Solve (sqrt{2}+sqrt{5})^2-(2+sqrt{10})^2+sqrt{90}+(2sqrt{2}-1)(2sqrt{2}+1)

$$( \sqrt { 2 } + \sqrt { 5 } ) ^ { 2 } - ( 2 + \sqrt { 10 } ) ^ { 2 } + \sqrt { 90 } + ( 2 \sqrt { 2 } - 1 ) ( 2 \sqrt { 2 } + 1 )$$

Evaluate

$$\sqrt{10}\approx 3.16227766$$

Solution Steps

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

$$\left(\sqrt{2}\right)^{2}+2\sqrt{2}\sqrt{5}+\left(\sqrt{5}\right)^{2}-\left(2+\sqrt{10}\right)^{2}+\sqrt{90}+\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)$$

The square of \sqrt{2} is 2.

$$2+2\sqrt{2}\sqrt{5}+\left(\sqrt{5}\right)^{2}-\left(2+\sqrt{10}\right)^{2}+\sqrt{90}+\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)$$

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

$$2+2\sqrt{10}+\left(\sqrt{5}\right)^{2}-\left(2+\sqrt{10}\right)^{2}+\sqrt{90}+\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)$$

The square of \sqrt{5} is 5.

$$2+2\sqrt{10}+5-\left(2+\sqrt{10}\right)^{2}+\sqrt{90}+\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)$$

Add 2 and 5 to get 7.

$$7+2\sqrt{10}-\left(2+\sqrt{10}\right)^{2}+\sqrt{90}+\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{10}\right)^{2}.

$$7+2\sqrt{10}-\left(4+4\sqrt{10}+\left(\sqrt{10}\right)^{2}\right)+\sqrt{90}+\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)$$

The square of \sqrt{10} is 10.

$$7+2\sqrt{10}-\left(4+4\sqrt{10}+10\right)+\sqrt{90}+\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)$$

Add 4 and 10 to get 14.

$$7+2\sqrt{10}-\left(14+4\sqrt{10}\right)+\sqrt{90}+\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)$$

To find the opposite of 14+4\sqrt{10}, find the opposite of each term.

$$7+2\sqrt{10}-14-4\sqrt{10}+\sqrt{90}+\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)$$

Subtract 14 from 7 to get -7.

$$-7+2\sqrt{10}-4\sqrt{10}+\sqrt{90}+\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)$$

Combine 2\sqrt{10} and -4\sqrt{10} to get -2\sqrt{10}.

$$-7-2\sqrt{10}+\sqrt{90}+\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)$$

Factor 90=3^{2}\times 10. Rewrite the square root of the product \sqrt{3^{2}\times 10} as the product of square roots \sqrt{3^{2}}\sqrt{10}. Take the square root of 3^{2}.

$$-7-2\sqrt{10}+3\sqrt{10}+\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)$$

Combine -2\sqrt{10} and 3\sqrt{10} to get \sqrt{10}.

$$-7+\sqrt{10}+\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)$$

Consider \left(2\sqrt{2}-1\right)\left(2\sqrt{2}+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}. Square 1.

$$-7+\sqrt{10}+\left(2\sqrt{2}\right)^{2}-1$$

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

$$-7+\sqrt{10}+2^{2}\left(\sqrt{2}\right)^{2}-1$$

Calculate 2 to the power of 2 and get 4.

$$-7+\sqrt{10}+4\left(\sqrt{2}\right)^{2}-1$$

The square of \sqrt{2} is 2.

$$-7+\sqrt{10}+4\times 2-1$$

Multiply 4 and 2 to get 8.

$$-7+\sqrt{10}+8-1$$

Subtract 1 from 8 to get 7.

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

Add -7 and 7 to get 0.

$$\sqrt{10}$$