Solve F(sqrt{5})=frac{2(sqrt{5})^2+1}{(sqrt{5})^2+(sqrt{5})-2}

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

Evaluate

$$\frac{33-11\sqrt{5}}{4}\approx 2.100813062$$

Solution Steps

The square of \sqrt{5} is 5.

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

Multiply 2 and 5 to get 10.

$$\frac{10+1}{\left(\sqrt{5}\right)^{2}+\sqrt{5}-2}$$

Add 10 and 1 to get 11.

$$\frac{11}{\left(\sqrt{5}\right)^{2}+\sqrt{5}-2}$$

The square of \sqrt{5} is 5.

$$\frac{11}{5+\sqrt{5}-2}$$

Subtract 2 from 5 to get 3.

$$\frac{11}{3+\sqrt{5}}$$

Rationalize the denominator of \frac{11}{3+\sqrt{5}} by multiplying numerator and denominator by 3-\sqrt{5}.

$$\frac{11\left(3-\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}$$

Consider \left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{11\left(3-\sqrt{5}\right)}{3^{2}-\left(\sqrt{5}\right)^{2}}$$

Square 3. Square \sqrt{5}.

$$\frac{11\left(3-\sqrt{5}\right)}{9-5}$$

Subtract 5 from 9 to get 4.

$$\frac{11\left(3-\sqrt{5}\right)}{4}$$

Use the distributive property to multiply 11 by 3-\sqrt{5}.

$$\frac{33-11\sqrt{5}}{4}$$