In the following question, two quadratic equations are given: (1) 5a^2 - 18a + 9 = 0 (2) 3b^2 + 5b - 2 = 0 You must solve both equations and determine the correct relationship between a and b.

Introduction / Context:
Quadratic comparison questions require you to solve two quadratic equations and then compare all possible values of the variables. The key point is that each quadratic can have up to two real roots. We must compare every possible root of a with every possible root of b and then decide if a consistent relationship like a > b or a < b always holds. If the comparison changes depending on which roots we choose, then no definite relation can be established.

Given Data / Assumptions:

• Equation (1): 5a^2 - 18a + 9 = 0.
• Equation (2): 3b^2 + 5b - 2 = 0.
• We assume real roots only, because the comparison of magnitudes is usually asked in the real number domain.
• We must find all real roots of each equation.

Concept / Approach:
To compare a and b, we first solve each quadratic using factorisation or the quadratic formula. Once we find the two possible values for a and the two possible values for b, we check their numerical values. If every value of a is greater than every value of b, then a > b always holds. Similarly, we test for other relations. Calculating the roots accurately is crucial for this type of problem.

Step-by-Step Solution:
Step 1: Solve 5a^2 - 18a + 9 = 0. Using the quadratic formula, a = [18 ± sqrt(18^2 - 4 * 5 * 9)] / (2 * 5).Step 2: Compute the discriminant: 18^2 = 324, and 4 * 5 * 9 = 180, so the discriminant is 324 - 180 = 144.Step 3: Square root of 144 is 12. So a = [18 ± 12] / 10.Step 4: That gives two roots: a1 = (18 + 12) / 10 = 30 / 10 = 3, and a2 = (18 - 12) / 10 = 6 / 10 = 3 / 5.Step 5: Now solve 3b^2 + 5b - 2 = 0. Using the quadratic formula, b = [-5 ± sqrt(5^2 - 4 * 3 * (-2))] / (2 * 3).Step 6: Compute the discriminant: 5^2 = 25, and -4 * 3 * (-2) = -24, so 25 - (-24) becomes 25 + 24 = 49.Step 7: Square root of 49 is 7. So b = [-5 ± 7] / 6.Step 8: The two roots are b1 = (-5 + 7) / 6 = 2 / 6 = 1 / 3, and b2 = (-5 - 7) / 6 = -12 / 6 = -2.Step 9: Compare each a root with each b root. The smaller value of a is 3 / 5 which is 0.6. The larger value of b is 1 / 3 which is approximately 0.33.Step 10: Clearly 0.6 is greater than 0.33 and also greater than -2. The other value of a is 3, which is greater than both 1 / 3 and -2 as well.Step 11: Therefore, in every possible pairing, a is always greater than b.

Verification / Alternative check:
List the sets explicitly: possible values of a are {3, 0.6}. Possible values of b are {1 / 3, -2}.Compare: 3 > 1 / 3 and 3 > -2. Also, 0.6 > 1 / 3 and 0.6 > -2.Because all four comparisons give a > b, the relation a > b holds for all possible roots.

Why Other Options Are Wrong:
Option a, a >= b, is true but weaker than a > b. Typically, the test expects the strongest always true relation, which is a > b.Option b and option c suggest that a can be less than or equal to b, which is contradicted by the actual numerical values.Option e claims that no definite relation can be established, but our comparison clearly shows a consistent ordering.

Common Pitfalls:
Solving one equation incorrectly and then drawing conclusions from wrong roots.Comparing just one pair of roots instead of every possible combination.Assuming that if both equations are quadratic, the relation will always be indeterminate, which is not true.

Final Answer:
Since each possible value of a is greater than each possible value of b, the correct relationship is a > b.