Solve (sqrt{5}+1)^2-(sqrt{5}+1)sqrt{5}-1

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

Evaluate

$$\sqrt{5}\approx 2.236067977$$

Solution Steps

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

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

The square of \sqrt{5} is 5.

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

Add 5 and 1 to get 6.

$$6+2\sqrt{5}-\left(\sqrt{5}+1\right)\sqrt{5}-1$$

Use the distributive property to multiply \sqrt{5}+1 by \sqrt{5}.

$$6+2\sqrt{5}-\left(\left(\sqrt{5}\right)^{2}+\sqrt{5}\right)-1$$

The square of \sqrt{5} is 5.

$$6+2\sqrt{5}-\left(5+\sqrt{5}\right)-1$$

To find the opposite of 5+\sqrt{5}, find the opposite of each term.

$$6+2\sqrt{5}-5-\sqrt{5}-1$$

Subtract 5 from 6 to get 1.

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

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

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

Subtract 1 from 1 to get 0.

$$\sqrt{5}$$