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

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

Evaluate

$$\sqrt{14}+3\sqrt{7}-13\approx -1.32108868$$

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}.

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

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

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

The square of \sqrt{7} is 7.

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

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

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

The square of \sqrt{3} is 3.

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

Add 7 and 3 to get 10.

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

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

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

The square of \sqrt{21} is 21.

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

Add 21 and 4 to get 25.

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

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

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

Subtract 25 from 10 to get -15.

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

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

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

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}.

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

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

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

Use the distributive property to multiply \sqrt{2}+\sqrt{7} by \sqrt{2}.

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

The square of \sqrt{2} is 2.

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

To multiply \sqrt{7} and \sqrt{2}, multiply the numbers under the square root.

$$3\sqrt{7}-15+2+\sqrt{14}$$

Add -15 and 2 to get -13.

$$3\sqrt{7}-13+\sqrt{14}$$