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The function g is defined by \(\mathrm{g(x) = (x - 12)^2 + (x - 18)^2}\). For what value of x...

GMAT Advanced Math : (Adv_Math) Questions

Source: Prism
Advanced Math
Nonlinear functions
HARD
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Notes
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The function \(\mathrm{g}\) is defined by \(\mathrm{g(x) = (x - 12)^2 + (x - 18)^2}\). For what value of \(\mathrm{x}\) does \(\mathrm{g(x)}\) attain its minimum?

A
12
B
14
C
15
D
18
Solution

1. TRANSLATE the problem information

  • Given: \(\mathrm{g(x) = (x - 12)^2 + (x - 18)^2}\)
  • Find: Value of x where \(\mathrm{g(x)}\) is minimized
  • Answer choices: 12, 14, 15, or 18

2. INFER the solution approach

  • This is asking for the minimum of a quadratic function
  • Two main approaches available:
    • Algebraic: Expand and complete the square to find vertex
    • Geometric: Recognize this as sum of squared distances to two points

3. SIMPLIFY using the algebraic method

  • Expand both squared terms:
    • \(\mathrm{(x - 12)^2 = x^2 - 24x + 144}\)
    • \(\mathrm{(x - 18)^2 = x^2 - 36x + 324}\)
  • Combine: \(\mathrm{g(x) = 2x^2 - 60x + 468}\)

4. SIMPLIFY by completing the square

  • Factor out the 2: \(\mathrm{g(x) = 2(x^2 - 30x) + 468}\)
  • Complete the square inside parentheses: \(\mathrm{x^2 - 30x = (x - 15)^2 - 225}\)
  • Substitute back: \(\mathrm{g(x) = 2((x - 15)^2 - 225) + 468 = 2(x - 15)^2 + 18}\)

5. INFER the minimum location

  • Since \(\mathrm{(x - 15)^2 \geq 0}\) for all real x, the minimum value occurs when \(\mathrm{(x - 15)^2 = 0}\)
  • This happens when \(\mathrm{x = 15}\)

Answer: C) 15




Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make algebraic errors when expanding the squared terms or combining like terms, leading to incorrect coefficients in the quadratic expression.

For example, they might incorrectly expand \(\mathrm{(x - 12)^2}\) as \(\mathrm{x^2 - 12x + 144}\) (missing the factor of 2 in the middle term), or make sign errors when combining terms. This leads to a wrong quadratic expression and therefore an incorrect vertex location. This may cause them to guess among the given options or select Choice B (14) if they get close to the correct answer.

Second Most Common Error:

Poor INFER reasoning: Students don't recognize the geometric interpretation that this represents squared distances to two points, missing the elegant shortcut that the minimum occurs at the midpoint \(\mathrm{(12 + 18)/2 = 15}\).

Instead, they might test each answer choice by substitution, which is time-consuming and error-prone, or they might confuse this with finding where the derivative equals zero without properly setting up the problem. This leads to confusion and potentially random guessing among the choices.

The Bottom Line:

This problem rewards students who can either execute algebraic manipulation cleanly or recognize the underlying geometric pattern. The key insight is that both approaches lead to the same answer - the midpoint of the two reference points.

Answer Choices Explained
A
12
B
14
C
15
D
18
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