If$$(a+b\sqrt{2})^2=19+6\sqrt{2}$$​ then a is equal to:

Correct option is C

Given:
​$$(a + b\sqrt{2})^2 = 19 + 6\sqrt{2}$$​​
Formula Used:
​$$(a + b)^2 = a^2 + b^2 + 2ab$$​​
Solution:
​$$=> (a + b\sqrt{2})^2 = 19 + 6\sqrt{2}\\=> (a + b\sqrt{2})^2 = 1 + 18 + 2 × 3\sqrt{2}\\=> (a + b\sqrt{2})^2 = 12 + (3\sqrt{2})^2 + 2 × 1 × 3\sqrt{2}\\$$​​
By using $$(a + b)^2 = a^2 + b^2 + 2ab$$​​
​$$=> (a + b\sqrt{2})^2 = (1 + 3\sqrt{2})^2\\=> (a + b\sqrt{2}) = 1 + (3\sqrt{2})$$​​
Now comparing both sides we get,
=> a = 1 and b = 3
Hence option 3 is the correct answer.