Solve and compare: Let x satisfy (4/√x) + (7/√x) = √x with x > 0, and let y satisfy y^2 − (11)^(5/2)/√y = 0 with y > 0. Compare x and y.

Difficulty: Medium

if x > y
if x ≥ y
if x < y
if x ≤ y
if x = y

Correct Answer: if x ≥ y

Explanation:

Introduction / Context:
Two radical equations are posed with positivity constraints (because of square roots). We solve each exactly by appropriate substitutions and then compare the resulting values.

Given Data / Assumptions:

(4/√x) + (7/√x) = √x with x > 0.
y^2 − (11)^(5/2)/√y = 0 with y > 0.

Concept / Approach:
For x, combine like terms: (11/√x) = √x ⇒ cross-multiply to get x. For y, substitute t = √y (t > 0) to reduce to a simple power equation.

Step-by-Step Solution:

(11/√x) = √x ⇒ multiply both sides by √x: 11 = x ⇒ x = 11.For y: let t = √y ⇒ y = t^2. Equation: t^4 − (11)^(5/2)/t = 0.Multiply by t: t^5 = (11)^(5/2).Take positive fifth root: t = (11)^(1/2) ⇒ √y = √11 ⇒ y = 11.Thus x = y = 11, which satisfies x ≥ y.

Verification / Alternative check:
Substitute back to confirm each equation balances at 11.

Why Other Options Are Wrong:
Strict inequalities are incorrect; equality holds exactly.

Common Pitfalls:
Mishandling fractional exponents, or allowing negative square-root values contrary to domain constraints.

Solve and compare: Let x satisfy (4/√x) + (7/√x) = √x with x > 0, and let y satisfy y^2 − (11)^(5/2)/√y = 0 with y > 0. Compare x and y.

More Questions from Quadratic Equation

Discussion & Comments