Solve 3sqrt{96}-2sqrt{50}+sqrt{24}

$$3 \sqrt{ 96 } -2 \sqrt{ 50 } + \sqrt{ 24 }$$

Evaluate

$$14\sqrt{6}-10\sqrt{2}\approx 20.150720775$$

Solution Steps

Factor 96=4^{2}\times 6. Rewrite the square root of the product \sqrt{4^{2}\times 6} as the product of square roots \sqrt{4^{2}}\sqrt{6}. Take the square root of 4^{2}.

$$3\times 4\sqrt{6}-2\sqrt{50}+\sqrt{24}$$

Multiply 3 and 4 to get 12.

$$12\sqrt{6}-2\sqrt{50}+\sqrt{24}$$

Factor 50=5^{2}\times 2. Rewrite the square root of the product \sqrt{5^{2}\times 2} as the product of square roots \sqrt{5^{2}}\sqrt{2}. Take the square root of 5^{2}.

$$12\sqrt{6}-2\times 5\sqrt{2}+\sqrt{24}$$

Multiply -2 and 5 to get -10.

$$12\sqrt{6}-10\sqrt{2}+\sqrt{24}$$

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$12\sqrt{6}-10\sqrt{2}+2\sqrt{6}$$

Combine 12\sqrt{6} and 2\sqrt{6} to get 14\sqrt{6}.

$$14\sqrt{6}-10\sqrt{2}$$