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

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

Evaluate

$$\sqrt{5}\approx 2.236067977$$

Short Solution Steps

To raise \frac{\sqrt{5}+1}{2} to a power, raise both numerator and denominator to the power and then divide.

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

To raise \frac{\sqrt{5}-1}{2} to a power, raise both numerator and denominator to the power and then divide.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-1\right)^{2}.

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

The square of \sqrt{5} is 5.

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

Add 5 and 1 to get 6.

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

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{2^{2}}-\frac{6-2\sqrt{5}}{4}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 2^{2}.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{4}-\frac{6-2\sqrt{5}}{4}$$

Since \frac{\left(\sqrt{5}+1\right)^{2}}{4} and \frac{6-2\sqrt{5}}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(\sqrt{5}+1\right)^{2}-\left(6-2\sqrt{5}\right)}{4}$$

Do the multiplications in \left(\sqrt{5}+1\right)^{2}-\left(6-2\sqrt{5}\right).

$$\frac{\left(\sqrt{5}\right)^{2}+2\sqrt{5}+1-6+2\sqrt{5}}{4}$$

Do the calculations in \left(\sqrt{5}\right)^{2}+2\sqrt{5}+1-6+2\sqrt{5}.

$$\frac{4\sqrt{5}}{4}$$

Cancel out 4 and 4.

$$\sqrt{5}$$

Expand

$$\sqrt{5} = 2.236067977$$

Short Solution Steps

To raise \frac{\sqrt{5}+1}{2} to a power, raise both numerator and denominator to the power and then divide.

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

To raise \frac{\sqrt{5}-1}{2} to a power, raise both numerator and denominator to the power and then divide.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-1\right)^{2}.

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

The square of \sqrt{5} is 5.

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

Add 5 and 1 to get 6.

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

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{2^{2}}-\frac{6-2\sqrt{5}}{4}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 2^{2}.

$$\frac{\left(\sqrt{5}+1\right)^{2}}{4}-\frac{6-2\sqrt{5}}{4}$$

Since \frac{\left(\sqrt{5}+1\right)^{2}}{4} and \frac{6-2\sqrt{5}}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(\sqrt{5}+1\right)^{2}-\left(6-2\sqrt{5}\right)}{4}$$

Do the multiplications in \left(\sqrt{5}+1\right)^{2}-\left(6-2\sqrt{5}\right).

$$\frac{\left(\sqrt{5}\right)^{2}+2\sqrt{5}+1-6+2\sqrt{5}}{4}$$

Do the calculations in \left(\sqrt{5}\right)^{2}+2\sqrt{5}+1-6+2\sqrt{5}.

$$\frac{4\sqrt{5}}{4}$$

Cancel out 4 and 4.

$$\sqrt{5}$$