Solve frac{sqrt{5}+2}{sqrt{5-2}}*frac{sqrt{5}-2}{sqrt{5}+2}

$$\frac{ \sqrt{ 5 } +2 }{ \sqrt{ 5-2 } } \times \frac{ \sqrt{ 5 } -2 }{ \sqrt{ 5 } +2 }$$

Evaluate

$$\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{3}\approx 0.13629391$$

Solution Steps

Subtract 2 from 5 to get 3.

$$\frac{\sqrt{5}+2}{\sqrt{3}}\times \frac{\sqrt{5}-2}{\sqrt{5}+2}$$

Rationalize the denominator of \frac{\sqrt{5}+2}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

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

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

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

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}.

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

Square \sqrt{5}. Square 2.

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

Subtract 4 from 5 to get 1.

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

Anything divided by one gives itself.

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

Multiply \sqrt{5}-2 and \sqrt{5}-2 to get \left(\sqrt{5}-2\right)^{2}.

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

Express \frac{\left(\sqrt{5}+2\right)\sqrt{3}}{3}\left(\sqrt{5}-2\right)^{2} as a single fraction.

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

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

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

The square of \sqrt{5} is 5.

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

Add 5 and 4 to get 9.

$$\frac{\left(\sqrt{5}+2\right)\sqrt{3}\left(9-4\sqrt{5}\right)}{3}$$

Use the distributive property to multiply \sqrt{5}+2 by \sqrt{3}.

$$\frac{\left(\sqrt{5}\sqrt{3}+2\sqrt{3}\right)\left(9-4\sqrt{5}\right)}{3}$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$\frac{\left(\sqrt{15}+2\sqrt{3}\right)\left(9-4\sqrt{5}\right)}{3}$$

Apply the distributive property by multiplying each term of \sqrt{15}+2\sqrt{3} by each term of 9-4\sqrt{5}.

$$\frac{9\sqrt{15}-4\sqrt{15}\sqrt{5}+18\sqrt{3}-8\sqrt{3}\sqrt{5}}{3}$$

Factor 15=5\times 3. Rewrite the square root of the product \sqrt{5\times 3} as the product of square roots \sqrt{5}\sqrt{3}.

$$\frac{9\sqrt{15}-4\sqrt{5}\sqrt{3}\sqrt{5}+18\sqrt{3}-8\sqrt{3}\sqrt{5}}{3}$$

Multiply \sqrt{5} and \sqrt{5} to get 5.

$$\frac{9\sqrt{15}-4\times 5\sqrt{3}+18\sqrt{3}-8\sqrt{3}\sqrt{5}}{3}$$

Multiply -4 and 5 to get -20.

$$\frac{9\sqrt{15}-20\sqrt{3}+18\sqrt{3}-8\sqrt{3}\sqrt{5}}{3}$$

Combine -20\sqrt{3} and 18\sqrt{3} to get -2\sqrt{3}.

$$\frac{9\sqrt{15}-2\sqrt{3}-8\sqrt{3}\sqrt{5}}{3}$$

To multiply \sqrt{3} and \sqrt{5}, multiply the numbers under the square root.

$$\frac{9\sqrt{15}-2\sqrt{3}-8\sqrt{15}}{3}$$

Combine 9\sqrt{15} and -8\sqrt{15} to get \sqrt{15}.

$$\frac{\sqrt{15}-2\sqrt{3}}{3}$$