Solve sqrt{40}div2sqrt{10}-3sqrt{2}*sqrt{12}

$$\sqrt { 40 } \div 2 \sqrt { 10 } - 3 \sqrt { 2 } \times \sqrt { 12 }$$

Evaluate

$$10-6\sqrt{6}\approx -4.696938457$$

Solution Steps

Factor 40=2^{2}\times 10. Rewrite the square root of the product \sqrt{2^{2}\times 10} as the product of square roots \sqrt{2^{2}}\sqrt{10}. Take the square root of 2^{2}.

$$\frac{2\sqrt{10}}{2}\sqrt{10}-3\sqrt{2}\sqrt{12}$$

Cancel out 2 and 2.

$$\sqrt{10}\sqrt{10}-3\sqrt{2}\sqrt{12}$$

Multiply \sqrt{10} and \sqrt{10} to get 10.

$$10-3\sqrt{2}\sqrt{12}$$

Factor 12=2\times 6. Rewrite the square root of the product \sqrt{2\times 6} as the product of square roots \sqrt{2}\sqrt{6}.

$$10-3\sqrt{2}\sqrt{2}\sqrt{6}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$10-3\times 2\sqrt{6}$$

Multiply 3 and 2 to get 6.

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

Factor

$$2 {(5 - 3 \sqrt{6})} = -4.696938457$$