Solve 3sqrt{48}-9sqrt{1/3}+3sqrt{18}-4sqrt{1/8}

$$3 \sqrt{ 48 } -9 \sqrt{ \frac{ 1 }{ 3 } } +3 \sqrt{ 18 } -4 \sqrt{ \frac{ 1 }{ 8 } }$$

Evaluate

$$8\sqrt{2}+9\sqrt{3}\approx 26.902165767$$

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}.

$$3\times 4\sqrt{3}-9\sqrt{\frac{1}{3}}+3\sqrt{18}-4\sqrt{\frac{1}{8}}$$

Multiply 3 and 4 to get 12.

$$12\sqrt{3}-9\sqrt{\frac{1}{3}}+3\sqrt{18}-4\sqrt{\frac{1}{8}}$$

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

$$12\sqrt{3}-9\times \frac{\sqrt{1}}{\sqrt{3}}+3\sqrt{18}-4\sqrt{\frac{1}{8}}$$

Calculate the square root of 1 and get 1.

$$12\sqrt{3}-9\times \frac{1}{\sqrt{3}}+3\sqrt{18}-4\sqrt{\frac{1}{8}}$$

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$12\sqrt{3}-9\times \frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}+3\sqrt{18}-4\sqrt{\frac{1}{8}}$$

The square of \sqrt{3} is 3.

$$12\sqrt{3}-9\times \frac{\sqrt{3}}{3}+3\sqrt{18}-4\sqrt{\frac{1}{8}}$$

Cancel out 3, the greatest common factor in 9 and 3.

$$12\sqrt{3}-3\sqrt{3}+3\sqrt{18}-4\sqrt{\frac{1}{8}}$$

Combine 12\sqrt{3} and -3\sqrt{3} to get 9\sqrt{3}.

$$9\sqrt{3}+3\sqrt{18}-4\sqrt{\frac{1}{8}}$$

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}.

$$9\sqrt{3}+3\times 3\sqrt{2}-4\sqrt{\frac{1}{8}}$$

Multiply 3 and 3 to get 9.

$$9\sqrt{3}+9\sqrt{2}-4\sqrt{\frac{1}{8}}$$

Rewrite the square root of the division \sqrt{\frac{1}{8}} as the division of square roots \frac{\sqrt{1}}{\sqrt{8}}.

$$9\sqrt{3}+9\sqrt{2}-4\times \frac{\sqrt{1}}{\sqrt{8}}$$

Calculate the square root of 1 and get 1.

$$9\sqrt{3}+9\sqrt{2}-4\times \frac{1}{\sqrt{8}}$$

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}.

$$9\sqrt{3}+9\sqrt{2}-4\times \frac{1}{2\sqrt{2}}$$

Rationalize the denominator of \frac{1}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

$$9\sqrt{3}+9\sqrt{2}-4\times \frac{\sqrt{2}}{2\times 2}$$

Multiply 2 and 2 to get 4.

$$9\sqrt{3}+9\sqrt{2}-4\times \frac{\sqrt{2}}{4}$$

Cancel out 4 and 4.

$$9\sqrt{3}+9\sqrt{2}-\sqrt{2}$$

Combine 9\sqrt{2} and -\sqrt{2} to get 8\sqrt{2}.

$$9\sqrt{3}+8\sqrt{2}$$