The graphs of y = x^2 - 6x + t and y = 2x - 1 intersect at exactly one...
GMAT Advanced Math : (Adv_Math) Questions
The graphs of \(\mathrm{y = x^2 - 6x + t}\) and \(\mathrm{y = 2x - 1}\) intersect at exactly one point. What is the value of t?
\(\mathrm{12}\)
\(\mathrm{14}\)
\(\mathrm{15}\)
\(\mathrm{16}\)
\(\mathrm{18}\)
1. TRANSLATE the problem information
- Given information:
- Two functions: \(\mathrm{y = x² - 6x + t}\) and \(\mathrm{y = 2x - 1}\)
- They intersect at exactly one point
- Need to find value of t
2. INFER the mathematical approach
- Key insight: When two functions intersect at exactly one point, setting them equal creates a quadratic equation with exactly one real solution
- This happens when the discriminant equals zero
- Strategy: Set equations equal, rearrange to standard form, then use discriminant condition
3. SIMPLIFY to find intersection points
Set the equations equal:
\(\mathrm{x² - 6x + t = 2x - 1}\)
Rearrange to standard quadratic form:
\(\mathrm{x² - 6x + t - 2x + 1 = 0}\)
\(\mathrm{x² - 8x + (t + 1) = 0}\)
4. INFER the discriminant condition
For exactly one real solution, the discriminant must equal zero:
\(\mathrm{Δ = b² - 4ac = 0}\)
With \(\mathrm{a = 1, b = -8, c = (t + 1)}\):
\(\mathrm{(-8)² - 4(1)(t + 1) = 0}\)
5. SIMPLIFY to solve for t
\(\mathrm{64 - 4(t + 1) = 0}\)
\(\mathrm{64 - 4t - 4 = 0}\)
\(\mathrm{60 - 4t = 0}\)
\(\mathrm{4t = 60}\)
\(\mathrm{t = 15}\)
Answer: C. 15
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students don't connect "exactly one intersection point" to the discriminant condition. Instead, they might try to solve the system directly by substitution or elimination, not realizing they need to use the special condition about having exactly one solution.
This leads to confusion about what to do next, causing them to get stuck and guess randomly.
Second Most Common Error:
Poor SIMPLIFY execution: Students correctly set up the discriminant equation but make arithmetic errors, particularly when expanding -4(t + 1) or combining like terms in 64 - 4t - 4.
Common calculation mistakes include getting \(\mathrm{64 - 4t + 4 = 68 - 4t}\) instead of \(\mathrm{60 - 4t}\), leading to \(\mathrm{t = 17}\). This may lead them to select Choice E (18) as the closest option.
The Bottom Line:
This problem requires understanding the geometric meaning of intersection points in terms of algebraic solutions. The key breakthrough is recognizing that "exactly one intersection" is a constraint that determines the discriminant value.
\(\mathrm{12}\)
\(\mathrm{14}\)
\(\mathrm{15}\)
\(\mathrm{16}\)
\(\mathrm{18}\)