Solve sqrt{63}+(sqrt{7}-sqrt{3})^2-(sqrt{21}-2)^2-sqrt{84}+(sqrt{2}+sqrt{7})*(sqrt{2}-sqrt{7})+sqrt{400}

$$\sqrt { 63 } + ( \sqrt { 7 } - \sqrt { 3 } ) ^ { 2 } - ( \sqrt { 21 } - 2 ) ^ { 2 } - \sqrt { 84 } + ( \sqrt { 2 } + \sqrt { 7 } ) \cdot ( \sqrt { 2 } - \sqrt { 7 } ) + \sqrt { 400 }$$

Evaluate

$$3\sqrt{7}\approx 7.937253933$$

Solution Steps

Factor 63=3^{2}\times 7. Rewrite the square root of the product \sqrt{3^{2}\times 7} as the product of square roots \sqrt{3^{2}}\sqrt{7}. Take the square root of 3^{2}.

$$3\sqrt{7}+\left(\sqrt{7}-\sqrt{3}\right)^{2}-\left(\sqrt{21}-2\right)^{2}-\sqrt{84}+\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)+\sqrt{400}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{7}-\sqrt{3}\right)^{2}.

$$3\sqrt{7}+\left(\sqrt{7}\right)^{2}-2\sqrt{7}\sqrt{3}+\left(\sqrt{3}\right)^{2}-\left(\sqrt{21}-2\right)^{2}-\sqrt{84}+\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)+\sqrt{400}$$

The square of \sqrt{7} is 7.

$$3\sqrt{7}+7-2\sqrt{7}\sqrt{3}+\left(\sqrt{3}\right)^{2}-\left(\sqrt{21}-2\right)^{2}-\sqrt{84}+\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)+\sqrt{400}$$

To multiply \sqrt{7} and \sqrt{3}, multiply the numbers under the square root.

$$3\sqrt{7}+7-2\sqrt{21}+\left(\sqrt{3}\right)^{2}-\left(\sqrt{21}-2\right)^{2}-\sqrt{84}+\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)+\sqrt{400}$$

The square of \sqrt{3} is 3.

$$3\sqrt{7}+7-2\sqrt{21}+3-\left(\sqrt{21}-2\right)^{2}-\sqrt{84}+\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)+\sqrt{400}$$

Add 7 and 3 to get 10.

$$3\sqrt{7}+10-2\sqrt{21}-\left(\sqrt{21}-2\right)^{2}-\sqrt{84}+\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)+\sqrt{400}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{21}-2\right)^{2}.

$$3\sqrt{7}+10-2\sqrt{21}-\left(\left(\sqrt{21}\right)^{2}-4\sqrt{21}+4\right)-\sqrt{84}+\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)+\sqrt{400}$$

The square of \sqrt{21} is 21.

$$3\sqrt{7}+10-2\sqrt{21}-\left(21-4\sqrt{21}+4\right)-\sqrt{84}+\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)+\sqrt{400}$$

Add 21 and 4 to get 25.

$$3\sqrt{7}+10-2\sqrt{21}-\left(25-4\sqrt{21}\right)-\sqrt{84}+\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)+\sqrt{400}$$

To find the opposite of 25-4\sqrt{21}, find the opposite of each term.

$$3\sqrt{7}+10-2\sqrt{21}-25+4\sqrt{21}-\sqrt{84}+\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)+\sqrt{400}$$

Subtract 25 from 10 to get -15.

$$3\sqrt{7}-15-2\sqrt{21}+4\sqrt{21}-\sqrt{84}+\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)+\sqrt{400}$$

Combine -2\sqrt{21} and 4\sqrt{21} to get 2\sqrt{21}.

$$3\sqrt{7}-15+2\sqrt{21}-\sqrt{84}+\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)+\sqrt{400}$$

Factor 84=2^{2}\times 21. Rewrite the square root of the product \sqrt{2^{2}\times 21} as the product of square roots \sqrt{2^{2}}\sqrt{21}. Take the square root of 2^{2}.

$$3\sqrt{7}-15+2\sqrt{21}-2\sqrt{21}+\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)+\sqrt{400}$$

Combine 2\sqrt{21} and -2\sqrt{21} to get 0.

$$3\sqrt{7}-15+\left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right)+\sqrt{400}$$

Consider \left(\sqrt{2}+\sqrt{7}\right)\left(\sqrt{2}-\sqrt{7}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$3\sqrt{7}-15+\left(\sqrt{2}\right)^{2}-\left(\sqrt{7}\right)^{2}+\sqrt{400}$$

The square of \sqrt{2} is 2.

$$3\sqrt{7}-15+2-\left(\sqrt{7}\right)^{2}+\sqrt{400}$$

The square of \sqrt{7} is 7.

$$3\sqrt{7}-15+2-7+\sqrt{400}$$

Subtract 7 from 2 to get -5.

$$3\sqrt{7}-15-5+\sqrt{400}$$

Subtract 5 from -15 to get -20.

$$3\sqrt{7}-20+\sqrt{400}$$

Calculate the square root of 400 and get 20.

$$3\sqrt{7}-20+20$$

Add -20 and 20 to get 0.

$$3\sqrt{7}$$