Solve =frac{22+12sqrt{2}}{(3+sqrt{2})^2}

$$= \frac { 22 + 12 \sqrt { 2 } } { ( 3 + \sqrt { 2 } ) ^ { 2 } }$$

Evaluate

$$2$$

Solution Steps

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

$$\frac{22+12\sqrt{2}}{9+6\sqrt{2}+\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{22+12\sqrt{2}}{9+6\sqrt{2}+2}$$

Add 9 and 2 to get 11.

$$\frac{22+12\sqrt{2}}{11+6\sqrt{2}}$$

Rationalize the denominator of \frac{22+12\sqrt{2}}{11+6\sqrt{2}} by multiplying numerator and denominator by 11-6\sqrt{2}.

$$\frac{\left(22+12\sqrt{2}\right)\left(11-6\sqrt{2}\right)}{\left(11+6\sqrt{2}\right)\left(11-6\sqrt{2}\right)}$$

Consider \left(11+6\sqrt{2}\right)\left(11-6\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}.

$$\frac{\left(22+12\sqrt{2}\right)\left(11-6\sqrt{2}\right)}{11^{2}-\left(6\sqrt{2}\right)^{2}}$$

Calculate 11 to the power of 2 and get 121.

$$\frac{\left(22+12\sqrt{2}\right)\left(11-6\sqrt{2}\right)}{121-\left(6\sqrt{2}\right)^{2}}$$

Expand \left(6\sqrt{2}\right)^{2}.

$$\frac{\left(22+12\sqrt{2}\right)\left(11-6\sqrt{2}\right)}{121-6^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{\left(22+12\sqrt{2}\right)\left(11-6\sqrt{2}\right)}{121-36\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(22+12\sqrt{2}\right)\left(11-6\sqrt{2}\right)}{121-36\times 2}$$

Multiply 36 and 2 to get 72.

$$\frac{\left(22+12\sqrt{2}\right)\left(11-6\sqrt{2}\right)}{121-72}$$

Subtract 72 from 121 to get 49.

$$\frac{\left(22+12\sqrt{2}\right)\left(11-6\sqrt{2}\right)}{49}$$

Use the distributive property to multiply 22+12\sqrt{2} by 11-6\sqrt{2} and combine like terms.

$$\frac{242-72\left(\sqrt{2}\right)^{2}}{49}$$

The square of \sqrt{2} is 2.

$$\frac{242-72\times 2}{49}$$

Multiply -72 and 2 to get -144.

$$\frac{242-144}{49}$$

Subtract 144 from 242 to get 98.

$$\frac{98}{49}$$

Divide 98 by 49 to get 2.

$$2$$

Factor

$$2$$