Solve frac{sqrt{12}(sqrt{20}+sqrt{180})}{sqrt{60}-sqrt{15}}

$$\frac { \sqrt { 12 } ( \sqrt { 20 } + \sqrt { 180 } ) } { \sqrt { 60 } - \sqrt { 15 } }$$

Evaluate

$$16$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\frac{2\sqrt{3}\left(\sqrt{20}+\sqrt{180}\right)}{\sqrt{60}-\sqrt{15}}$$

Factor 20=2^{2}\times 5. Rewrite the square root of the product \sqrt{2^{2}\times 5} as the product of square roots \sqrt{2^{2}}\sqrt{5}. Take the square root of 2^{2}.

$$\frac{2\sqrt{3}\left(2\sqrt{5}+\sqrt{180}\right)}{\sqrt{60}-\sqrt{15}}$$

Factor 180=6^{2}\times 5. Rewrite the square root of the product \sqrt{6^{2}\times 5} as the product of square roots \sqrt{6^{2}}\sqrt{5}. Take the square root of 6^{2}.

$$\frac{2\sqrt{3}\left(2\sqrt{5}+6\sqrt{5}\right)}{\sqrt{60}-\sqrt{15}}$$

Combine 2\sqrt{5} and 6\sqrt{5} to get 8\sqrt{5}.

$$\frac{2\sqrt{3}\times 8\sqrt{5}}{\sqrt{60}-\sqrt{15}}$$

Multiply 2 and 8 to get 16.

$$\frac{16\sqrt{3}\sqrt{5}}{\sqrt{60}-\sqrt{15}}$$

To multiply \sqrt{3} and \sqrt{5}, multiply the numbers under the square root.

$$\frac{16\sqrt{15}}{\sqrt{60}-\sqrt{15}}$$

Factor 60=2^{2}\times 15. Rewrite the square root of the product \sqrt{2^{2}\times 15} as the product of square roots \sqrt{2^{2}}\sqrt{15}. Take the square root of 2^{2}.

$$\frac{16\sqrt{15}}{2\sqrt{15}-\sqrt{15}}$$

Combine 2\sqrt{15} and -\sqrt{15} to get \sqrt{15}.

$$\frac{16\sqrt{15}}{\sqrt{15}}$$

Rationalize the denominator of \frac{16\sqrt{15}}{\sqrt{15}} by multiplying numerator and denominator by \sqrt{15}.

$$\frac{16\sqrt{15}\sqrt{15}}{\left(\sqrt{15}\right)^{2}}$$

The square of \sqrt{15} is 15.

$$\frac{16\sqrt{15}\sqrt{15}}{15}$$

Multiply \sqrt{15} and \sqrt{15} to get 15.

$$\frac{16\times 15}{15}$$

Multiply 16 and 15 to get 240.

$$\frac{240}{15}$$

Divide 240 by 15 to get 16.

$$16$$

Factor

$$2^{4}$$