If $$x = \sqrt{13 + 2\sqrt{40}}$$​ and $$y = \sqrt{13 - 2\sqrt{40}}$$​ , then what is the value of (x - y) ?

Correct option is C

Given:

​$$x = \sqrt{13 + 2\sqrt{40}} ​​$$​​

​$$y = \sqrt{13 - 2\sqrt{40}} ​$$​​

We need to find the value of(x - y)

Formula Used:

​$$(a + b)(a - b) = a^2 - b^2$$​​

$$(a - b)^2 = a^2 + b^2 – 2ab$$​​

Solution:

​$$x^2 + y^2 = (13 + 2\sqrt{40}) + (13 - 2\sqrt{40}) = 26$$​​

​$$xy = \sqrt{(13 + 2\sqrt{40})(13 - 2\sqrt{40})} ​$$​​

Using the identity $$(a + b)(a - b) = a^2 - b^2:$$​​

​$$xy = \sqrt{13^2 - (2\sqrt{40})^2} = \sqrt{169 - 4 \times 40} = \sqrt{169 - 160} = \sqrt{9} = 3$$​​

Now, using the formula for $$(x - y)^2:$$ 

​$$​​(x - y)^2 = x^2 + y^2 - 2xy = 26 - 2 \times 3 = 26 - 6 = 20$$​​

Therefore, $$x - y = \sqrt{20} = 2\sqrt{5}$$

Alternate Solution: ​

​$$x = \sqrt{13 + 2\sqrt{40}} ​​ \\ \ \\ = \sqrt{(\sqrt6)^2+(\sqrt5)^2 + 2(\sqrt{5}\sqrt6)} \\ \ \\ = \sqrt{(\sqrt 6+\sqrt5)^2} \\ \ \\ =\sqrt6 +\sqrt5$$

$$y = \sqrt{13 - 2\sqrt{40}} ​​ \\ \ \\ = \sqrt{(\sqrt6)^2+(\sqrt5)^2 - 2(\sqrt{5}\sqrt6)} \\ \ \\ = \sqrt{(\sqrt 6-\sqrt5)^2} \\ \ \\ =\sqrt6 -\sqrt5$$ 

Now, 

​$$x-y = (\sqrt6 +\sqrt5) - (\sqrt6 -\sqrt5) = \sqrt6 +\sqrt5 -\sqrt6 + \sqrt5 = 2\sqrt5$$​​

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