Solve frac{(sqrt{12}+sqrt{8})}{10+sqrt{96}}

$$\frac { ( \sqrt { 12 } + \sqrt { 8 } ) } { 10 + \sqrt { 96 } }$$

Evaluate

$$\sqrt{3}-\sqrt{2}\approx 0.317837245$$

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}.

$$\frac{2\sqrt{3}+\sqrt{8}}{10+\sqrt{96}}$$

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}.

$$\frac{2\sqrt{3}+2\sqrt{2}}{10+\sqrt{96}}$$

Factor 96=4^{2}\times 6. Rewrite the square root of the product \sqrt{4^{2}\times 6} as the product of square roots \sqrt{4^{2}}\sqrt{6}. Take the square root of 4^{2}.

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

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

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

Consider \left(10+4\sqrt{6}\right)\left(10-4\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 10 to the power of 2 and get 100.

$$\frac{\left(2\sqrt{3}+2\sqrt{2}\right)\left(10-4\sqrt{6}\right)}{100-\left(4\sqrt{6}\right)^{2}}$$

Expand \left(4\sqrt{6}\right)^{2}.

$$\frac{\left(2\sqrt{3}+2\sqrt{2}\right)\left(10-4\sqrt{6}\right)}{100-4^{2}\left(\sqrt{6}\right)^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{\left(2\sqrt{3}+2\sqrt{2}\right)\left(10-4\sqrt{6}\right)}{100-16\left(\sqrt{6}\right)^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{\left(2\sqrt{3}+2\sqrt{2}\right)\left(10-4\sqrt{6}\right)}{100-16\times 6}$$

Multiply 16 and 6 to get 96.

$$\frac{\left(2\sqrt{3}+2\sqrt{2}\right)\left(10-4\sqrt{6}\right)}{100-96}$$

Subtract 96 from 100 to get 4.

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

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

$$\frac{20\sqrt{3}-8\sqrt{3}\sqrt{6}+20\sqrt{2}-8\sqrt{2}\sqrt{6}}{4}$$

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}.

$$\frac{20\sqrt{3}-8\sqrt{3}\sqrt{3}\sqrt{2}+20\sqrt{2}-8\sqrt{2}\sqrt{6}}{4}$$

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

$$\frac{20\sqrt{3}-8\times 3\sqrt{2}+20\sqrt{2}-8\sqrt{2}\sqrt{6}}{4}$$

Multiply -8 and 3 to get -24.

$$\frac{20\sqrt{3}-24\sqrt{2}+20\sqrt{2}-8\sqrt{2}\sqrt{6}}{4}$$

Combine -24\sqrt{2} and 20\sqrt{2} to get -4\sqrt{2}.

$$\frac{20\sqrt{3}-4\sqrt{2}-8\sqrt{2}\sqrt{6}}{4}$$

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}.

$$\frac{20\sqrt{3}-4\sqrt{2}-8\sqrt{2}\sqrt{2}\sqrt{3}}{4}$$

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

$$\frac{20\sqrt{3}-4\sqrt{2}-8\times 2\sqrt{3}}{4}$$

Multiply -8 and 2 to get -16.

$$\frac{20\sqrt{3}-4\sqrt{2}-16\sqrt{3}}{4}$$

Combine 20\sqrt{3} and -16\sqrt{3} to get 4\sqrt{3}.

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

Divide each term of 4\sqrt{3}-4\sqrt{2} by 4 to get \sqrt{3}-\sqrt{2}.

$$\sqrt{3}-\sqrt{2}$$