Solve {left(1/3-sqrt{2}right)}^2

$${ \left( \frac{ 1 }{ 3- \sqrt{ 2 } } \right) }^{ 2 }$$

Evaluate

$$\frac{6\sqrt{2}+11}{49}\approx 0.397658804$$

Solution Steps

Rationalize the denominator of \frac{1}{3-\sqrt{2}} by multiplying numerator and denominator by 3+\sqrt{2}.

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

Consider \left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 3. Square \sqrt{2}.

$$\left(\frac{3+\sqrt{2}}{9-2}\right)^{2}$$

Subtract 2 from 9 to get 7.

$$\left(\frac{3+\sqrt{2}}{7}\right)^{2}$$

To raise \frac{3+\sqrt{2}}{7} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3+\sqrt{2}\right)^{2}}{7^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+\sqrt{2}\right)^{2}.

$$\frac{9+6\sqrt{2}+\left(\sqrt{2}\right)^{2}}{7^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{9+6\sqrt{2}+2}{7^{2}}$$

Add 9 and 2 to get 11.

$$\frac{11+6\sqrt{2}}{7^{2}}$$

Calculate 7 to the power of 2 and get 49.

$$\frac{11+6\sqrt{2}}{49}$$