Problem 4 · Compare discriminants and count distinct roots · 4 points
Made by AI and not reviewed by a real person.
For which real value of t does (x² − 2tx + 1)(x² − 2x + t²) = 0 have exactly one distinct real solution for x?
- t = −1 or t = 1
- No real value of t
- t = 1
- −1 < t < 1
- t = −1
Show a hint
- The product collects solutions from both quadratics. Compare when each can have two real roots before counting the solutions together.
- The two discriminants have opposite signs unless both are zero. In that special case, check whether the repeated roots coincide.
Show the answer and solution
Answer: C — t = 1
The discriminants are 4(t² − 1) and 4(1 − t²). Unless t² = 1, one is positive, so the product has at least two distinct real solutions. At t = 1, both factors are (x − 1)², giving just x = 1. At t = −1, the factors are (x + 1)² and (x − 1)², giving two solutions. Thus t = 1; the tempting answer t = −1 or t = 1 overlooks that the repeated roots differ when t = −1.