What is the value of $$\frac{4 + \sqrt2}{\sqrt2 + 1}$$ ?​

Correct option is B

​Given:

$$\frac{4 + \sqrt{2}}{\sqrt{2} + 1}$$​

Solution:

Rationalize the denominator by multiplying numerator and denominator by $$\sqrt{2} - 1$$​​

​$$\frac{4 + \sqrt{2}}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{(4 + \sqrt{2})(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)}$$​​

Denominator:

​$$(\sqrt{2} + 1)(\sqrt{2} - 1) = (\sqrt{2})^2 - (1)^2 = 2 - 1 = 1$$​​

Expand the numerator:

​$$(4 + \sqrt{2})(\sqrt{2} - 1) = 4\sqrt{2} - 4 + \sqrt{2} \times \sqrt{2} - \sqrt{2}\\ \ \\ = 4\sqrt{2} - 4 + 2 - \sqrt{2}= 4\sqrt{2} - \sqrt{2} - 4 + 2 = 3\sqrt{2} - 2$$​​

thus the value of the expression is $$\bf 3\sqrt{2} - 2$$​​