Solve (2sqrt{3}+3sqrt{2})div(sqrt{3}+sqrt{2})

$$(2 \sqrt{ 3 } +3 \sqrt{ 2 } ) \div ( \sqrt{ 3 } + \sqrt{ 2 } )$$

Evaluate

$$\sqrt{6}\approx 2.449489743$$

Solution Steps

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

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

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

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

Square \sqrt{3}. Square \sqrt{2}.

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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

Apply the distributive property by multiplying each term of 2\sqrt{3}+3\sqrt{2} by each term of \sqrt{3}-\sqrt{2}.

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

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

Combine -2\sqrt{6} and 3\sqrt{6} to get \sqrt{6}.

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

The square of \sqrt{2} is 2.

$$6+\sqrt{6}-3\times 2$$

Multiply -3 and 2 to get -6.

$$6+\sqrt{6}-6$$

Subtract 6 from 6 to get 0.

$$\sqrt{6}$$