Solve frac{3sqrt{5}-2}{3sqrt{5}+2}

$$\frac { 3 \sqrt { 5 } - 2 } { 3 \sqrt { 5 } + 2 }$$

Evaluate

$$\frac{49-12\sqrt{5}}{41}\approx 0.540663031$$

Solution Steps

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

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

Consider \left(3\sqrt{5}+2\right)\left(3\sqrt{5}-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}.

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

Multiply 3\sqrt{5}-2 and 3\sqrt{5}-2 to get \left(3\sqrt{5}-2\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{5}-2\right)^{2}.

$$\frac{9\left(\sqrt{5}\right)^{2}-12\sqrt{5}+4}{\left(3\sqrt{5}\right)^{2}-2^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{9\times 5-12\sqrt{5}+4}{\left(3\sqrt{5}\right)^{2}-2^{2}}$$

Multiply 9 and 5 to get 45.

$$\frac{45-12\sqrt{5}+4}{\left(3\sqrt{5}\right)^{2}-2^{2}}$$

Add 45 and 4 to get 49.

$$\frac{49-12\sqrt{5}}{\left(3\sqrt{5}\right)^{2}-2^{2}}$$

Expand \left(3\sqrt{5}\right)^{2}.

$$\frac{49-12\sqrt{5}}{3^{2}\left(\sqrt{5}\right)^{2}-2^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{49-12\sqrt{5}}{9\left(\sqrt{5}\right)^{2}-2^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{49-12\sqrt{5}}{9\times 5-2^{2}}$$

Multiply 9 and 5 to get 45.

$$\frac{49-12\sqrt{5}}{45-2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{49-12\sqrt{5}}{45-4}$$

Subtract 4 from 45 to get 41.

$$\frac{49-12\sqrt{5}}{41}$$