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

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

Evaluate

$$-\frac{7}{2}=-3.5$$

Short Solution Steps

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

$$\left(2+\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)\left(\frac{1}{\sqrt{2}}-2\right)$$

The square of \sqrt{2} is 2.

$$\left(2+\frac{\sqrt{2}}{2}\right)\left(\frac{1}{\sqrt{2}}-2\right)$$

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

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

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

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

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

$$\frac{4+\sqrt{2}}{2}\left(\frac{1}{\sqrt{2}}-2\right)$$

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

$$\frac{4+\sqrt{2}}{2}\left(\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}-2\right)$$

The square of \sqrt{2} is 2.

$$\frac{4+\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}-2\right)$$

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

$$\frac{4+\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}-\frac{2\times 2}{2}\right)$$

Since \frac{\sqrt{2}}{2} and \frac{2\times 2}{2} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \sqrt{2}-2\times 2.

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

Multiply \frac{4+\sqrt{2}}{2} times \frac{\sqrt{2}-4}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(4+\sqrt{2}\right)\left(\sqrt{2}-4\right)}{2\times 2}$$

Multiply 2 and 2 to get 4.

$$\frac{\left(4+\sqrt{2}\right)\left(\sqrt{2}-4\right)}{4}$$

Apply the distributive property by multiplying each term of 4+\sqrt{2} by each term of \sqrt{2}-4.

$$\frac{4\sqrt{2}-16+\left(\sqrt{2}\right)^{2}-4\sqrt{2}}{4}$$

The square of \sqrt{2} is 2.

$$\frac{4\sqrt{2}-16+2-4\sqrt{2}}{4}$$

Add -16 and 2 to get -14.

$$\frac{4\sqrt{2}-14-4\sqrt{2}}{4}$$

Combine 4\sqrt{2} and -4\sqrt{2} to get 0.

$$\frac{-14}{4}$$

Reduce the fraction \frac{-14}{4} to lowest terms by extracting and canceling out 2.

$$-\frac{7}{2}$$

Factor

$$-\frac{7}{2} = -3\frac{1}{2} = -3.5$$