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

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

Evaluate

$$5-2\sqrt{5}\approx 0.527864045$$

Solution Steps

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

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

The square of \sqrt{5} is 5.

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

Subtract 5 from 4 to get -1.

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

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

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

The square of \sqrt{5} is 5.

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

Add 5 and 1 to get 6.

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

Add -1 and 6 to get 5.

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