Solve frac{sqrt{48}+sqrt{32}}{sqrt{27}-sqrt{18}}

$$\frac { \sqrt { 48 } + \sqrt { 32 } } { \sqrt { 27 } - \sqrt { 18 } }$$

Evaluate

$$\frac{8\sqrt{6}+20}{3}\approx 13.198639314$$

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}.

$$\frac{4\sqrt{3}+\sqrt{32}}{\sqrt{27}-\sqrt{18}}$$

Factor 32=4^{2}\times 2. Rewrite the square root of the product \sqrt{4^{2}\times 2} as the product of square roots \sqrt{4^{2}}\sqrt{2}. Take the square root of 4^{2}.

$$\frac{4\sqrt{3}+4\sqrt{2}}{\sqrt{27}-\sqrt{18}}$$

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}.

$$\frac{4\sqrt{3}+4\sqrt{2}}{3\sqrt{3}-\sqrt{18}}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

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

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

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

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

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

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

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

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(4\sqrt{3}+4\sqrt{2}\right)\left(3\sqrt{3}+3\sqrt{2}\right)}{9\left(\sqrt{3}\right)^{2}-\left(-3\sqrt{2}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(4\sqrt{3}+4\sqrt{2}\right)\left(3\sqrt{3}+3\sqrt{2}\right)}{9\times 3-\left(-3\sqrt{2}\right)^{2}}$$

Multiply 9 and 3 to get 27.

$$\frac{\left(4\sqrt{3}+4\sqrt{2}\right)\left(3\sqrt{3}+3\sqrt{2}\right)}{27-\left(-3\sqrt{2}\right)^{2}}$$

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

$$\frac{\left(4\sqrt{3}+4\sqrt{2}\right)\left(3\sqrt{3}+3\sqrt{2}\right)}{27-\left(-3\right)^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate -3 to the power of 2 and get 9.

$$\frac{\left(4\sqrt{3}+4\sqrt{2}\right)\left(3\sqrt{3}+3\sqrt{2}\right)}{27-9\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(4\sqrt{3}+4\sqrt{2}\right)\left(3\sqrt{3}+3\sqrt{2}\right)}{27-9\times 2}$$

Multiply 9 and 2 to get 18.

$$\frac{\left(4\sqrt{3}+4\sqrt{2}\right)\left(3\sqrt{3}+3\sqrt{2}\right)}{27-18}$$

Subtract 18 from 27 to get 9.

$$\frac{\left(4\sqrt{3}+4\sqrt{2}\right)\left(3\sqrt{3}+3\sqrt{2}\right)}{9}$$

Apply the distributive property by multiplying each term of 4\sqrt{3}+4\sqrt{2} by each term of 3\sqrt{3}+3\sqrt{2}.

$$\frac{12\left(\sqrt{3}\right)^{2}+12\sqrt{3}\sqrt{2}+12\sqrt{3}\sqrt{2}+12\left(\sqrt{2}\right)^{2}}{9}$$

The square of \sqrt{3} is 3.

$$\frac{12\times 3+12\sqrt{3}\sqrt{2}+12\sqrt{3}\sqrt{2}+12\left(\sqrt{2}\right)^{2}}{9}$$

Multiply 12 and 3 to get 36.

$$\frac{36+12\sqrt{3}\sqrt{2}+12\sqrt{3}\sqrt{2}+12\left(\sqrt{2}\right)^{2}}{9}$$

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

$$\frac{36+12\sqrt{6}+12\sqrt{3}\sqrt{2}+12\left(\sqrt{2}\right)^{2}}{9}$$

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

$$\frac{36+12\sqrt{6}+12\sqrt{6}+12\left(\sqrt{2}\right)^{2}}{9}$$

Combine 12\sqrt{6} and 12\sqrt{6} to get 24\sqrt{6}.

$$\frac{36+24\sqrt{6}+12\left(\sqrt{2}\right)^{2}}{9}$$

The square of \sqrt{2} is 2.

$$\frac{36+24\sqrt{6}+12\times 2}{9}$$

Multiply 12 and 2 to get 24.

$$\frac{36+24\sqrt{6}+24}{9}$$

Add 36 and 24 to get 60.

$$\frac{60+24\sqrt{6}}{9}$$