Solve frac{-2-sqrt{3}}{4+3sqrt{2}}

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

Evaluate

$$-\frac{3\sqrt{6}}{2}+2\sqrt{3}+4-3\sqrt{2}\approx -0.452773686$$

Solution Steps

Rationalize the denominator of \frac{-2-\sqrt{3}}{4+3\sqrt{2}} by multiplying numerator and denominator by 4-3\sqrt{2}.

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

Consider \left(4+3\sqrt{2}\right)\left(4-3\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}.

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

Calculate 4 to the power of 2 and get 16.

$$\frac{\left(-2-\sqrt{3}\right)\left(4-3\sqrt{2}\right)}{16-\left(3\sqrt{2}\right)^{2}}$$

Expand \left(3\sqrt{2}\right)^{2}.

$$\frac{\left(-2-\sqrt{3}\right)\left(4-3\sqrt{2}\right)}{16-3^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(-2-\sqrt{3}\right)\left(4-3\sqrt{2}\right)}{16-9\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(-2-\sqrt{3}\right)\left(4-3\sqrt{2}\right)}{16-9\times 2}$$

Multiply 9 and 2 to get 18.

$$\frac{\left(-2-\sqrt{3}\right)\left(4-3\sqrt{2}\right)}{16-18}$$

Subtract 18 from 16 to get -2.

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

Apply the distributive property by multiplying each term of -2-\sqrt{3} by each term of 4-3\sqrt{2}.

$$\frac{-8+6\sqrt{2}-4\sqrt{3}+3\sqrt{3}\sqrt{2}}{-2}$$

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

$$\frac{-8+6\sqrt{2}-4\sqrt{3}+3\sqrt{6}}{-2}$$

Multiply both numerator and denominator by -1.

$$\frac{8-6\sqrt{2}+4\sqrt{3}-3\sqrt{6}}{2}$$