Solve 3sqrt{12}-2sqrt{48}+sqrt{8}

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

Evaluate

$$2\sqrt{2}-2\sqrt{3}\approx -0.63567449$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

Multiply 3 and 2 to get 6.

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

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}.

$$6\sqrt{3}-2\times 4\sqrt{3}+\sqrt{8}$$

Multiply -2 and 4 to get -8.

$$6\sqrt{3}-8\sqrt{3}+\sqrt{8}$$

Combine 6\sqrt{3} and -8\sqrt{3} to get -2\sqrt{3}.

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

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}.

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