Solve frac{6-sqrt{2}}{3+sqrt{2}}

$$\frac { 6 - \sqrt { 2 } } { 3 + \sqrt { 2 } }$$

Evaluate

$$\frac{20-9\sqrt{2}}{7}\approx 1.038868277$$

Solution Steps

Rationalize the denominator of \frac{6-\sqrt{2}}{3+\sqrt{2}} by multiplying numerator and denominator by 3-\sqrt{2}.

$$\frac{\left(6-\sqrt{2}\right)\left(3-\sqrt{2}\right)}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}$$

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}.

$$\frac{\left(6-\sqrt{2}\right)\left(3-\sqrt{2}\right)}{3^{2}-\left(\sqrt{2}\right)^{2}}$$

Square 3. Square \sqrt{2}.

$$\frac{\left(6-\sqrt{2}\right)\left(3-\sqrt{2}\right)}{9-2}$$

Subtract 2 from 9 to get 7.

$$\frac{\left(6-\sqrt{2}\right)\left(3-\sqrt{2}\right)}{7}$$

Apply the distributive property by multiplying each term of 6-\sqrt{2} by each term of 3-\sqrt{2}.

$$\frac{18-6\sqrt{2}-3\sqrt{2}+\left(\sqrt{2}\right)^{2}}{7}$$

Combine -6\sqrt{2} and -3\sqrt{2} to get -9\sqrt{2}.

$$\frac{18-9\sqrt{2}+\left(\sqrt{2}\right)^{2}}{7}$$

The square of \sqrt{2} is 2.

$$\frac{18-9\sqrt{2}+2}{7}$$

Add 18 and 2 to get 20.

$$\frac{20-9\sqrt{2}}{7}$$