Solve (sqrt{2}-2)^2+sqrt{12/3}divsqrt{5/24}

$$( \sqrt { 2 } - 2 ) ^ { 2 } + \sqrt { 1 \frac { 2 } { 3 } } \div \sqrt { \frac { 5 } { 24 } }$$

Evaluate

$$6-2\sqrt{2}\approx 3.171572875$$

Solution Steps

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

$$\left(\sqrt{2}\right)^{2}-4\sqrt{2}+4+\frac{\sqrt{\frac{1\times 3+2}{3}}}{\sqrt{\frac{5}{24}}}$$

The square of \sqrt{2} is 2.

$$2-4\sqrt{2}+4+\frac{\sqrt{\frac{1\times 3+2}{3}}}{\sqrt{\frac{5}{24}}}$$

Add 2 and 4 to get 6.

$$6-4\sqrt{2}+\frac{\sqrt{\frac{1\times 3+2}{3}}}{\sqrt{\frac{5}{24}}}$$

Multiply 1 and 3 to get 3.

$$6-4\sqrt{2}+\frac{\sqrt{\frac{3+2}{3}}}{\sqrt{\frac{5}{24}}}$$

Add 3 and 2 to get 5.

$$6-4\sqrt{2}+\frac{\sqrt{\frac{5}{3}}}{\sqrt{\frac{5}{24}}}$$

Rewrite the square root of the division \sqrt{\frac{5}{3}} as the division of square roots \frac{\sqrt{5}}{\sqrt{3}}.

$$6-4\sqrt{2}+\frac{\frac{\sqrt{5}}{\sqrt{3}}}{\sqrt{\frac{5}{24}}}$$

Rationalize the denominator of \frac{\sqrt{5}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$6-4\sqrt{2}+\frac{\frac{\sqrt{5}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}}{\sqrt{\frac{5}{24}}}$$

The square of \sqrt{3} is 3.

$$6-4\sqrt{2}+\frac{\frac{\sqrt{5}\sqrt{3}}{3}}{\sqrt{\frac{5}{24}}}$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$6-4\sqrt{2}+\frac{\frac{\sqrt{15}}{3}}{\sqrt{\frac{5}{24}}}$$

Rewrite the square root of the division \sqrt{\frac{5}{24}} as the division of square roots \frac{\sqrt{5}}{\sqrt{24}}.

$$6-4\sqrt{2}+\frac{\frac{\sqrt{15}}{3}}{\frac{\sqrt{5}}{\sqrt{24}}}$$

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$6-4\sqrt{2}+\frac{\frac{\sqrt{15}}{3}}{\frac{\sqrt{5}}{2\sqrt{6}}}$$

Rationalize the denominator of \frac{\sqrt{5}}{2\sqrt{6}} by multiplying numerator and denominator by \sqrt{6}.

$$6-4\sqrt{2}+\frac{\frac{\sqrt{15}}{3}}{\frac{\sqrt{5}\sqrt{6}}{2\left(\sqrt{6}\right)^{2}}}$$

The square of \sqrt{6} is 6.

$$6-4\sqrt{2}+\frac{\frac{\sqrt{15}}{3}}{\frac{\sqrt{5}\sqrt{6}}{2\times 6}}$$

To multiply \sqrt{5} and \sqrt{6}, multiply the numbers under the square root.

$$6-4\sqrt{2}+\frac{\frac{\sqrt{15}}{3}}{\frac{\sqrt{30}}{2\times 6}}$$

Multiply 2 and 6 to get 12.

$$6-4\sqrt{2}+\frac{\frac{\sqrt{15}}{3}}{\frac{\sqrt{30}}{12}}$$

Divide \frac{\sqrt{15}}{3} by \frac{\sqrt{30}}{12} by multiplying \frac{\sqrt{15}}{3} by the reciprocal of \frac{\sqrt{30}}{12}.

$$6-4\sqrt{2}+\frac{\sqrt{15}\times 12}{3\sqrt{30}}$$

Cancel out 3 in both numerator and denominator.

$$6-4\sqrt{2}+\frac{4\sqrt{15}}{\sqrt{30}}$$

Rationalize the denominator of \frac{4\sqrt{15}}{\sqrt{30}} by multiplying numerator and denominator by \sqrt{30}.

$$6-4\sqrt{2}+\frac{4\sqrt{15}\sqrt{30}}{\left(\sqrt{30}\right)^{2}}$$

The square of \sqrt{30} is 30.

$$6-4\sqrt{2}+\frac{4\sqrt{15}\sqrt{30}}{30}$$

Factor 30=15\times 2. Rewrite the square root of the product \sqrt{15\times 2} as the product of square roots \sqrt{15}\sqrt{2}.

$$6-4\sqrt{2}+\frac{4\sqrt{15}\sqrt{15}\sqrt{2}}{30}$$

Multiply \sqrt{15} and \sqrt{15} to get 15.

$$6-4\sqrt{2}+\frac{4\times 15\sqrt{2}}{30}$$

Multiply 4 and 15 to get 60.

$$6-4\sqrt{2}+\frac{60\sqrt{2}}{30}$$

Divide 60\sqrt{2} by 30 to get 2\sqrt{2}.

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

Combine -4\sqrt{2} and 2\sqrt{2} to get -2\sqrt{2}.

$$6-2\sqrt{2}$$