Solve (sqrt{5}-2-5/sqrt{5-2})divfrac{(sqrt{5}+3)^2}{sqrt{5}+2}

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

Evaluate

$$-\sqrt{5}-1\approx -3.236067977$$

Solution Steps

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

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

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}.

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

Square \sqrt{5}. Square 2.

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

Subtract 4 from 5 to get 1.

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

Anything divided by one gives itself.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{5}+3\right)^{2}.

$$\frac{\sqrt{5}-2-5\left(\sqrt{5}+2\right)}{\frac{\left(\sqrt{5}\right)^{2}+6\sqrt{5}+9}{\sqrt{5}+2}}$$

The square of \sqrt{5} is 5.

$$\frac{\sqrt{5}-2-5\left(\sqrt{5}+2\right)}{\frac{5+6\sqrt{5}+9}{\sqrt{5}+2}}$$

Add 5 and 9 to get 14.

$$\frac{\sqrt{5}-2-5\left(\sqrt{5}+2\right)}{\frac{14+6\sqrt{5}}{\sqrt{5}+2}}$$

Rationalize the denominator of \frac{14+6\sqrt{5}}{\sqrt{5}+2} by multiplying numerator and denominator by \sqrt{5}-2.

$$\frac{\sqrt{5}-2-5\left(\sqrt{5}+2\right)}{\frac{\left(14+6\sqrt{5}\right)\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}$$

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}.

$$\frac{\sqrt{5}-2-5\left(\sqrt{5}+2\right)}{\frac{\left(14+6\sqrt{5}\right)\left(\sqrt{5}-2\right)}{\left(\sqrt{5}\right)^{2}-2^{2}}}$$

Square \sqrt{5}. Square 2.

$$\frac{\sqrt{5}-2-5\left(\sqrt{5}+2\right)}{\frac{\left(14+6\sqrt{5}\right)\left(\sqrt{5}-2\right)}{5-4}}$$

Subtract 4 from 5 to get 1.

$$\frac{\sqrt{5}-2-5\left(\sqrt{5}+2\right)}{\frac{\left(14+6\sqrt{5}\right)\left(\sqrt{5}-2\right)}{1}}$$

Anything divided by one gives itself.

$$\frac{\sqrt{5}-2-5\left(\sqrt{5}+2\right)}{\left(14+6\sqrt{5}\right)\left(\sqrt{5}-2\right)}$$

Use the distributive property to multiply 5 by \sqrt{5}+2.

$$\frac{\sqrt{5}-2-\left(5\sqrt{5}+10\right)}{\left(14+6\sqrt{5}\right)\left(\sqrt{5}-2\right)}$$

To find the opposite of 5\sqrt{5}+10, find the opposite of each term.

$$\frac{\sqrt{5}-2-5\sqrt{5}-10}{\left(14+6\sqrt{5}\right)\left(\sqrt{5}-2\right)}$$

Combine \sqrt{5} and -5\sqrt{5} to get -4\sqrt{5}.

$$\frac{-4\sqrt{5}-2-10}{\left(14+6\sqrt{5}\right)\left(\sqrt{5}-2\right)}$$

Subtract 10 from -2 to get -12.

$$\frac{-4\sqrt{5}-12}{\left(14+6\sqrt{5}\right)\left(\sqrt{5}-2\right)}$$

Use the distributive property to multiply 14+6\sqrt{5} by \sqrt{5}-2 and combine like terms.

$$\frac{-4\sqrt{5}-12}{2\sqrt{5}-28+6\left(\sqrt{5}\right)^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{-4\sqrt{5}-12}{2\sqrt{5}-28+6\times 5}$$

Multiply 6 and 5 to get 30.

$$\frac{-4\sqrt{5}-12}{2\sqrt{5}-28+30}$$

Add -28 and 30 to get 2.

$$\frac{-4\sqrt{5}-12}{2\sqrt{5}+2}$$

Rationalize the denominator of \frac{-4\sqrt{5}-12}{2\sqrt{5}+2} by multiplying numerator and denominator by 2\sqrt{5}-2.

$$\frac{\left(-4\sqrt{5}-12\right)\left(2\sqrt{5}-2\right)}{\left(2\sqrt{5}+2\right)\left(2\sqrt{5}-2\right)}$$

Consider \left(2\sqrt{5}+2\right)\left(2\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}.

$$\frac{\left(-4\sqrt{5}-12\right)\left(2\sqrt{5}-2\right)}{\left(2\sqrt{5}\right)^{2}-2^{2}}$$

Expand \left(2\sqrt{5}\right)^{2}.

$$\frac{\left(-4\sqrt{5}-12\right)\left(2\sqrt{5}-2\right)}{2^{2}\left(\sqrt{5}\right)^{2}-2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(-4\sqrt{5}-12\right)\left(2\sqrt{5}-2\right)}{4\left(\sqrt{5}\right)^{2}-2^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{\left(-4\sqrt{5}-12\right)\left(2\sqrt{5}-2\right)}{4\times 5-2^{2}}$$

Multiply 4 and 5 to get 20.

$$\frac{\left(-4\sqrt{5}-12\right)\left(2\sqrt{5}-2\right)}{20-2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(-4\sqrt{5}-12\right)\left(2\sqrt{5}-2\right)}{20-4}$$

Subtract 4 from 20 to get 16.

$$\frac{\left(-4\sqrt{5}-12\right)\left(2\sqrt{5}-2\right)}{16}$$

Use the distributive property to multiply -4\sqrt{5}-12 by 2\sqrt{5}-2 and combine like terms.

$$\frac{-8\left(\sqrt{5}\right)^{2}-16\sqrt{5}+24}{16}$$

The square of \sqrt{5} is 5.

$$\frac{-8\times 5-16\sqrt{5}+24}{16}$$

Multiply -8 and 5 to get -40.

$$\frac{-40-16\sqrt{5}+24}{16}$$

Add -40 and 24 to get -16.

$$\frac{-16-16\sqrt{5}}{16}$$

Divide each term of -16-16\sqrt{5} by 16 to get -1-\sqrt{5}.

$$-1-\sqrt{5}$$