Solve (5)[sqrt{2}(sqrt{12}-3sqrt{8})-(2sqrt{2}-1)^2]divsqrt{132}

$$( 5 ) [ \sqrt { 2 } ( \sqrt { 12 } - 3 \sqrt { 8 } ) - ( 2 \sqrt { 2 } - 1 ) ^ { 2 } ] \div \sqrt { 132 }$$

Evaluate

$$\frac{10\sqrt{66}}{33}+\frac{15\sqrt{22}}{33}-\frac{35\sqrt{33}}{22}\approx -4.545239955$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\frac{5\left(\sqrt{2}\left(2\sqrt{3}-3\sqrt{8}\right)-\left(2\sqrt{2}-1\right)^{2}\right)}{\sqrt{132}}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\frac{5\left(\sqrt{2}\left(2\sqrt{3}-3\times 2\sqrt{2}\right)-\left(2\sqrt{2}-1\right)^{2}\right)}{\sqrt{132}}$$

Multiply -3 and 2 to get -6.

$$\frac{5\left(\sqrt{2}\left(2\sqrt{3}-6\sqrt{2}\right)-\left(2\sqrt{2}-1\right)^{2}\right)}{\sqrt{132}}$$

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

$$\frac{5\left(\sqrt{2}\left(2\sqrt{3}-6\sqrt{2}\right)-\left(4\left(\sqrt{2}\right)^{2}-4\sqrt{2}+1\right)\right)}{\sqrt{132}}$$

The square of \sqrt{2} is 2.

$$\frac{5\left(\sqrt{2}\left(2\sqrt{3}-6\sqrt{2}\right)-\left(4\times 2-4\sqrt{2}+1\right)\right)}{\sqrt{132}}$$

Multiply 4 and 2 to get 8.

$$\frac{5\left(\sqrt{2}\left(2\sqrt{3}-6\sqrt{2}\right)-\left(8-4\sqrt{2}+1\right)\right)}{\sqrt{132}}$$

Add 8 and 1 to get 9.

$$\frac{5\left(\sqrt{2}\left(2\sqrt{3}-6\sqrt{2}\right)-\left(9-4\sqrt{2}\right)\right)}{\sqrt{132}}$$

To find the opposite of 9-4\sqrt{2}, find the opposite of each term.

$$\frac{5\left(\sqrt{2}\left(2\sqrt{3}-6\sqrt{2}\right)-9+4\sqrt{2}\right)}{\sqrt{132}}$$

Factor 132=2^{2}\times 33. Rewrite the square root of the product \sqrt{2^{2}\times 33} as the product of square roots \sqrt{2^{2}}\sqrt{33}. Take the square root of 2^{2}.

$$\frac{5\left(\sqrt{2}\left(2\sqrt{3}-6\sqrt{2}\right)-9+4\sqrt{2}\right)}{2\sqrt{33}}$$

Rationalize the denominator of \frac{5\left(\sqrt{2}\left(2\sqrt{3}-6\sqrt{2}\right)-9+4\sqrt{2}\right)}{2\sqrt{33}} by multiplying numerator and denominator by \sqrt{33}.

$$\frac{5\left(\sqrt{2}\left(2\sqrt{3}-6\sqrt{2}\right)-9+4\sqrt{2}\right)\sqrt{33}}{2\left(\sqrt{33}\right)^{2}}$$

The square of \sqrt{33} is 33.

$$\frac{5\left(\sqrt{2}\left(2\sqrt{3}-6\sqrt{2}\right)-9+4\sqrt{2}\right)\sqrt{33}}{2\times 33}$$

Multiply 2 and 33 to get 66.

$$\frac{5\left(\sqrt{2}\left(2\sqrt{3}-6\sqrt{2}\right)-9+4\sqrt{2}\right)\sqrt{33}}{66}$$

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

$$\frac{5\left(2\sqrt{2}\sqrt{3}-6\left(\sqrt{2}\right)^{2}-9+4\sqrt{2}\right)\sqrt{33}}{66}$$

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

$$\frac{5\left(2\sqrt{6}-6\left(\sqrt{2}\right)^{2}-9+4\sqrt{2}\right)\sqrt{33}}{66}$$

The square of \sqrt{2} is 2.

$$\frac{5\left(2\sqrt{6}-6\times 2-9+4\sqrt{2}\right)\sqrt{33}}{66}$$

Multiply -6 and 2 to get -12.

$$\frac{5\left(2\sqrt{6}-12-9+4\sqrt{2}\right)\sqrt{33}}{66}$$

Subtract 9 from -12 to get -21.

$$\frac{5\left(2\sqrt{6}-21+4\sqrt{2}\right)\sqrt{33}}{66}$$

Use the distributive property to multiply 5 by 2\sqrt{6}-21+4\sqrt{2}.

$$\frac{\left(10\sqrt{6}-105+20\sqrt{2}\right)\sqrt{33}}{66}$$

Use the distributive property to multiply 10\sqrt{6}-105+20\sqrt{2} by \sqrt{33}.

$$\frac{10\sqrt{6}\sqrt{33}-105\sqrt{33}+20\sqrt{2}\sqrt{33}}{66}$$

To multiply \sqrt{6} and \sqrt{33}, multiply the numbers under the square root.

$$\frac{10\sqrt{198}-105\sqrt{33}+20\sqrt{2}\sqrt{33}}{66}$$

To multiply \sqrt{2} and \sqrt{33}, multiply the numbers under the square root.

$$\frac{10\sqrt{198}-105\sqrt{33}+20\sqrt{66}}{66}$$

Factor 198=3^{2}\times 22. Rewrite the square root of the product \sqrt{3^{2}\times 22} as the product of square roots \sqrt{3^{2}}\sqrt{22}. Take the square root of 3^{2}.

$$\frac{10\times 3\sqrt{22}-105\sqrt{33}+20\sqrt{66}}{66}$$

Multiply 10 and 3 to get 30.

$$\frac{30\sqrt{22}-105\sqrt{33}+20\sqrt{66}}{66}$$