Solve sqrt{12}+1/2-sqrt{3}*(2+sqrt{3})^2=

$$\sqrt { 12 } + \frac { 1 } { 2 - \sqrt { 3 } } \cdot ( 2 + \sqrt { 3 } ) ^ { 2 } =$$

Evaluate

$$17\sqrt{3}+26\approx 55.444863729$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

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

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

Consider \left(2-\sqrt{3}\right)\left(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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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

Use the distributive property to multiply 2+\sqrt{3} by 7+4\sqrt{3} and combine like terms.

$$2\sqrt{3}+14+15\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$2\sqrt{3}+14+15\sqrt{3}+4\times 3$$

Multiply 4 and 3 to get 12.

$$2\sqrt{3}+14+15\sqrt{3}+12$$

Add 14 and 12 to get 26.

$$2\sqrt{3}+26+15\sqrt{3}$$

Combine 2\sqrt{3} and 15\sqrt{3} to get 17\sqrt{3}.

$$17\sqrt{3}+26$$