Problem 4 · Split into regions, then verify roots · 4 points
Made by AI and not reviewed by a real person.
For which positive value of a does |x² − 4x + 3| = ax have exactly three distinct real solutions?
- 4 + 2√3
- 1/2
- 2√3 − 3
- 4 − 2√3
- 1
Show a hint
- As a changes, where could two solutions merge? The quadratic’s zeros divide the graph into regions worth considering separately.
- For x > 0, compare |x² − 4x + 3|/x with a, and check that any candidate lies in the region where you found it.
Show the answer and solution
Answer: D — 4 − 2√3
Since a > 0, any solution has x > 0. On (0, 1) and (3, ∞), the equation is x² − (4 + a)x + 3 = 0. This quadratic changes sign between 0 and 1 and again between 3 and ∞, so it has exactly one root in each interval. For 1 < x < 3, the equation becomes a = 4 − x − 3/x. This expression rises from 0 to its maximum 4 − 2√3 at x = √3, then falls to 0; it equals a at exactly one point in this interval only when a = 4 − 2√3. That gives three solutions total. The tempting 4 + 2√3 also makes the middle quadratic’s discriminant zero, but its double root is −√3, outside the middle interval.