Solve {left(sqrt{5}-1right)}^2+20/5-sqrt{5}

$${ \left( \sqrt{ 5 } -1 \right) }^{ 2 } + \frac{ 20 }{ 5- \sqrt{ 5 } }$$

Evaluate

$$11-\sqrt{5}\approx 8.763932023$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-1\right)^{2}.

$$\left(\sqrt{5}\right)^{2}-2\sqrt{5}+1+\frac{20}{5-\sqrt{5}}$$

The square of \sqrt{5} is 5.

$$5-2\sqrt{5}+1+\frac{20}{5-\sqrt{5}}$$

Add 5 and 1 to get 6.

$$6-2\sqrt{5}+\frac{20}{5-\sqrt{5}}$$

Rationalize the denominator of \frac{20}{5-\sqrt{5}} by multiplying numerator and denominator by 5+\sqrt{5}.

$$6-2\sqrt{5}+\frac{20\left(5+\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}$$

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

$$6-2\sqrt{5}+\frac{20\left(5+\sqrt{5}\right)}{5^{2}-\left(\sqrt{5}\right)^{2}}$$

Square 5. Square \sqrt{5}.

$$6-2\sqrt{5}+\frac{20\left(5+\sqrt{5}\right)}{25-5}$$

Subtract 5 from 25 to get 20.

$$6-2\sqrt{5}+\frac{20\left(5+\sqrt{5}\right)}{20}$$

Cancel out 20 and 20.

$$6-2\sqrt{5}+5+\sqrt{5}$$

Add 6 and 5 to get 11.

$$11-2\sqrt{5}+\sqrt{5}$$

Combine -2\sqrt{5} and \sqrt{5} to get -\sqrt{5}.

$$11-\sqrt{5}$$