In AMC 12 preparation, a quadratic equation becomes more demanding when a coefficient is an unknown parameter. Instead of solving for one numerical answer, you must identify all parameter values that make a condition true. The discriminant provides a useful route, but the boundary cases still need attention.
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An original parameter question
For which real values of k does x² − 2kx + k + 2 = 0 have at least one real solution for x? This is an original teaching problem, not an official AMC 12 question.
The coefficient of x² is always 1, so the equation remains quadratic for every real k. This check is important in parameter problems: if the leading coefficient could become zero, that special case would need separate treatment.
Translate the condition into an inequality
For a quadratic ax² + bx + c = 0 with real coefficients and a ≠ 0, real roots exist when b² − 4ac ≥ 0. Here a = 1, b = −2k and c = k + 2.
The discriminant is (−2k)² − 4(k + 2) = 4k² − 4k − 8 = 4(k − 2)(k + 1). Therefore we need (k − 2)(k + 1) ≥ 0.
Solve the inequality by signs
The critical values are −1 and 2. Below −1, both factors are negative, so their product is positive. Between −1 and 2, the factors have opposite signs, so the product is negative. Above 2, both factors are positive.
At either critical value the product is zero, which is allowed by the phrase “at least one real solution”. The complete answer is k ≤ −1 or k ≥ 2.
Do not write −1 ≤ k ≤ 2 simply because the factors vanish at the endpoints. The sign between the endpoints must be checked. Substituting k = 0 gives x² + 2 = 0, which has no real roots and immediately exposes that incorrect interval.
Check what happens at the boundaries
At k = −1, the equation is x² + 2x + 1 = (x + 1)² = 0. There is one repeated real root, x = −1. At k = 2, it becomes x² − 4x + 4 = (x − 2)² = 0, with repeated root x = 2.
If the question instead required two distinct real roots, the discriminant would need to be strictly positive. The answer would then be k < −1 or k > 2. A small change in wording changes whether the boundary values are included.
Confirm the result by completing the square
Rewrite the original equation as (x − k)² = k² − k − 2 = (k − 2)(k + 1). A square of a real number cannot be negative, so the right side must be nonnegative. This gives the same condition and explains its meaning without relying only on a memorised discriminant rule.
For your next review, record the leading-coefficient check, the required inequality, the sign intervals and the boundary interpretation. Use MAA's official resources for additional AMC practice. Understanding why endpoints are included is more reliable than memorising a final interval.
New to the subject? Read our AMC 12 Preparation: What to Study and How to Practise Effectively.
Official resources
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