Solve 1/4-sqrt{5}+1/4+sqrt{5}

$$\frac { 1 } { 4 - \sqrt { 5 } } + \frac { 1 } { 4 + \sqrt { 5 } }$$

Evaluate

$$\frac{8}{11}\approx 0.727272727$$

Short Solution Steps

Rationalize the denominator of \frac{1}{4-\sqrt{5}} by multiplying numerator and denominator by 4+\sqrt{5}.

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

Consider \left(4-\sqrt{5}\right)\left(4+\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}.

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

Square 4. Square \sqrt{5}.

$$\frac{4+\sqrt{5}}{16-5}+\frac{1}{4+\sqrt{5}}$$

Subtract 5 from 16 to get 11.

$$\frac{4+\sqrt{5}}{11}+\frac{1}{4+\sqrt{5}}$$

Rationalize the denominator of \frac{1}{4+\sqrt{5}} by multiplying numerator and denominator by 4-\sqrt{5}.

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

Consider \left(4+\sqrt{5}\right)\left(4-\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}.

$$\frac{4+\sqrt{5}}{11}+\frac{4-\sqrt{5}}{4^{2}-\left(\sqrt{5}\right)^{2}}$$

Square 4. Square \sqrt{5}.

$$\frac{4+\sqrt{5}}{11}+\frac{4-\sqrt{5}}{16-5}$$

Subtract 5 from 16 to get 11.

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

Since \frac{4+\sqrt{5}}{11} and \frac{4-\sqrt{5}}{11} have the same denominator, add them by adding their numerators.

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

Do the calculations in 4+\sqrt{5}+4-\sqrt{5}.

$$\frac{8}{11}$$

Factor

$$\frac{2 ^ {3}}{11} = 0.7272727272727273$$