Solve (sqrt{18}-sqrt{27})-sqrt{6}+8sqrt{1/2}

$$( \sqrt { 18 } - \sqrt { 27 } ) - \sqrt { 6 } + 8 \sqrt { \frac { 1 } { 2 } }$$

Evaluate

$$7\sqrt{2}-\sqrt{6}-3\sqrt{3}\approx 2.253852771$$

Solution Steps

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}.

$$3\sqrt{2}-\sqrt{27}-\sqrt{6}+8\sqrt{\frac{1}{2}}$$

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}.

$$3\sqrt{2}-3\sqrt{3}-\sqrt{6}+8\sqrt{\frac{1}{2}}$$

Rewrite the square root of the division \sqrt{\frac{1}{2}} as the division of square roots \frac{\sqrt{1}}{\sqrt{2}}.

$$3\sqrt{2}-3\sqrt{3}-\sqrt{6}+8\times \frac{\sqrt{1}}{\sqrt{2}}$$

Calculate the square root of 1 and get 1.

$$3\sqrt{2}-3\sqrt{3}-\sqrt{6}+8\times \frac{1}{\sqrt{2}}$$

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

$$3\sqrt{2}-3\sqrt{3}-\sqrt{6}+8\times \frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$3\sqrt{2}-3\sqrt{3}-\sqrt{6}+8\times \frac{\sqrt{2}}{2}$$

Cancel out 2, the greatest common factor in 8 and 2.

$$3\sqrt{2}-3\sqrt{3}-\sqrt{6}+4\sqrt{2}$$

Combine 3\sqrt{2} and 4\sqrt{2} to get 7\sqrt{2}.

$$7\sqrt{2}-3\sqrt{3}-\sqrt{6}$$