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Question:The functions f and g are defined by \(\mathrm{f(x) = 3x^2 + 4x - 15}\) and \(\mathrm{g(x) = 10x +...

GMAT Advanced Math : (Adv_Math) Questions

Source: Prism
Advanced Math
Nonlinear equations in 1 variable
MEDIUM
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Notes
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Question:

The functions f and g are defined by \(\mathrm{f(x) = 3x^2 + 4x - 15}\) and \(\mathrm{g(x) = 10x + 9}\).


For what positive value of \(\mathrm{c}\) is \(\mathrm{f(c) = g(c)}\)?

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Solution

1. TRANSLATE the problem information

  • Given information:
    • \(\mathrm{f(x) = 3x^2 + 4x - 15}\)
    • \(\mathrm{g(x) = 10x + 9}\)
    • Need to find positive c where \(\mathrm{f(c) = g(c)}\)
  • This means we need: \(\mathrm{3c^2 + 4c - 15 = 10c + 9}\)

2. SIMPLIFY to standard quadratic form

  • Move all terms to one side:
    \(\mathrm{3c^2 + 4c - 15 - 10c - 9 = 0}\)
  • Combine like terms:
    \(\mathrm{3c^2 - 6c - 24 = 0}\)
  • Factor out the common factor of 3:
    \(\mathrm{c^2 - 2c - 8 = 0}\)

3. SIMPLIFY by factoring the quadratic

  • Need two numbers that multiply to -8 and add to -2
  • Those numbers are -4 and 2: \(\mathrm{(-4)(2) = -8}\) and \(\mathrm{(-4) + 2 = -2}\)
  • Factor: \(\mathrm{(c - 4)(c + 2) = 0}\)

4. Solve using zero product property

  • If \(\mathrm{(c - 4)(c + 2) = 0}\), then either:
    • \(\mathrm{c - 4 = 0}\), so \(\mathrm{c = 4}\)
    • \(\mathrm{c + 2 = 0}\), so \(\mathrm{c = -2}\)

5. APPLY CONSTRAINTS to select final answer

  • The problem asks for the positive value of c
  • Since \(\mathrm{c = 4}\) is positive and \(\mathrm{c = -2}\) is negative: \(\mathrm{c = 4}\)

Answer: 4




Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make algebraic errors when rearranging the initial equation or combining like terms.

For example, they might incorrectly get \(\mathrm{3c^2 + 14c - 6 = 0}\) instead of \(\mathrm{3c^2 - 6c - 24 = 0}\), leading to a completely different quadratic that doesn't factor neatly. This leads to confusion and guessing.

Second Most Common Error:

Missing APPLY CONSTRAINTS reasoning: Students correctly solve the quadratic to get \(\mathrm{c = 4}\) or \(\mathrm{c = -2}\), but fail to notice the question specifically asks for the positive value.

They might randomly select either solution or get confused about which one to choose. This may lead them to select -2 if that were an answer choice, or cause them to second-guess their correct work.

The Bottom Line:

This problem tests whether students can systematically work through setting functions equal and solving the resulting quadratic, while carefully reading what the question is asking for. The algebra isn't complex, but it requires attention to detail at multiple steps.

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