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

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

Evaluate

$$\frac{3\sqrt{2}+3\sqrt{3}-\sqrt{6}-9}{7}\approx -0.287242376$$

Solution Steps

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

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

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

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

Square \sqrt{2}. Square 3.

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

Subtract 9 from 2 to get -7.

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

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

$$\frac{\sqrt{3}\sqrt{2}-3\sqrt{3}-3\sqrt{2}+9}{-7}$$

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

$$\frac{\sqrt{6}-3\sqrt{3}-3\sqrt{2}+9}{-7}$$

Multiply both numerator and denominator by -1.

$$\frac{-\sqrt{6}+3\sqrt{3}+3\sqrt{2}-9}{7}$$