Quadratic with rational coefficients: If a quadratic equation with rational coefficients has one root equal to √5, then which equation below can it be?

Introduction / Context:
This question checks your understanding of conjugate roots for polynomials with rational (or real) coefficients. If an irrational root like √5 occurs, its conjugate −√5 must also be a root so that the product expands to a polynomial with rational coefficients.

Given Data / Assumptions:

• One root is √5.
• Coefficients are rational numbers.
• We need a quadratic (degree 2) that admits √5 as a root and has only rational coefficients.

Concept / Approach:
For quadratics with rational coefficients, irrational roots appear in conjugate pairs. So if √5 is a root, then −√5 must also be a root. The minimal monic polynomial having roots ±√5 is (x − √5)(x + √5) = x^2 − 5.

Step-by-Step Solution:

Verification / Alternative check:
Plug x = √5 into x^2 − 5: (√5)^2 − 5 = 5 − 5 = 0. Works. Plug x = −√5: (−√5)^2 − 5 = 5 − 5 = 0. Also works.

Why Other Options Are Wrong:
x^2 + 5 = 0 has imaginary roots ±i√5; x^2 − 10 = 0 has roots ±√10 (not √5); the extra option with √5 in the coefficient is not rational-coefficient; “None of these” is incorrect because x^2 − 5 = 0 fits perfectly.

Common Pitfalls:
Assuming √5 alone can be a root without its conjugate for rational coefficients, or confusing √5 with √10 or i√5.

Final Answer:
x^2 − 5 = 0