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

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

Evaluate

$$-\frac{\sqrt{2}}{2}\approx -0.707106781$$

Short Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{2}+1\right)^{2}.

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

The square of \sqrt{2} is 2.

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

Add 2 and 1 to get 3.

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

Rationalize the denominator of \frac{3+2\sqrt{2}}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

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

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

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

Since \frac{2\left(\sqrt{2}+2\right)}{2} and \frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{2} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2\left(\sqrt{2}+2\right)-\left(3+2\sqrt{2}\right)\sqrt{2}.

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

Do the calculations in 2\sqrt{2}+4-3\sqrt{2}-4.

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