Find the lowest integer x?

Question

x is a 2-digit integer such that when you take the square root of the sum of x and its square, you get a number that's in the form a* sqrt(a+1). Find the least possible value of x

Answer 1

I'm assuming you need a to be an integer...

So let's try various values of a:
We need values where (a + 1) will be a square:

a = 0 --> 0
a = 3 --> 6
a = 8 --> 24
a = 15 --> 60
a = 24 --> 120
a = 35 --> 210
a = 48 --> 336

Let's make a similar list for values of x:
x = 10 --> 110
x = 11 --> 132
x = 12 --> 156
x = 13 --> 182
x = 14 --> 210

Bingo! We didn't have to go far...
It works when x = 14 (and a = 35)

14 + 196 = 210
35 * sqrt (36) = 210

So the answer is 14

Edit:

Actually, I revise my answer. I missed that it is the *square root* of x + x².

If you square both sides you get:
x ( x + 1 ) = a² ( a + 1 )

Table for values of a:
2
12
36
80
150
252
392
576
810
1100
1452
1872
2366
2940
3600
4352
5202
6156
7220
8400
9702

Table for values of x:
110
132
156
182
210
240
272
....
8742
8930
9120
9312
9506
9702
9900

They overlap at 9702 (x = 98, a = 21)

98 + (98²) = 9702
21² (22) = 9702

So I agree with the person above that the answer is x = 98

Answer 2

Because you have

sqrt( x + x^2) = a * sqrt( a+ 1)

square both sides

x + x^2 = a^2 + a

notice that the "x" and "a" are identical forms. So for all 2 digit integers you can find an "a" in fact it is the same so the smallest 2 digit integer would be 10.

Answer 3

98

Answer 4

I agree with bandf though not with his method.