Solve frac{2sqrt{8}+sqrt{3}}{2sqrt{3}-sqrt{8}}

$$\frac { 2 \sqrt { 8 } + \sqrt { 3 } } { 2 \sqrt { 3 } - \sqrt { 8 } }$$

Evaluate

$$\frac{5\sqrt{6}+11}{2}\approx 11.623724357$$

Solution Steps

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}.

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

Multiply 2 and 2 to get 4.

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

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}.

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

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

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

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

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

Calculate -2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

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

Subtract 8 from 12 to get 4.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Multiply 8 and 2 to get 16.

$$\frac{8\sqrt{6}+16+2\left(\sqrt{3}\right)^{2}+2\sqrt{3}\sqrt{2}}{4}$$

The square of \sqrt{3} is 3.

$$\frac{8\sqrt{6}+16+2\times 3+2\sqrt{3}\sqrt{2}}{4}$$

Multiply 2 and 3 to get 6.

$$\frac{8\sqrt{6}+16+6+2\sqrt{3}\sqrt{2}}{4}$$

Add 16 and 6 to get 22.

$$\frac{8\sqrt{6}+22+2\sqrt{3}\sqrt{2}}{4}$$

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

$$\frac{8\sqrt{6}+22+2\sqrt{6}}{4}$$

Combine 8\sqrt{6} and 2\sqrt{6} to get 10\sqrt{6}.

$$\frac{10\sqrt{6}+22}{4}$$