Solve (sqrt{6}+sqrt{48})div(sqrt{108}-sqrt{15})

$$( \sqrt { 6 } + \sqrt { 48 } ) \div ( \sqrt { 108 } - \sqrt { 15 } )$$

Evaluate

$$\frac{\sqrt{10}+4\sqrt{5}+6\sqrt{2}+24}{31}\approx 1.438446159$$

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}.

$$\frac{\sqrt{6}+4\sqrt{3}}{\sqrt{108}-\sqrt{15}}$$

Factor 108=6^{2}\times 3. Rewrite the square root of the product \sqrt{6^{2}\times 3} as the product of square roots \sqrt{6^{2}}\sqrt{3}. Take the square root of 6^{2}.

$$\frac{\sqrt{6}+4\sqrt{3}}{6\sqrt{3}-\sqrt{15}}$$

Rationalize the denominator of \frac{\sqrt{6}+4\sqrt{3}}{6\sqrt{3}-\sqrt{15}} by multiplying numerator and denominator by 6\sqrt{3}+\sqrt{15}.

$$\frac{\left(\sqrt{6}+4\sqrt{3}\right)\left(6\sqrt{3}+\sqrt{15}\right)}{\left(6\sqrt{3}-\sqrt{15}\right)\left(6\sqrt{3}+\sqrt{15}\right)}$$

Consider \left(6\sqrt{3}-\sqrt{15}\right)\left(6\sqrt{3}+\sqrt{15}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(\sqrt{6}+4\sqrt{3}\right)\left(6\sqrt{3}+\sqrt{15}\right)}{\left(6\sqrt{3}\right)^{2}-\left(\sqrt{15}\right)^{2}}$$

Expand \left(6\sqrt{3}\right)^{2}.

$$\frac{\left(\sqrt{6}+4\sqrt{3}\right)\left(6\sqrt{3}+\sqrt{15}\right)}{6^{2}\left(\sqrt{3}\right)^{2}-\left(\sqrt{15}\right)^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{\left(\sqrt{6}+4\sqrt{3}\right)\left(6\sqrt{3}+\sqrt{15}\right)}{36\left(\sqrt{3}\right)^{2}-\left(\sqrt{15}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(\sqrt{6}+4\sqrt{3}\right)\left(6\sqrt{3}+\sqrt{15}\right)}{36\times 3-\left(\sqrt{15}\right)^{2}}$$

Multiply 36 and 3 to get 108.

$$\frac{\left(\sqrt{6}+4\sqrt{3}\right)\left(6\sqrt{3}+\sqrt{15}\right)}{108-\left(\sqrt{15}\right)^{2}}$$

The square of \sqrt{15} is 15.

$$\frac{\left(\sqrt{6}+4\sqrt{3}\right)\left(6\sqrt{3}+\sqrt{15}\right)}{108-15}$$

Subtract 15 from 108 to get 93.

$$\frac{\left(\sqrt{6}+4\sqrt{3}\right)\left(6\sqrt{3}+\sqrt{15}\right)}{93}$$

Apply the distributive property by multiplying each term of \sqrt{6}+4\sqrt{3} by each term of 6\sqrt{3}+\sqrt{15}.

$$\frac{6\sqrt{6}\sqrt{3}+\sqrt{6}\sqrt{15}+24\left(\sqrt{3}\right)^{2}+4\sqrt{3}\sqrt{15}}{93}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$\frac{6\sqrt{3}\sqrt{2}\sqrt{3}+\sqrt{6}\sqrt{15}+24\left(\sqrt{3}\right)^{2}+4\sqrt{3}\sqrt{15}}{93}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{6\times 3\sqrt{2}+\sqrt{6}\sqrt{15}+24\left(\sqrt{3}\right)^{2}+4\sqrt{3}\sqrt{15}}{93}$$

Multiply 6 and 3 to get 18.

$$\frac{18\sqrt{2}+\sqrt{6}\sqrt{15}+24\left(\sqrt{3}\right)^{2}+4\sqrt{3}\sqrt{15}}{93}$$

To multiply \sqrt{6} and \sqrt{15}, multiply the numbers under the square root.

$$\frac{18\sqrt{2}+\sqrt{90}+24\left(\sqrt{3}\right)^{2}+4\sqrt{3}\sqrt{15}}{93}$$

The square of \sqrt{3} is 3.

$$\frac{18\sqrt{2}+\sqrt{90}+24\times 3+4\sqrt{3}\sqrt{15}}{93}$$

Multiply 24 and 3 to get 72.

$$\frac{18\sqrt{2}+\sqrt{90}+72+4\sqrt{3}\sqrt{15}}{93}$$

Factor 15=3\times 5. Rewrite the square root of the product \sqrt{3\times 5} as the product of square roots \sqrt{3}\sqrt{5}.

$$\frac{18\sqrt{2}+\sqrt{90}+72+4\sqrt{3}\sqrt{3}\sqrt{5}}{93}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{18\sqrt{2}+\sqrt{90}+72+4\times 3\sqrt{5}}{93}$$

Multiply 4 and 3 to get 12.

$$\frac{18\sqrt{2}+\sqrt{90}+72+12\sqrt{5}}{93}$$

Factor 90=3^{2}\times 10. Rewrite the square root of the product \sqrt{3^{2}\times 10} as the product of square roots \sqrt{3^{2}}\sqrt{10}. Take the square root of 3^{2}.

$$\frac{18\sqrt{2}+3\sqrt{10}+72+12\sqrt{5}}{93}$$