Solve sqrt{6}*sqrt{3}-4sqrt{50}

$$\sqrt { 6 } \cdot \sqrt { 3 } - 4 \sqrt { 50 }$$

Evaluate

$$-17\sqrt{2}\approx -24.04163056$$

Solution Steps

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

$$\sqrt{3}\sqrt{2}\sqrt{3}-4\sqrt{50}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$3\sqrt{2}-4\sqrt{50}$$

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}.

$$3\sqrt{2}-4\times 5\sqrt{2}$$

Multiply -4 and 5 to get -20.

$$3\sqrt{2}-20\sqrt{2}$$

Combine 3\sqrt{2} and -20\sqrt{2} to get -17\sqrt{2}.

$$-17\sqrt{2}$$