Solve frac{sqrt{54}+sqrt{160}}{2sqrt{2}+2sqrt{3}}

$$\frac{ \sqrt{ 54 } + \sqrt{ 160 } }{ 2 \sqrt{ 2 } +2 \sqrt{ 3 } }$$

Evaluate

$$\frac{9\sqrt{2}}{2}+2\sqrt{30}-3\sqrt{3}-4\sqrt{5}\approx 3.177987848$$

Solution Steps

Factor 54=3^{2}\times 6. Rewrite the square root of the product \sqrt{3^{2}\times 6} as the product of square roots \sqrt{3^{2}}\sqrt{6}. Take the square root of 3^{2}.

$$\frac{3\sqrt{6}+\sqrt{160}}{2\sqrt{2}+2\sqrt{3}}$$

Factor 160=4^{2}\times 10. Rewrite the square root of the product \sqrt{4^{2}\times 10} as the product of square roots \sqrt{4^{2}}\sqrt{10}. Take the square root of 4^{2}.

$$\frac{3\sqrt{6}+4\sqrt{10}}{2\sqrt{2}+2\sqrt{3}}$$

Rationalize the denominator of \frac{3\sqrt{6}+4\sqrt{10}}{2\sqrt{2}+2\sqrt{3}} by multiplying numerator and denominator by 2\sqrt{2}-2\sqrt{3}.

$$\frac{\left(3\sqrt{6}+4\sqrt{10}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{\left(2\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}$$

Consider \left(2\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(3\sqrt{6}+4\sqrt{10}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{\left(2\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(3\sqrt{6}+4\sqrt{10}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{2^{2}\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(3\sqrt{6}+4\sqrt{10}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{4\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(3\sqrt{6}+4\sqrt{10}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{4\times 2-\left(2\sqrt{3}\right)^{2}}$$

Multiply 4 and 2 to get 8.

$$\frac{\left(3\sqrt{6}+4\sqrt{10}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-\left(2\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(3\sqrt{6}+4\sqrt{10}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-2^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(3\sqrt{6}+4\sqrt{10}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(3\sqrt{6}+4\sqrt{10}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-4\times 3}$$

Multiply 4 and 3 to get 12.

$$\frac{\left(3\sqrt{6}+4\sqrt{10}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-12}$$

Subtract 12 from 8 to get -4.

$$\frac{\left(3\sqrt{6}+4\sqrt{10}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{-4}$$

Apply the distributive property by multiplying each term of 3\sqrt{6}+4\sqrt{10} by each term of 2\sqrt{2}-2\sqrt{3}.

$$\frac{6\sqrt{6}\sqrt{2}-6\sqrt{3}\sqrt{6}+8\sqrt{2}\sqrt{10}-8\sqrt{3}\sqrt{10}}{-4}$$

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

$$\frac{6\sqrt{2}\sqrt{3}\sqrt{2}-6\sqrt{3}\sqrt{6}+8\sqrt{2}\sqrt{10}-8\sqrt{3}\sqrt{10}}{-4}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{6\times 2\sqrt{3}-6\sqrt{3}\sqrt{6}+8\sqrt{2}\sqrt{10}-8\sqrt{3}\sqrt{10}}{-4}$$

Multiply 6 and 2 to get 12.

$$\frac{12\sqrt{3}-6\sqrt{3}\sqrt{6}+8\sqrt{2}\sqrt{10}-8\sqrt{3}\sqrt{10}}{-4}$$

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}.

$$\frac{12\sqrt{3}-6\sqrt{3}\sqrt{3}\sqrt{2}+8\sqrt{2}\sqrt{10}-8\sqrt{3}\sqrt{10}}{-4}$$

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

$$\frac{12\sqrt{3}-6\times 3\sqrt{2}+8\sqrt{2}\sqrt{10}-8\sqrt{3}\sqrt{10}}{-4}$$

Multiply -6 and 3 to get -18.

$$\frac{12\sqrt{3}-18\sqrt{2}+8\sqrt{2}\sqrt{10}-8\sqrt{3}\sqrt{10}}{-4}$$

Factor 10=2\times 5. Rewrite the square root of the product \sqrt{2\times 5} as the product of square roots \sqrt{2}\sqrt{5}.

$$\frac{12\sqrt{3}-18\sqrt{2}+8\sqrt{2}\sqrt{2}\sqrt{5}-8\sqrt{3}\sqrt{10}}{-4}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{12\sqrt{3}-18\sqrt{2}+8\times 2\sqrt{5}-8\sqrt{3}\sqrt{10}}{-4}$$

Multiply 8 and 2 to get 16.

$$\frac{12\sqrt{3}-18\sqrt{2}+16\sqrt{5}-8\sqrt{3}\sqrt{10}}{-4}$$

To multiply \sqrt{3} and \sqrt{10}, multiply the numbers under the square root.

$$\frac{12\sqrt{3}-18\sqrt{2}+16\sqrt{5}-8\sqrt{30}}{-4}$$