Surd decomposition and parameter sum Given that √28 − 6√3 is attempted to be written in the form √3·a + b with a, b rational numbers, determine the value of (a + b).

Introduction / Context:
This question explores expressing a radical combination in a restricted “linear in √3” form. We test whether √28 − 6√3 can be written as √3·a + b where a and b are rational numbers, and then interpret what (a + b) would be. The core idea is to compare like radicals and rational parts consistently.

Given Data / Assumptions:

• Expression: √28 − 6√3.
• Target form: √3·a + b with a, b ∈ rationals.
• We treat √7 and √3 as linearly independent over the rationals (they cannot be converted into each other with rational coefficients).

Concept / Approach:
Break √28 as 2√7. Then √28 − 6√3 = 2√7 − 6√3. If the target is √3·a + b, then any √7 term must vanish or be representable using √3 and rationals, which is not possible since √7 is irrational and independent from √3 over Q.

Step-by-Step Solution:
Rewrite √28 = 2√7, so the expression is 2√7 − 6√3.Target form is a√3 + b with a, b rational.There is no way to express 2√7 as q√3 + r for rational q, r, because √7 and √3 are distinct quadratic irrationals.Therefore, the proposed representation does not exist with rational a and b.

Verification / Alternative check:
Assume, for contradiction, 2√7 = a√3 + b. Squaring gives 28 = 3a^2 + b^2 + 2ab√3. The left side has no √3 term; hence 2ab must be 0. If ab = 0, then either a = 0 or b = 0, which still cannot yield 28 from 3a^2 + b^2 without contradicting the absence of √3 and the presence of √7 upon taking square roots. Hence impossible.

Why Other Options Are Wrong:
Numerical choices like −2, 2, 1, −1 assume a valid (a, b) exists; but no rational pair exists, so those sums are meaningless for this constraint.

Common Pitfalls:
Confusing √28 with 2√7 and then trying to blend √7 into a combination of √3 and rationals; these radicals cannot be interchanged with rational coefficients.

Final Answer:
Not possible (no such rational a, b)