Solve (5-3sqrt{5})div(5+3sqrt{5})

$$(5-3 \sqrt{ 5 } ) \div (5+3 \sqrt{ 5 } )$$

Evaluate

$$\frac{3\sqrt{5}-7}{2}\approx -0.145898034$$

Solution Steps

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

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

Consider \left(5+3\sqrt{5}\right)\left(5-3\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

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

$$\frac{25-30\sqrt{5}+9\left(\sqrt{5}\right)^{2}}{5^{2}-\left(3\sqrt{5}\right)^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{25-30\sqrt{5}+9\times 5}{5^{2}-\left(3\sqrt{5}\right)^{2}}$$

Multiply 9 and 5 to get 45.

$$\frac{25-30\sqrt{5}+45}{5^{2}-\left(3\sqrt{5}\right)^{2}}$$

Add 25 and 45 to get 70.

$$\frac{70-30\sqrt{5}}{5^{2}-\left(3\sqrt{5}\right)^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{70-30\sqrt{5}}{25-\left(3\sqrt{5}\right)^{2}}$$

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

$$\frac{70-30\sqrt{5}}{25-3^{2}\left(\sqrt{5}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{70-30\sqrt{5}}{25-9\left(\sqrt{5}\right)^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{70-30\sqrt{5}}{25-9\times 5}$$

Multiply 9 and 5 to get 45.

$$\frac{70-30\sqrt{5}}{25-45}$$

Subtract 45 from 25 to get -20.

$$\frac{70-30\sqrt{5}}{-20}$$

Factor

$$\frac{3 \sqrt{5} - 7}{2} = -0.1458980337503153$$