Solve frac{4-3sqrt{2}}{3-3sqrt{2}}

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

Evaluate

$$\frac{2-\sqrt{2}}{3}\approx 0.195262146$$

Solution Steps

Rationalize the denominator of \frac{4-3\sqrt{2}}{3-3\sqrt{2}} by multiplying numerator and denominator by 3+3\sqrt{2}.

$$\frac{\left(4-3\sqrt{2}\right)\left(3+3\sqrt{2}\right)}{\left(3-3\sqrt{2}\right)\left(3+3\sqrt{2}\right)}$$

Consider \left(3-3\sqrt{2}\right)\left(3+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}.

$$\frac{\left(4-3\sqrt{2}\right)\left(3+3\sqrt{2}\right)}{3^{2}-\left(-3\sqrt{2}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(4-3\sqrt{2}\right)\left(3+3\sqrt{2}\right)}{9-\left(-3\sqrt{2}\right)^{2}}$$

Expand \left(-3\sqrt{2}\right)^{2}.

$$\frac{\left(4-3\sqrt{2}\right)\left(3+3\sqrt{2}\right)}{9-\left(-3\right)^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate -3 to the power of 2 and get 9.

$$\frac{\left(4-3\sqrt{2}\right)\left(3+3\sqrt{2}\right)}{9-9\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(4-3\sqrt{2}\right)\left(3+3\sqrt{2}\right)}{9-9\times 2}$$

Multiply 9 and 2 to get 18.

$$\frac{\left(4-3\sqrt{2}\right)\left(3+3\sqrt{2}\right)}{9-18}$$

Subtract 18 from 9 to get -9.

$$\frac{\left(4-3\sqrt{2}\right)\left(3+3\sqrt{2}\right)}{-9}$$

Apply the distributive property by multiplying each term of 4-3\sqrt{2} by each term of 3+3\sqrt{2}.

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

Combine 12\sqrt{2} and -9\sqrt{2} to get 3\sqrt{2}.

$$\frac{12+3\sqrt{2}-9\left(\sqrt{2}\right)^{2}}{-9}$$

The square of \sqrt{2} is 2.

$$\frac{12+3\sqrt{2}-9\times 2}{-9}$$

Multiply -9 and 2 to get -18.

$$\frac{12+3\sqrt{2}-18}{-9}$$

Subtract 18 from 12 to get -6.

$$\frac{-6+3\sqrt{2}}{-9}$$