take (x+2y)=a
then the given equation reduces to
a2 + a + 2y=710.
clearly x+2y is an even number which implies x is even.
now we if we have to make sum equal to 710.
taking a=26 we get 2y=8.
now (x+2y)=26 → x=18.
Hence the answer is (c)
If x and y are integers such that
(x+2y)2 + (x+4y) = 710
The value of x is
(A) 13
(B) 15
(C) 18
(D) 26
take (x+2y)=a
then the given equation reduces to
a2 + a + 2y=710.
clearly x+2y is an even number which implies x is even.
now we if we have to make sum equal to 710.
taking a=26 we get 2y=8.
now (x+2y)=26 → x=18.
Hence the answer is (c)
x2 + 4xy + 4y2 + x + 4y = 710
x2 + (2y)2 + 12 + 2(x)(2y) + 2(2y)(1) + 2(1)(x) = 711 + x
(x+2y+1)2 = 711 + x
so, 711+x must be a perfect square for which x=18 from given options
(x,y) = (18,4)
How did you take a=26 to get 2y=8?
Is it possible to solve without using the options?
if x+2y is not even then what would hav been the problem in @dil sirs solution .