Solve (1/sqrt{5-sqrt{2}}+1/sqrt{5+sqrt{4}})^2

$$( \frac { 1 } { \sqrt { 5 } - \sqrt { 2 } } + \frac { 1 } { \sqrt { 5 } + \sqrt { 4 } } ) ^ { 2 }$$

Evaluate

$$\frac{8\sqrt{10}}{9}-\frac{4\sqrt{2}}{3}-\frac{16\sqrt{5}}{3}+\frac{118}{9}\approx 2.110710624$$

Short Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{5}-\sqrt{2}} by multiplying numerator and denominator by \sqrt{5}+\sqrt{2}.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}+\frac{1}{\sqrt{5}+\sqrt{4}}\right)^{2}$$

Consider \left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{2}\right)^{2}}+\frac{1}{\sqrt{5}+\sqrt{4}}\right)^{2}$$

Square \sqrt{5}. Square \sqrt{2}.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{5-2}+\frac{1}{\sqrt{5}+\sqrt{4}}\right)^{2}$$

Subtract 2 from 5 to get 3.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{1}{\sqrt{5}+\sqrt{4}}\right)^{2}$$

Calculate the square root of 4 and get 2.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{1}{\sqrt{5}+2}\right)^{2}$$

Rationalize the denominator of \frac{1}{\sqrt{5}+2} by multiplying numerator and denominator by \sqrt{5}-2.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{\sqrt{5}-2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\right)^{2}$$

Consider \left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{\sqrt{5}-2}{\left(\sqrt{5}\right)^{2}-2^{2}}\right)^{2}$$

Square \sqrt{5}. Square 2.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{\sqrt{5}-2}{5-4}\right)^{2}$$

Subtract 4 from 5 to get 1.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{\sqrt{5}-2}{1}\right)^{2}$$

Anything divided by one gives itself.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\sqrt{5}-2\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{5}-2 times \frac{3}{3}.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{3\left(\sqrt{5}-2\right)}{3}\right)^{2}$$

Since \frac{\sqrt{5}+\sqrt{2}}{3} and \frac{3\left(\sqrt{5}-2\right)}{3} have the same denominator, add them by adding their numerators.

$$\left(\frac{\sqrt{5}+\sqrt{2}+3\left(\sqrt{5}-2\right)}{3}\right)^{2}$$

Do the multiplications in \sqrt{5}+\sqrt{2}+3\left(\sqrt{5}-2\right).

$$\left(\frac{\sqrt{5}+\sqrt{2}+3\sqrt{5}-6}{3}\right)^{2}$$

Do the calculations in \sqrt{5}+\sqrt{2}+3\sqrt{5}-6.

$$\left(\frac{4\sqrt{5}+\sqrt{2}-6}{3}\right)^{2}$$

To raise \frac{4\sqrt{5}+\sqrt{2}-6}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(4\sqrt{5}+\sqrt{2}-6\right)^{2}}{3^{2}}$$

Square 4\sqrt{5}+\sqrt{2}-6.

$$\frac{8\sqrt{2}\sqrt{5}+16\left(\sqrt{5}\right)^{2}+\left(\sqrt{2}\right)^{2}-48\sqrt{5}-12\sqrt{2}+36}{3^{2}}$$

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

$$\frac{8\sqrt{10}+16\left(\sqrt{5}\right)^{2}+\left(\sqrt{2}\right)^{2}-48\sqrt{5}-12\sqrt{2}+36}{3^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{8\sqrt{10}+16\times 5+\left(\sqrt{2}\right)^{2}-48\sqrt{5}-12\sqrt{2}+36}{3^{2}}$$

Multiply 16 and 5 to get 80.

$$\frac{8\sqrt{10}+80+\left(\sqrt{2}\right)^{2}-48\sqrt{5}-12\sqrt{2}+36}{3^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{8\sqrt{10}+80+2-48\sqrt{5}-12\sqrt{2}+36}{3^{2}}$$

Add 80 and 2 to get 82.

$$\frac{8\sqrt{10}+82-48\sqrt{5}-12\sqrt{2}+36}{3^{2}}$$

Add 82 and 36 to get 118.

$$\frac{8\sqrt{10}+118-48\sqrt{5}-12\sqrt{2}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{8\sqrt{10}+118-48\sqrt{5}-12\sqrt{2}}{9}$$

Expand

$$\frac{8 \sqrt{10}}{9} - \frac{4 \sqrt{2}}{3} - \frac{16 \sqrt{5}}{3} + \frac{118}{9} = 2.110710624$$

Short Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{5}-\sqrt{2}} by multiplying numerator and denominator by \sqrt{5}+\sqrt{2}.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}+\frac{1}{\sqrt{5}+\sqrt{4}}\right)^{2}$$

Consider \left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{2}\right)^{2}}+\frac{1}{\sqrt{5}+\sqrt{4}}\right)^{2}$$

Square \sqrt{5}. Square \sqrt{2}.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{5-2}+\frac{1}{\sqrt{5}+\sqrt{4}}\right)^{2}$$

Subtract 2 from 5 to get 3.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{1}{\sqrt{5}+\sqrt{4}}\right)^{2}$$

Calculate the square root of 4 and get 2.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{1}{\sqrt{5}+2}\right)^{2}$$

Rationalize the denominator of \frac{1}{\sqrt{5}+2} by multiplying numerator and denominator by \sqrt{5}-2.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{\sqrt{5}-2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\right)^{2}$$

Consider \left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{\sqrt{5}-2}{\left(\sqrt{5}\right)^{2}-2^{2}}\right)^{2}$$

Square \sqrt{5}. Square 2.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{\sqrt{5}-2}{5-4}\right)^{2}$$

Subtract 4 from 5 to get 1.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{\sqrt{5}-2}{1}\right)^{2}$$

Anything divided by one gives itself.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\sqrt{5}-2\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{5}-2 times \frac{3}{3}.

$$\left(\frac{\sqrt{5}+\sqrt{2}}{3}+\frac{3\left(\sqrt{5}-2\right)}{3}\right)^{2}$$

Since \frac{\sqrt{5}+\sqrt{2}}{3} and \frac{3\left(\sqrt{5}-2\right)}{3} have the same denominator, add them by adding their numerators.

$$\left(\frac{\sqrt{5}+\sqrt{2}+3\left(\sqrt{5}-2\right)}{3}\right)^{2}$$

Do the multiplications in \sqrt{5}+\sqrt{2}+3\left(\sqrt{5}-2\right).

$$\left(\frac{\sqrt{5}+\sqrt{2}+3\sqrt{5}-6}{3}\right)^{2}$$

Do the calculations in \sqrt{5}+\sqrt{2}+3\sqrt{5}-6.

$$\left(\frac{4\sqrt{5}+\sqrt{2}-6}{3}\right)^{2}$$

To raise \frac{4\sqrt{5}+\sqrt{2}-6}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(4\sqrt{5}+\sqrt{2}-6\right)^{2}}{3^{2}}$$

Square 4\sqrt{5}+\sqrt{2}-6.

$$\frac{8\sqrt{2}\sqrt{5}+16\left(\sqrt{5}\right)^{2}+\left(\sqrt{2}\right)^{2}-48\sqrt{5}-12\sqrt{2}+36}{3^{2}}$$

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

$$\frac{8\sqrt{10}+16\left(\sqrt{5}\right)^{2}+\left(\sqrt{2}\right)^{2}-48\sqrt{5}-12\sqrt{2}+36}{3^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{8\sqrt{10}+16\times 5+\left(\sqrt{2}\right)^{2}-48\sqrt{5}-12\sqrt{2}+36}{3^{2}}$$

Multiply 16 and 5 to get 80.

$$\frac{8\sqrt{10}+80+\left(\sqrt{2}\right)^{2}-48\sqrt{5}-12\sqrt{2}+36}{3^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{8\sqrt{10}+80+2-48\sqrt{5}-12\sqrt{2}+36}{3^{2}}$$

Add 80 and 2 to get 82.

$$\frac{8\sqrt{10}+82-48\sqrt{5}-12\sqrt{2}+36}{3^{2}}$$

Add 82 and 36 to get 118.

$$\frac{8\sqrt{10}+118-48\sqrt{5}-12\sqrt{2}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{8\sqrt{10}+118-48\sqrt{5}-12\sqrt{2}}{9}$$