Solve (sqrt{5}-sqrt{2})^2-(2sqrt{3}+sqrt{5})(2sqrt{3}-sqrt{5})

$$( \sqrt { 5 } - \sqrt { 2 } ) ^ { 2 } - ( 2 \sqrt { 3 } + \sqrt { 5 } ) ( 2 \sqrt { 3 } - \sqrt { 5 } )$$

Evaluate

$$-2\sqrt{10}\approx -6.32455532$$

Solution Steps

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

$$\left(\sqrt{5}\right)^{2}-2\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{3}-\sqrt{5}\right)$$

The square of \sqrt{5} is 5.

$$5-2\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{3}-\sqrt{5}\right)$$

To multiply \sqrt{5} and \sqrt{2}, multiply the numbers under the square root.

$$5-2\sqrt{10}+\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{3}-\sqrt{5}\right)$$

The square of \sqrt{2} is 2.

$$5-2\sqrt{10}+2-\left(2\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{3}-\sqrt{5}\right)$$

Add 5 and 2 to get 7.

$$7-2\sqrt{10}-\left(2\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{3}-\sqrt{5}\right)$$

Consider \left(2\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{3}-\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}.

$$7-2\sqrt{10}-\left(\left(2\sqrt{3}\right)^{2}-\left(\sqrt{5}\right)^{2}\right)$$

Expand \left(2\sqrt{3}\right)^{2}.

$$7-2\sqrt{10}-\left(2^{2}\left(\sqrt{3}\right)^{2}-\left(\sqrt{5}\right)^{2}\right)$$

Calculate 2 to the power of 2 and get 4.

$$7-2\sqrt{10}-\left(4\left(\sqrt{3}\right)^{2}-\left(\sqrt{5}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$7-2\sqrt{10}-\left(4\times 3-\left(\sqrt{5}\right)^{2}\right)$$

Multiply 4 and 3 to get 12.

$$7-2\sqrt{10}-\left(12-\left(\sqrt{5}\right)^{2}\right)$$

The square of \sqrt{5} is 5.

$$7-2\sqrt{10}-\left(12-5\right)$$

Subtract 5 from 12 to get 7.

$$7-2\sqrt{10}-7$$

Subtract 7 from 7 to get 0.

$$-2\sqrt{10}$$