Solve sqrt{49}+(sqrt{7}-2)^2+[10-(sqrt{7}-sqrt{3})^2]:sqrt{3}

$$\sqrt { 49 } + ( \sqrt { 7 } - 2 ) ^ { 2 } + [ 10 - ( \sqrt { 7 } - \sqrt { 3 } ) ^ { 2 } ] : \sqrt { 3 }$$

Evaluate

$$18-2\sqrt{7}\approx 12.708497378$$

Solution Steps

Calculate the square root of 49 and get 7.

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

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

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

The square of \sqrt{7} is 7.

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

Add 7 and 4 to get 11.

$$7+11-4\sqrt{7}+\frac{10-\left(\sqrt{7}-\sqrt{3}\right)^{2}}{\sqrt{3}}$$

Add 7 and 11 to get 18.

$$18-4\sqrt{7}+\frac{10-\left(\sqrt{7}-\sqrt{3}\right)^{2}}{\sqrt{3}}$$

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

$$18-4\sqrt{7}+\frac{10-\left(\left(\sqrt{7}\right)^{2}-2\sqrt{7}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)}{\sqrt{3}}$$

The square of \sqrt{7} is 7.

$$18-4\sqrt{7}+\frac{10-\left(7-2\sqrt{7}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)}{\sqrt{3}}$$

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

$$18-4\sqrt{7}+\frac{10-\left(7-2\sqrt{21}+\left(\sqrt{3}\right)^{2}\right)}{\sqrt{3}}$$

The square of \sqrt{3} is 3.

$$18-4\sqrt{7}+\frac{10-\left(7-2\sqrt{21}+3\right)}{\sqrt{3}}$$

Add 7 and 3 to get 10.

$$18-4\sqrt{7}+\frac{10-\left(10-2\sqrt{21}\right)}{\sqrt{3}}$$

To find the opposite of 10-2\sqrt{21}, find the opposite of each term.

$$18-4\sqrt{7}+\frac{10-10+2\sqrt{21}}{\sqrt{3}}$$

Subtract 10 from 10 to get 0.

$$18-4\sqrt{7}+\frac{2\sqrt{21}}{\sqrt{3}}$$

Rationalize the denominator of \frac{2\sqrt{21}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$18-4\sqrt{7}+\frac{2\sqrt{21}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$18-4\sqrt{7}+\frac{2\sqrt{21}\sqrt{3}}{3}$$

Factor 21=3\times 7. Rewrite the square root of the product \sqrt{3\times 7} as the product of square roots \sqrt{3}\sqrt{7}.

$$18-4\sqrt{7}+\frac{2\sqrt{3}\sqrt{7}\sqrt{3}}{3}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$18-4\sqrt{7}+\frac{2\times 3\sqrt{7}}{3}$$

Cancel out 3 and 3.

$$18-4\sqrt{7}+2\sqrt{7}$$

Combine -4\sqrt{7} and 2\sqrt{7} to get -2\sqrt{7}.

$$18-2\sqrt{7}$$