Imagine that it is a square of a binomial.
It would have to be of the form (X+Y)^2, where Y is some real number.
Using the FOIL method, expanding the square should get you...
X^2+XY+YX+Y^2, which you can simplify to X^2+2YX+Y^2.
So, X^2+50X+100 = X^2+2YX+Y^2
Y^2 must equal 100, and 2Y must equal 50.
But, there's no value for Y like that, is there?
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It is the square of an integer: that is what a perfect square is.
The square root of 1/100 is 1/10 which is a perfect square but perfect square are usually integers.
100 is a perfect squarebecause 100 = 10*10
Yes
No, it is not. For it to be a perfect square, the square root of it would have to be an integer, but √120 ≈ 10.954. 100 and 121 are perfect squares nearest to 120. ■