Solve 7sqrt{3}-4sqrt{6}+sqrt{48}-sqrt{54}

$$7 \sqrt { 3 } - 4 \sqrt { 6 } + \sqrt { 48 } - \sqrt { 54 }$$

Evaluate

$$11\sqrt{3}-7\sqrt{6}\approx 1.906130684$$

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}.

$$7\sqrt{3}-4\sqrt{6}+4\sqrt{3}-\sqrt{54}$$

Combine 7\sqrt{3} and 4\sqrt{3} to get 11\sqrt{3}.

$$11\sqrt{3}-4\sqrt{6}-\sqrt{54}$$

Factor 54=3^{2}\times 6. Rewrite the square root of the product \sqrt{3^{2}\times 6} as the product of square roots \sqrt{3^{2}}\sqrt{6}. Take the square root of 3^{2}.

$$11\sqrt{3}-4\sqrt{6}-3\sqrt{6}$$

Combine -4\sqrt{6} and -3\sqrt{6} to get -7\sqrt{6}.

$$11\sqrt{3}-7\sqrt{6}$$