Solve frac{sqrt{3}+sqrt{2}}{sqrt{8}+sqrt{6}}*frac{sqrt{8}+sqrt{6}}{sqrt{8}+sqrt{6}}

$$\frac { \sqrt { 3 } + \sqrt { 2 } } { \sqrt { 8 } + \sqrt { 6 } } \cdot \frac { \sqrt { 8 } + \sqrt { 6 } } { \sqrt { 8 } + \sqrt { 6 } }$$

Evaluate

$$-\frac{3\sqrt{2}}{2}+\sqrt{6}+2-\sqrt{3}\approx 0.596118592$$

Solution Steps

Divide \sqrt{8}+\sqrt{6} by \sqrt{8}+\sqrt{6} to get 1.

$$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{8}+\sqrt{6}}\times 1$$

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}.

$$\frac{\sqrt{3}+\sqrt{2}}{2\sqrt{2}+\sqrt{6}}\times 1$$

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

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

Consider \left(2\sqrt{2}+\sqrt{6}\right)\left(2\sqrt{2}-\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}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

$$\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(2\sqrt{2}-\sqrt{6}\right)}{8-\left(\sqrt{6}\right)^{2}}\times 1$$

The square of \sqrt{6} is 6.

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

Subtract 6 from 8 to get 2.

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

Express \frac{\left(\sqrt{3}+\sqrt{2}\right)\left(2\sqrt{2}-\sqrt{6}\right)}{2}\times 1 as a single fraction.

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

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

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

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

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

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}.

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

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

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

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

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

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}.

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

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

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

Factor

$$\frac{2 \sqrt{6} + 4 - 2 \sqrt{3} - 3 \sqrt{2}}{2} = 0.5961185916546579$$