Solve (sqrt{3}+1)^2+1/(sqrt{3+1)^2}

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

Evaluate

$$\frac{3\sqrt{3}}{2}+5\approx 7.598076211$$

Short Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

Rationalize the denominator of \frac{1}{4+2\sqrt{3}} by multiplying numerator and denominator by 4-2\sqrt{3}.

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

Consider \left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$4+2\sqrt{3}+\frac{4-2\sqrt{3}}{16-\left(2\sqrt{3}\right)^{2}}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$4+2\sqrt{3}+\frac{4-2\sqrt{3}}{16-2^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$4+2\sqrt{3}+\frac{4-2\sqrt{3}}{16-4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$4+2\sqrt{3}+\frac{4-2\sqrt{3}}{16-12}$$

Subtract 12 from 16 to 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, add them by adding their numerators.

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

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

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

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

$$\frac{20+6\sqrt{3}}{4}$$

Expand

$$\frac{3 \sqrt{3}}{2} + 5 = 7.598076211$$

Short Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

Rationalize the denominator of \frac{1}{4+2\sqrt{3}} by multiplying numerator and denominator by 4-2\sqrt{3}.

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

Consider \left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$4+2\sqrt{3}+\frac{4-2\sqrt{3}}{16-\left(2\sqrt{3}\right)^{2}}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$4+2\sqrt{3}+\frac{4-2\sqrt{3}}{16-2^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$4+2\sqrt{3}+\frac{4-2\sqrt{3}}{16-4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$4+2\sqrt{3}+\frac{4-2\sqrt{3}}{16-12}$$

Subtract 12 from 16 to 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, add them by adding their numerators.

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

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

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

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

$$\frac{20+6\sqrt{3}}{4}$$