Solve sqrt{2}*(3+1/sqrt{2})-2sqrt{2}

$$\sqrt { 2 } \times ( 3 + \frac { 1 } { \sqrt { 2 } } ) - 2 \sqrt { 2 }$$

Evaluate

$$\sqrt{2}+1\approx 2.414213562$$

Short Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\sqrt{2}\left(3+\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)-2\sqrt{2}$$

The square of \sqrt{2} is 2.

$$\sqrt{2}\left(3+\frac{\sqrt{2}}{2}\right)-2\sqrt{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{2}{2}.

$$\sqrt{2}\left(\frac{3\times 2}{2}+\frac{\sqrt{2}}{2}\right)-2\sqrt{2}$$

Since \frac{3\times 2}{2} and \frac{\sqrt{2}}{2} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 3\times 2+\sqrt{2}.

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

Express \sqrt{2}\times \frac{6+\sqrt{2}}{2} as a single fraction.

$$\frac{\sqrt{2}\left(6+\sqrt{2}\right)}{2}-2\sqrt{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -2\sqrt{2} times \frac{2}{2}.

$$\frac{\sqrt{2}\left(6+\sqrt{2}\right)}{2}+\frac{2\left(-2\right)\sqrt{2}}{2}$$

Since \frac{\sqrt{2}\left(6+\sqrt{2}\right)}{2} and \frac{2\left(-2\right)\sqrt{2}}{2} have the same denominator, add them by adding their numerators.

$$\frac{\sqrt{2}\left(6+\sqrt{2}\right)+2\left(-2\right)\sqrt{2}}{2}$$

Do the multiplications in \sqrt{2}\left(6+\sqrt{2}\right)+2\left(-2\right)\sqrt{2}.

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

Do the calculations in 6\sqrt{2}+2-4\sqrt{2}.

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

Divide each term of 2\sqrt{2}+2 by 2 to get \sqrt{2}+1.

$$\sqrt{2}+1$$

Factor

$$\sqrt{2} + 1 = 2.414213562$$