Solve (sqrt{3}+1)^2-(frac{sqrt{3}+1}{2})^2=

$$( \sqrt { 3 } + 1 ) ^ { 2 } - ( \frac { \sqrt { 3 } + 1 } { 2 } ) ^ { 2 } =$$

Evaluate

$$\frac{3\left(\sqrt{3}+2\right)}{2}\approx 5.598076211$$

Short Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+1\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

$$4+2\sqrt{3}-\left(\frac{\sqrt{3}+1}{2}\right)^{2}$$

To raise \frac{\sqrt{3}+1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$4+2\sqrt{3}-\frac{\left(\sqrt{3}+1\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+1\right)^{2}.

$$4+2\sqrt{3}-\frac{\left(\sqrt{3}\right)^{2}+2\sqrt{3}+1}{2^{2}}$$

The square of \sqrt{3} is 3.

$$4+2\sqrt{3}-\frac{3+2\sqrt{3}+1}{2^{2}}$$

Add 3 and 1 to get 4.

$$4+2\sqrt{3}-\frac{4+2\sqrt{3}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$4+2\sqrt{3}-\frac{4+2\sqrt{3}}{4}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4+2\sqrt{3} times \frac{4}{4}.

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

Since \frac{4\left(4+2\sqrt{3}\right)}{4} and \frac{4+2\sqrt{3}}{4} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 4\left(4+2\sqrt{3}\right)-\left(4+2\sqrt{3}\right).

$$\frac{16+8\sqrt{3}-4-2\sqrt{3}}{4}$$

Do the calculations in 16+8\sqrt{3}-4-2\sqrt{3}.

$$\frac{12+6\sqrt{3}}{4}$$

Expand

$$\frac{3 \sqrt{3}}{2} + 3 = 5.598076211$$

Short Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+1\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

$$4+2\sqrt{3}-\left(\frac{\sqrt{3}+1}{2}\right)^{2}$$

To raise \frac{\sqrt{3}+1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$4+2\sqrt{3}-\frac{\left(\sqrt{3}+1\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+1\right)^{2}.

$$4+2\sqrt{3}-\frac{\left(\sqrt{3}\right)^{2}+2\sqrt{3}+1}{2^{2}}$$

The square of \sqrt{3} is 3.

$$4+2\sqrt{3}-\frac{3+2\sqrt{3}+1}{2^{2}}$$

Add 3 and 1 to get 4.

$$4+2\sqrt{3}-\frac{4+2\sqrt{3}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$4+2\sqrt{3}-\frac{4+2\sqrt{3}}{4}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4+2\sqrt{3} times \frac{4}{4}.

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

Since \frac{4\left(4+2\sqrt{3}\right)}{4} and \frac{4+2\sqrt{3}}{4} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 4\left(4+2\sqrt{3}\right)-\left(4+2\sqrt{3}\right).

$$\frac{16+8\sqrt{3}-4-2\sqrt{3}}{4}$$

Do the calculations in 16+8\sqrt{3}-4-2\sqrt{3}.

$$\frac{12+6\sqrt{3}}{4}$$