Solve frac{2sqrt{2}}{sqrt{48}-sqrt{8}-sqrt{27}}

$$\frac { 2 \sqrt { 2 } } { \sqrt { 48 } - \sqrt { 8 } - \sqrt { 27 } }$$

Evaluate

$$\frac{-2\sqrt{6}-8}{5}\approx -2.579795897$$

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}.

$$\frac{2\sqrt{2}}{4\sqrt{3}-\sqrt{8}-\sqrt{27}}$$

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}.

$$\frac{2\sqrt{2}}{4\sqrt{3}-2\sqrt{2}-\sqrt{27}}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

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

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

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

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

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

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

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

The square of \sqrt{3} is 3.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

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

Subtract 8 from 3 to get -5.

$$\frac{2\sqrt{2}\left(\sqrt{3}+2\sqrt{2}\right)}{-5}$$

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

$$\frac{2\sqrt{2}\sqrt{3}+4\left(\sqrt{2}\right)^{2}}{-5}$$

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

$$\frac{2\sqrt{6}+4\left(\sqrt{2}\right)^{2}}{-5}$$

The square of \sqrt{2} is 2.

$$\frac{2\sqrt{6}+4\times 2}{-5}$$

Multiply 4 and 2 to get 8.

$$\frac{2\sqrt{6}+8}{-5}$$