Solve sqrt{2}sqrt{4}+4-2div6

$$\sqrt{ 2 } \sqrt{ 4 } +4-2 \div 6$$

Evaluate

$$2\sqrt{2}+\frac{11}{3}\approx 6.495093791$$

Solution Steps

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

$$\sqrt{2}\sqrt{2}\sqrt{2}+4-\frac{2}{6}$$

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

$$2\sqrt{2}+4-\frac{2}{6}$$

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

$$2\sqrt{2}+4-\frac{1}{3}$$

Convert 4 to fraction \frac{12}{3}.

$$2\sqrt{2}+\frac{12}{3}-\frac{1}{3}$$

Since \frac{12}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.

$$2\sqrt{2}+\frac{12-1}{3}$$

Subtract 1 from 12 to get 11.

$$2\sqrt{2}+\frac{11}{3}$$

Factor

$$\frac{6 \sqrt{2} + 11}{3} = 6.495093791412857$$