Solve 1/2sqrt{48}div(3sqrt{2}-sqrt{3})

$$\frac { 1 } { 2 } \sqrt { 48 } \div ( 3 \sqrt { 2 } - \sqrt { 3 } )$$

Evaluate

$$\frac{2\left(\sqrt{6}+1\right)}{5}\approx 1.379795897$$

Solution Steps

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$\frac{\frac{1}{2}\times 4\sqrt{3}}{3\sqrt{2}-\sqrt{3}}$$

Multiply \frac{1}{2} and 4 to get \frac{4}{2}.

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

Divide 4 by 2 to get 2.

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

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

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

Consider \left(3\sqrt{2}-\sqrt{3}\right)\left(3\sqrt{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}.

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

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

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

Calculate 3 to the power of 2 and get 9.

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

The square of \sqrt{2} is 2.

$$\frac{2\sqrt{3}\left(3\sqrt{2}+\sqrt{3}\right)}{9\times 2-\left(\sqrt{3}\right)^{2}}$$

Multiply 9 and 2 to get 18.

$$\frac{2\sqrt{3}\left(3\sqrt{2}+\sqrt{3}\right)}{18-\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{2\sqrt{3}\left(3\sqrt{2}+\sqrt{3}\right)}{18-3}$$

Subtract 3 from 18 to get 15.

$$\frac{2\sqrt{3}\left(3\sqrt{2}+\sqrt{3}\right)}{15}$$

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

$$\frac{6\sqrt{3}\sqrt{2}+2\left(\sqrt{3}\right)^{2}}{15}$$

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

$$\frac{6\sqrt{6}+2\left(\sqrt{3}\right)^{2}}{15}$$

The square of \sqrt{3} is 3.

$$\frac{6\sqrt{6}+2\times 3}{15}$$

Multiply 2 and 3 to get 6.

$$\frac{6\sqrt{6}+6}{15}$$