right)}{4}\approx 0.965925826$$

Solution Steps

Add 4 and 2 to get 6.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

To find the opposite of 4-2\sqrt{3}, find the opposite of each term.

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

Subtract 4 from 6 to get 2.

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

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

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

$$\frac{\left(2+2\sqrt{3}\right)\sqrt{2}}{8}$$

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

$$\frac{2\sqrt{2}+2\sqrt{3}\sqrt{2}}{8}$$

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

$$\frac{2\sqrt{2}+2\sqrt{6}}{8}$$