In aptitude (simplification: quadratic comparison), two quadratic equations in x and y are given. Solve both equations to obtain all real roots, then compare every possible pair of x and y values and choose the option that best describes the relationship between x and y. I. x^2 + 2x - 195 = 0 II. y^2 + 30y + 225 = 0

Introduction / Context:
This question again involves comparison of roots of two quadratic equations, which is a common pattern in banking and management aptitude exams. Instead of computing a single numeric answer, we must determine an inequality relation between x and y that holds for every possible root of the two equations. Such questions test understanding of quadratic roots, multiplicity, and careful logical comparison rather than simple arithmetic alone.

Given Data / Assumptions:
Equation I: x^2 + 2x - 195 = 0.
Equation II: y^2 + 30y + 225 = 0.
Both equations are standard quadratics in real variables x and y. All their real roots are considered possible values.

Concept / Approach:
The plan is to factorise each quadratic to obtain its roots. If equation II factors into a perfect square, then y will take only one repeated value. Once all roots are known, we compare every possible x value with that single y value. If all comparisons give x greater than or equal to y, the correct relation is x >= y. We must be careful to include equality cases as well as strict inequality cases while choosing the option.

Step-by-Step Solution:
For equation I: x^2 + 2x - 195 = 0. Factorise: x^2 + 2x - 195 = (x + 15)(x - 13) = 0. So x = -15 or x = 13. For equation II: y^2 + 30y + 225 = 0. Recognise a perfect square: y^2 + 30y + 225 = (y + 15)^2 = 0. So y = -15 (a repeated root). Now compare: when x = -15, y = -15 gives x = y. When x = 13, y = -15 gives x > y. Thus for all possible roots, x is always greater than or equal to y, and never less than y.

Verification / Alternative check:
We can verify using the sum and product of roots. For equation II, sum of roots is -30 and product is 225. This is consistent with a double root at -15 since -15 plus -15 is -30 and the product is 225. For equation I, the roots -15 and 13 clearly satisfy the factorisation found. Comparing numerically, the smallest x value equals y and the larger x is greater than y, so the relationship x >= y is firmly established.

Why Other Options Are Wrong:
Option x > y ignores the equality case when both x and y are -15. Option x < y is impossible because y is never greater than any x root. Option x = y fails because x can also be 13, which is strictly greater than y. The option relationship cannot be determined would be correct only if different combinations gave conflicting orderings, which does not happen here.

Common Pitfalls:
Many learners forget to consider both roots of the first quadratic and only compare a single pair. Others may misfactor the quadratics or overlook that equation II has a repeated root. Some students also confuse x <= y with x >= y when selecting options, so it is important to read the inequality direction carefully. Writing down all root values and comparing them systematically helps avoid such errors.

Final Answer:
For every possible pair, x is never less than y and is sometimes equal and sometimes greater. Therefore the correct relationship is x >= y.