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

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

Evaluate

$$-4\sqrt{5}-9\approx -17.94427191$$

Solution Steps

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

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

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

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

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

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

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

$$\frac{4\left(\sqrt{5}\right)^{2}+20\sqrt{5}+25}{\left(2\sqrt{5}\right)^{2}-5^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{4\times 5+20\sqrt{5}+25}{\left(2\sqrt{5}\right)^{2}-5^{2}}$$

Multiply 4 and 5 to get 20.

$$\frac{20+20\sqrt{5}+25}{\left(2\sqrt{5}\right)^{2}-5^{2}}$$

Add 20 and 25 to get 45.

$$\frac{45+20\sqrt{5}}{\left(2\sqrt{5}\right)^{2}-5^{2}}$$

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

$$\frac{45+20\sqrt{5}}{2^{2}\left(\sqrt{5}\right)^{2}-5^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{45+20\sqrt{5}}{4\left(\sqrt{5}\right)^{2}-5^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{45+20\sqrt{5}}{4\times 5-5^{2}}$$

Multiply 4 and 5 to get 20.

$$\frac{45+20\sqrt{5}}{20-5^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{45+20\sqrt{5}}{20-25}$$

Subtract 25 from 20 to get -5.

$$\frac{45+20\sqrt{5}}{-5}$$

Divide each term of 45+20\sqrt{5} by -5 to get -9-4\sqrt{5}.

$$-9-4\sqrt{5}$$

Factor

$$-\left(4\sqrt{5}+9\right)$$