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

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

Evaluate

$$-\sqrt{2}-2\approx -3.414213562$$

Solution Steps

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

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

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}.

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

Square \sqrt{2}. Square 3.

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

Subtract 9 from 2 to get -7.

$$\frac{\left(4+\sqrt{2}\right)\left(\sqrt{2}+3\right)}{-7}$$

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

$$\frac{4\sqrt{2}+12+\left(\sqrt{2}\right)^{2}+3\sqrt{2}}{-7}$$

The square of \sqrt{2} is 2.

$$\frac{4\sqrt{2}+12+2+3\sqrt{2}}{-7}$$

Add 12 and 2 to get 14.

$$\frac{4\sqrt{2}+14+3\sqrt{2}}{-7}$$

Combine 4\sqrt{2} and 3\sqrt{2} to get 7\sqrt{2}.

$$\frac{7\sqrt{2}+14}{-7}$$

Divide each term of 7\sqrt{2}+14 by -7 to get -\sqrt{2}-2.

$$-\sqrt{2}-2$$