If x and y are two positive numbers such that

​$$\sqrt{x} = 8$$ and $$x^2 + y = 4112,$$ then find the value of $$\sqrt{y}$$.​

Correct option is D

Given:

$$\sqrt{x} = 8$$ and $$x^2 + y = 4112$$​​

Solution:

​$$x^2 + y = 4112 \\\implies 64^2 + y = 4112 \quad (\sqrt{x} = 8, \, x = 64) \\\implies 4096 + y = 4112 \\\implies y = 16 \\\implies \sqrt{y} = 4$$

Hence, option (d) is the correct answer.