Solve (sqrt{5}+1)*4/sqrt{5-1}

$$( \sqrt{ 5 } +1) \times \frac{ 4 }{ \sqrt{ 5 } -1 }$$

Evaluate

$$2\left(\sqrt{5}+3\right)\approx 10.472135955$$

Solution Steps

Rationalize the denominator of \frac{4}{\sqrt{5}-1} by multiplying numerator and denominator by \sqrt{5}+1.

$$\left(\sqrt{5}+1\right)\times \frac{4\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}$$

Consider \left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\sqrt{5}+1\right)\times \frac{4\left(\sqrt{5}+1\right)}{\left(\sqrt{5}\right)^{2}-1^{2}}$$

Square \sqrt{5}. Square 1.

$$\left(\sqrt{5}+1\right)\times \frac{4\left(\sqrt{5}+1\right)}{5-1}$$

Subtract 1 from 5 to get 4.

$$\left(\sqrt{5}+1\right)\times \frac{4\left(\sqrt{5}+1\right)}{4}$$

Cancel out 4 and 4.

$$\left(\sqrt{5}+1\right)\left(\sqrt{5}+1\right)$$

Multiply \sqrt{5}+1 and \sqrt{5}+1 to get \left(\sqrt{5}+1\right)^{2}.

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

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

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

The square of \sqrt{5} is 5.

$$5+2\sqrt{5}+1$$

Add 5 and 1 to get 6.

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