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

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

Evaluate

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Solution Steps

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

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

Consider \left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Cancel out 2 and 2.

$$\sqrt{3}-1-\frac{3}{\sqrt{3}}$$

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

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

The square of \sqrt{3} is 3.

$$\sqrt{3}-1-\frac{3\sqrt{3}}{3}$$

Cancel out 3 and 3.

$$\sqrt{3}-1-\sqrt{3}$$

Subtract \sqrt{3} from \sqrt{3} to get 0.

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Factor

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