Form a quadratic when one root is 3 − √5 and the sum of roots is 6: Construct the quadratic equation with real coefficients.

Introduction / Context:
When a quadratic with real coefficients has an irrational root involving √5, its conjugate is also a root. Given one root 3 − √5 and the sum of roots 6, the other root must be 3 + √5, and we can write the quadratic from the sum and product of roots.

Given Data / Assumptions:

• Given root: 3 − √5.
• Sum of roots S = 6 ⇒ other root is 6 − (3 − √5) = 3 + √5.

Concept / Approach:
With roots r1 and r2, the monic quadratic is x^2 − Sx + P = 0 where P = r1*r2. Compute the product using (a − b)(a + b) = a^2 − b^2.

Step-by-Step Solution:

Verification / Alternative check:
Substitute x = 3 − √5: (3 − √5)^2 − 6(3 − √5) + 4 simplifies to 0. The conjugate also satisfies the equation.

Why Other Options Are Wrong:

• x^2 + 6x + 4 = 0: Wrong sign for the sum.
• x^2 − 6x − 4 = 0: Wrong product (−4 instead of 4).
• None of these: Incorrect since a correct option exists.

Common Pitfalls:
Forgetting conjugate pairing for real-coefficient polynomials or sign mistakes when forming x^2 − Sx + P.

Final Answer: