Solve frac{3sqrt{2}}{2sqrt{6}-4sqrt{2}}

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

Evaluate

$$-\frac{3\sqrt{3}}{2}-3\approx -5.598076211$$

Solution Steps

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

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

Consider \left(2\sqrt{6}-4\sqrt{2}\right)\left(2\sqrt{6}+4\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}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{6} is 6.

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

Multiply 4 and 6 to get 24.

$$\frac{3\sqrt{2}\left(2\sqrt{6}+4\sqrt{2}\right)}{24-\left(-4\sqrt{2}\right)^{2}}$$

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

$$\frac{3\sqrt{2}\left(2\sqrt{6}+4\sqrt{2}\right)}{24-\left(-4\right)^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate -4 to the power of 2 and get 16.

$$\frac{3\sqrt{2}\left(2\sqrt{6}+4\sqrt{2}\right)}{24-16\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{3\sqrt{2}\left(2\sqrt{6}+4\sqrt{2}\right)}{24-16\times 2}$$

Multiply 16 and 2 to get 32.

$$\frac{3\sqrt{2}\left(2\sqrt{6}+4\sqrt{2}\right)}{24-32}$$

Subtract 32 from 24 to get -8.

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

Use the distributive property to multiply 3\sqrt{2} by 2\sqrt{6}+4\sqrt{2}.

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

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}.

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

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

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

Multiply 6 and 2 to get 12.

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

The square of \sqrt{2} is 2.

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

Multiply 12 and 2 to get 24.

$$\frac{12\sqrt{3}+24}{-8}$$