Solve frac{sqrt{6}}{sqrt{2}}-frac{sqrt{24}}{sqrt{12}}+2sqrt{48}-3sqrt{8}

$$\frac { \sqrt { 6 } } { \sqrt { 2 } } - \frac { \sqrt { 24 } } { \sqrt { 12 } } + 2 \sqrt { 48 } - 3 \sqrt { 8 }$$

Evaluate

$$9\sqrt{3}-7\sqrt{2}\approx 5.688962332$$

Solution Steps

Rewrite the division of square roots \frac{\sqrt{6}}{\sqrt{2}} as the square root of the division \sqrt{\frac{6}{2}} and perform the division.

$$\sqrt{3}-\frac{\sqrt{24}}{\sqrt{12}}+2\sqrt{48}-3\sqrt{8}$$

Rewrite the division of square roots \frac{\sqrt{24}}{\sqrt{12}} as the square root of the division \sqrt{\frac{24}{12}} and perform the division.

$$\sqrt{3}-\sqrt{2}+2\sqrt{48}-3\sqrt{8}$$

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}.

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

Multiply 2 and 4 to get 8.

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

Combine \sqrt{3} and 8\sqrt{3} to get 9\sqrt{3}.

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

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}.

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

Multiply -3 and 2 to get -6.

$$9\sqrt{3}-\sqrt{2}-6\sqrt{2}$$

Combine -\sqrt{2} and -6\sqrt{2} to get -7\sqrt{2}.

$$9\sqrt{3}-7\sqrt{2}$$