PHALIN THAKKAR
Real roots are not the whole question.
Advanced extension · M3, M5 and M7 distinguish existence, location and nonnegative values over a specified domain.
Prerequisites: factorisation, the discriminant, a quadratic vertex and endpoint values. Start with the foundation quadratic graph.
| Requirement | Deciding check |
|---|---|
| Real distinct roots | Positive discriminant, after checking it is a quadratic |
| Two roots in an open interval | Vertex inside, negative minimum and positive endpoint values for a convex quadratic |
| Nonnegative everywhere | Positive leading coefficient and nonpositive discriminant; handle degeneration separately |
| Nonnegative on a closed interval | Minimum at the vertex if inside, otherwise at an endpoint |
M3. A quadratic for every real x
Find all real a for which:
holds for every real x. Include the case in which the quadratic coefficient vanishes.
Work through M3
If a < 1, the quadratic opens downward and becomes negative for large |x|.
If a = 1, the expression is −2x + 4, which also fails for all real x.
For a > 1, require a nonpositive discriminant:
Thus:
At a = 3/2 the expression is ½(x − 3)², so equality is valid.
Check: the discriminant condition was used only after verifying a positive leading coefficient.
M5. Roots inside a particular interval
Find every real k such that:
has two distinct roots, both strictly inside (0, 3).
Work through M5
For a convex quadratic to have two distinct roots inside (0, 3), its minimum must be inside that interval and below zero, with positive values at both endpoints.
The vertex is x = k. The discriminant is:
Together with 0 < k < 3 this gives k > 2.
At x = 0 the value k + 2 is positive. At x = 3:
Therefore:
At k = 2 the roots merge. At k = 11/5 one root is 3. Both boundaries must be excluded.
M7. Nonnegative on a restricted interval
Find all real a such that:
for every x in [0, 2]. Explain why requiring a nonpositive discriminant is unnecessarily restrictive here.
Work through M7
For a ≤ 0, the vertex is at or left of the interval. The minimum on [0, 2] is at x = 0 and equals a + 1. These values work exactly when −1 ≤ a ≤ 0.
For 0 < a < 2, the vertex lies inside the interval. Its value is:
It is nonnegative for:
For a ≥ 2, the minimum occurs at x = 2, where 5 − 3a < 0. These values fail.
Therefore:
At a = −1 the expression is x(x + 2), which is nonnegative on [0, 2] even though it has real roots and takes negative values outside that interval. At the upper boundary it is a perfect square.
Requiring a nonpositive discriminant would incorrectly exclude −1 ≤ a < (1 − √5)/2. Roots and negative values outside the requested interval need not invalidate the condition.