Solve 1/2-sqrt{3}+3/sqrt{3}-sqrt{27}

$$\frac { 1 } { 2 - \sqrt { 3 } } + \frac { 3 } { \sqrt { 3 } } - \sqrt { 27 }$$

Evaluate

$$2-\sqrt{3}\approx 0.267949192$$

Solution Steps

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

$$\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\frac{3}{\sqrt{3}}-\sqrt{27}$$

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}.

$$\frac{2+\sqrt{3}}{2^{2}-\left(\sqrt{3}\right)^{2}}+\frac{3}{\sqrt{3}}-\sqrt{27}$$

Square 2. Square \sqrt{3}.

$$\frac{2+\sqrt{3}}{4-3}+\frac{3}{\sqrt{3}}-\sqrt{27}$$

Subtract 3 from 4 to get 1.

$$\frac{2+\sqrt{3}}{1}+\frac{3}{\sqrt{3}}-\sqrt{27}$$

Anything divided by one gives itself.

$$2+\sqrt{3}+\frac{3}{\sqrt{3}}-\sqrt{27}$$

Rationalize the denominator of \frac{3}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$2+\sqrt{3}+\frac{3\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\sqrt{27}$$

The square of \sqrt{3} is 3.

$$2+\sqrt{3}+\frac{3\sqrt{3}}{3}-\sqrt{27}$$

Cancel out 3 and 3.

$$2+\sqrt{3}+\sqrt{3}-\sqrt{27}$$

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

$$2+2\sqrt{3}-\sqrt{27}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

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

Combine 2\sqrt{3} and -3\sqrt{3} to get -\sqrt{3}.

$$2-\sqrt{3}$$