Solve (2+sqrt{5})(2-sqrt{5})+(sqrt{5}+1^2)-sqrt{20}=

$$( 2 + \sqrt { 5 } ) ( 2 - \sqrt { 5 } ) + ( \sqrt { 5 } + 1 ^ { 2 } ) - \sqrt { 20 } =$$

Evaluate

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

Solution Steps

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

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

Calculate 2 to the power of 2 and get 4.

$$4-\left(\sqrt{5}\right)^{2}+\sqrt{5}+1^{2}-\sqrt{20}$$

The square of \sqrt{5} is 5.

$$4-5+\sqrt{5}+1^{2}-\sqrt{20}$$

Subtract 5 from 4 to get -1.

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

Calculate 1 to the power of 2 and get 1.

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

Add -1 and 1 to get 0.

$$\sqrt{5}-\sqrt{20}$$

Factor 20=2^{2}\times 5. Rewrite the square root of the product \sqrt{2^{2}\times 5} as the product of square roots \sqrt{2^{2}}\sqrt{5}. Take the square root of 2^{2}.

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

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

$$-\sqrt{5}$$