Finding the root of the problem

$\text{Let 2015 = a . Now the given expression can be written as :}$

$\huge{\sqrt{a^2+4a+4} \\ = \sqrt{(a+2)^2} \\ = a+2 \\ = 2015+2 \\ = \boxed{2017}}$

√(2015 ×2015+8064)

√(2015 ×2015+4×2016)

√((2017-2)×(2017-2)+4×(2017-1))

√(2017×2017+4-2×2017+4×(2017-1))

√(2017×2017+4-2×2×2017+4×2017-4)

√(2017×2017+4-4×2017+4×2017-4)

√(2017×2017)

2017

Grab a Calculator, Work it out

Let c = the answer. Let a = 2015 and b = 8064. So, c^2 - a^2 = 8064. Factor to get (c+a) (c-a) = 8064. Substituting back we get, c = some positive number as the terms are all positive in the addition and multiplication. (c>0 + 2015) (c>0 - 2015) = 8064. c < 2015 leads to the product being negative. c = 2015 makes the product 0. Neither of these work as 8064 is positive. Thus, we know quickly that 8064 factors into a big number greater than 2015. c = 2017 works by the first few factorings, since (2017 + 2015 = 4032) (2017 - 2015= 2) = 8064.

Nihar you should have mentioned this first: (2015²+4*2016)½ ={2015²+4(2015+1)}½ Now taking 2015=a we get, ={a²+4(a+1)}½ =(a²+4a+4)½ ={(a+2)²}½ =a+2 =2015+2 =2017

√(2015 ×2015+8064) √(2015 ×2015+4×2016) √((2017-2)×(2017-2)+4×(2017-1)) √(2017×2017+4-2×2017+4×(2017-1)) √(2017×2017+4-2×2×2017+4×2017-4) √(2017×2017+4-4×2017+4×2017-4) √(2017×2017) 2017

The trick is convert the given expression into a perfect square. Looking at 2015 x 2015, lets say we have a^2. Now evaluate the given constant so that we can get 'b' and proceed with a^2 + b^2 + 2.a.b

8064 - 2x2015 = 2017 , can be written as (2015 + 2)

Ok lets get back to original expression

2015x2015 + 2(2015 + 2017) => 2015 x 2015 + 2(2015 + 2015 + 2) we have (2015 + 2) ^2 i.e. sqrt(2017^2)