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

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

Evaluate

$$14$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 4 to get 7.

$$7+4\sqrt{3}+\frac{1}{\left(\sqrt{3}+2\right)^{2}}$$

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

$$7+4\sqrt{3}+\frac{1}{\left(\sqrt{3}\right)^{2}+4\sqrt{3}+4}$$

The square of \sqrt{3} is 3.

$$7+4\sqrt{3}+\frac{1}{3+4\sqrt{3}+4}$$

Add 3 and 4 to get 7.

$$7+4\sqrt{3}+\frac{1}{7+4\sqrt{3}}$$

Rationalize the denominator of \frac{1}{7+4\sqrt{3}} by multiplying numerator and denominator by 7-4\sqrt{3}.

$$7+4\sqrt{3}+\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}$$

Consider \left(7+4\sqrt{3}\right)\left(7-4\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}.

$$7+4\sqrt{3}+\frac{7-4\sqrt{3}}{7^{2}-\left(4\sqrt{3}\right)^{2}}$$

Calculate 7 to the power of 2 and get 49.

$$7+4\sqrt{3}+\frac{7-4\sqrt{3}}{49-\left(4\sqrt{3}\right)^{2}}$$

Expand \left(4\sqrt{3}\right)^{2}.

$$7+4\sqrt{3}+\frac{7-4\sqrt{3}}{49-4^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$7+4\sqrt{3}+\frac{7-4\sqrt{3}}{49-16\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$7+4\sqrt{3}+\frac{7-4\sqrt{3}}{49-16\times 3}$$

Multiply 16 and 3 to get 48.

$$7+4\sqrt{3}+\frac{7-4\sqrt{3}}{49-48}$$

Subtract 48 from 49 to get 1.

$$7+4\sqrt{3}+\frac{7-4\sqrt{3}}{1}$$

Anything divided by one gives itself.

$$7+4\sqrt{3}+7-4\sqrt{3}$$

Add 7 and 7 to get 14.

$$14+4\sqrt{3}-4\sqrt{3}$$

Combine 4\sqrt{3} and -4\sqrt{3} to get 0.

$$14$$

Factor

$$2\times 7$$