Which equation has real roots? Among the following equations, identify the one that definitely has real roots.

Difficulty: Easy

Correct Answer: (x − 1)(2x − 5) = 0

Explanation:


Introduction / Context:
To decide whether a quadratic has real roots, we can use factoring or the discriminant test (b^2 − 4ac). Factored forms immediately reveal roots; otherwise, compute the discriminant to infer the nature of roots.


Given Data / Assumptions:

  • We examine each option for factorization or discriminant sign.
  • Real roots exist if discriminant ≥ 0.


Concept / Approach:
Option (b) is already factored into two linear terms, giving real roots directly. For (a) and (c), we check discriminants. Option (d) is not an equation and is irrelevant as an answer. Option (e) is a known quadratic with negative discriminant.


Step-by-Step Solution:

(b): (x − 1)(2x − 5) = 0 ⇒ roots x = 1 and x = 2.5 (real)(a): 2x^2 − 3x + 4 ⇒ D = (−3)^2 − 4*2*4 = 9 − 32 = −23 (no real roots)(c): 3x^2 + 4x + 5 ⇒ D = 4^2 − 4*3*5 = 16 − 60 = −44 (no real roots)(e): x^2 + x + 1 ⇒ D = 1 − 4 = −3 (no real roots)


Verification / Alternative check:
Factored forms guarantee real roots because each linear factor equals zero at a real x-value.


Why Other Options Are Wrong:

  • (a), (c), (e): Negative discriminants imply complex (non-real) roots.
  • (d): Not a mathematical equation; cannot be “real roots.”


Common Pitfalls:
Ignoring signs in b^2 − 4ac or miscomputing products. Keep coefficients clear.


Final Answer:
(x − 1)(2x − 5) = 0

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion