Solve frac{2sqrt{3}}{7+sqrt{6}}

$$\frac { 2 \sqrt { 3 } } { 7 + \sqrt { 6 } }$$

Evaluate

$$\frac{14\sqrt{3}-6\sqrt{2}}{43}\approx 0.366591394$$

Solution Steps

Rationalize the denominator of \frac{2\sqrt{3}}{7+\sqrt{6}} by multiplying numerator and denominator by 7-\sqrt{6}.

$$\frac{2\sqrt{3}\left(7-\sqrt{6}\right)}{\left(7+\sqrt{6}\right)\left(7-\sqrt{6}\right)}$$

Consider \left(7+\sqrt{6}\right)\left(7-\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{2\sqrt{3}\left(7-\sqrt{6}\right)}{7^{2}-\left(\sqrt{6}\right)^{2}}$$

Square 7. Square \sqrt{6}.

$$\frac{2\sqrt{3}\left(7-\sqrt{6}\right)}{49-6}$$

Subtract 6 from 49 to get 43.

$$\frac{2\sqrt{3}\left(7-\sqrt{6}\right)}{43}$$

Use the distributive property to multiply 2\sqrt{3} by 7-\sqrt{6}.

$$\frac{14\sqrt{3}-2\sqrt{3}\sqrt{6}}{43}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$\frac{14\sqrt{3}-2\sqrt{3}\sqrt{3}\sqrt{2}}{43}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{14\sqrt{3}-2\times 3\sqrt{2}}{43}$$

Multiply -2 and 3 to get -6.

$$\frac{14\sqrt{3}-6\sqrt{2}}{43}$$