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

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

Evaluate

$$\sqrt{2}+7\sqrt{3}\approx 13.538569215$$

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}.

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

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}.

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

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

$$7\sqrt{3}+\sqrt{32}-\sqrt{18}$$

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}.

$$7\sqrt{3}+4\sqrt{2}-\sqrt{18}$$

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}.

$$7\sqrt{3}+4\sqrt{2}-3\sqrt{2}$$

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

$$7\sqrt{3}+\sqrt{2}$$