Question:Let \(\mathrm{R(t) = \frac{t^2 - 169}{t - 13}}\).For what real value of t does \(\mathrm{R(t)}\) equal 0?Enter your answer as...
GMAT Advanced Math : (Adv_Math) Questions
- Let \(\mathrm{R(t) = \frac{t^2 - 169}{t - 13}}\).
- For what real value of \(\mathrm{t}\) does \(\mathrm{R(t)}\) equal 0?
- Enter your answer as an integer.
Answer Format Instructions: Provide a single integer (no units).
1. TRANSLATE the problem information
- Given: \(\mathrm{R(t) = \frac{t^2 - 169}{t - 13}}\)
- Find: Value of t where \(\mathrm{R(t) = 0}\)
2. INFER the key principle for rational functions
- A rational function equals zero exactly when:
- Its numerator equals zero, AND
- Its denominator does not equal zero
- This means we need two conditions: \(\mathrm{t^2 - 169 = 0}\) and \(\mathrm{t - 13 \neq 0}\)
3. SIMPLIFY the numerator equation
- Solve \(\mathrm{t^2 - 169 = 0}\)
- Recognize this as a difference of squares: \(\mathrm{t^2 - 13^2 = 0}\)
- Factor: \(\mathrm{(t - 13)(t + 13) = 0}\)
- Solutions: \(\mathrm{t = 13}\) or \(\mathrm{t = -13}\)
4. APPLY CONSTRAINTS from the denominator
- Check each candidate solution against \(\mathrm{t - 13 \neq 0}\)
- For \(\mathrm{t = 13}\): This makes denominator = \(\mathrm{13 - 13 = 0}\) ✗ Invalid
- For \(\mathrm{t = -13}\): This makes denominator = \(\mathrm{-13 - 13 = -26 \neq 0}\) ✓ Valid
5. State the final answer
- Only \(\mathrm{t = -13}\) satisfies both conditions
Answer: -13
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students solve only \(\mathrm{t^2 - 169 = 0}\) and report both solutions \(\mathrm{t = 13}\) and \(\mathrm{t = -13}\) without checking domain restrictions.
They correctly factor and find both values but don't realize that rational functions have domain restrictions. When \(\mathrm{t = 13}\), the original function becomes \(\mathrm{\frac{0}{0}}\), which is undefined, not zero. This leads to confusion about whether to include both answers or just one.
This causes them to either guess or incorrectly include \(\mathrm{t = 13}\) as a valid solution.
The Bottom Line:
The key insight is that zeros of rational functions require satisfying two conditions simultaneously - not just making the numerator zero, but also ensuring the denominator stays nonzero. Students who focus only on algebraic manipulation without considering the function's domain will miss the constraint that eliminates one solution.