Solve sqrt[4]{49}+(sqrt{7}-2)^2+[10-(sqrt{7}-sqrt{3})^2]:sqrt{3}

$$\sqrt[ 4 ] { 49 } + ( \sqrt { 7 } - 2 ) ^ { 2 } + [ 10 - ( \sqrt { 7 } - \sqrt { 3 } ) ^ { 2 } ] : \sqrt { 3 }$$

Evaluate

$$11-\sqrt{7}\approx 8.354248689$$

Solution Steps

Rewrite \sqrt[4]{49} as \sqrt[4]{7^{2}}. Convert from radical to exponential form and cancel out 2 in the exponent. Convert back to radical form.

$$\sqrt[4]{49}=\sqrt[4]{7^{2}}=7^{\frac{2}{4}}=7^{\frac{1}{2}}=\sqrt{7}$$

Insert the obtained value back in the expression.

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

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

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

The square of \sqrt{7} is 7.

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

Add 7 and 4 to get 11.

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

Combine \sqrt{7} and -4\sqrt{7} to get -3\sqrt{7}.

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

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

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

The square of \sqrt{7} is 7.

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

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

$$-3\sqrt{7}+11+\frac{10-\left(7-2\sqrt{21}+\left(\sqrt{3}\right)^{2}\right)}{\sqrt{3}}$$

The square of \sqrt{3} is 3.

$$-3\sqrt{7}+11+\frac{10-\left(7-2\sqrt{21}+3\right)}{\sqrt{3}}$$

Add 7 and 3 to get 10.

$$-3\sqrt{7}+11+\frac{10-\left(10-2\sqrt{21}\right)}{\sqrt{3}}$$

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

$$-3\sqrt{7}+11+\frac{10-10+2\sqrt{21}}{\sqrt{3}}$$

Subtract 10 from 10 to get 0.

$$-3\sqrt{7}+11+\frac{2\sqrt{21}}{\sqrt{3}}$$

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

$$-3\sqrt{7}+11+\frac{2\sqrt{21}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$-3\sqrt{7}+11+\frac{2\sqrt{21}\sqrt{3}}{3}$$

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}.

$$-3\sqrt{7}+11+\frac{2\sqrt{3}\sqrt{7}\sqrt{3}}{3}$$

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

$$-3\sqrt{7}+11+\frac{2\times 3\sqrt{7}}{3}$$

Cancel out 3 and 3.

$$-3\sqrt{7}+11+2\sqrt{7}$$

Combine -3\sqrt{7} and 2\sqrt{7} to get -\sqrt{7}.

$$-\sqrt{7}+11$$