Solve frac{6+5sqrt{2}}{4+3sqrt{2}}

$$\frac { 6 + 5 \sqrt { 2 } } { 4 + 3 \sqrt { 2 } }$$

Evaluate

$$3-\sqrt{2}\approx 1.585786438$$

Solution Steps

Rationalize the denominator of \frac{6+5\sqrt{2}}{4+3\sqrt{2}} by multiplying numerator and denominator by 4-3\sqrt{2}.

$$\frac{\left(6+5\sqrt{2}\right)\left(4-3\sqrt{2}\right)}{\left(4+3\sqrt{2}\right)\left(4-3\sqrt{2}\right)}$$

Consider \left(4+3\sqrt{2}\right)\left(4-3\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}.

$$\frac{\left(6+5\sqrt{2}\right)\left(4-3\sqrt{2}\right)}{4^{2}-\left(3\sqrt{2}\right)^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{\left(6+5\sqrt{2}\right)\left(4-3\sqrt{2}\right)}{16-\left(3\sqrt{2}\right)^{2}}$$

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

$$\frac{\left(6+5\sqrt{2}\right)\left(4-3\sqrt{2}\right)}{16-3^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(6+5\sqrt{2}\right)\left(4-3\sqrt{2}\right)}{16-9\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(6+5\sqrt{2}\right)\left(4-3\sqrt{2}\right)}{16-9\times 2}$$

Multiply 9 and 2 to get 18.

$$\frac{\left(6+5\sqrt{2}\right)\left(4-3\sqrt{2}\right)}{16-18}$$

Subtract 18 from 16 to get -2.

$$\frac{\left(6+5\sqrt{2}\right)\left(4-3\sqrt{2}\right)}{-2}$$

Apply the distributive property by multiplying each term of 6+5\sqrt{2} by each term of 4-3\sqrt{2}.

$$\frac{24-18\sqrt{2}+20\sqrt{2}-15\left(\sqrt{2}\right)^{2}}{-2}$$

Combine -18\sqrt{2} and 20\sqrt{2} to get 2\sqrt{2}.

$$\frac{24+2\sqrt{2}-15\left(\sqrt{2}\right)^{2}}{-2}$$

The square of \sqrt{2} is 2.

$$\frac{24+2\sqrt{2}-15\times 2}{-2}$$

Multiply -15 and 2 to get -30.

$$\frac{24+2\sqrt{2}-30}{-2}$$

Subtract 30 from 24 to get -6.

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

Divide each term of -6+2\sqrt{2} by -2 to get 3-\sqrt{2}.

$$3-\sqrt{2}$$