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JEE questions often ask "for what values of k is this expression positive for all x?" The answer takes two conditions, and most lost marks come from forgetting one of them or from a parameter that can make the leading coefficient zero. This post shows why the test works, the four sign cases, and three worked problems with checks.
Is this in NCERT and the JEE Main syllabus?
The JEE (Main) 2026 syllabus lists "Quadratic equations in real and complex number systems and their solutions; Relations between roots and coefficients, nature of roots". NCERT introduces the discriminant in Class 10 Chapter 4, section 4.4 "Nature of Roots", and Class 11 Chapter 4 opens by pointing out that ax² + bx + c = 0 with D = b² − 4ac < 0 has no real solution. Neither book states the "positive for all x" rule as a result, and the Class 11 chapter on inequalities covers only linear ones. The rule below is the standard way JEE problems use those two NCERT facts together.
Why do a > 0 and D < 0 make it positive everywhere?
Complete the square:
ax² + bx + c = a[(x + b/2a)² − D/4a²].
The squared term is never negative. If D < 0, then −D/4a² is positive, so the bracket is always positive. Multiply by a > 0 and the whole expression stays positive. Its smallest value is −D/4a, reached at x = −b/2a.
Example. For 2x² − 8x + 11, D = 64 − 88 = −24. The minimum is 24/8 = 3 at x = 2. Check: 2(4) − 16 + 11 = 3.
What are the four sign cases?
Positive for all x: a > 0 and D < 0.
Non-negative for all x: a > 0 and D ≤ 0 (it touches zero at one point when D = 0).
Negative for all x: a < 0 and D < 0.
Non-positive for all x: a < 0 and D ≤ 0.
If a contains a parameter, test a = 0 on its own. A linear expression bx + c with b ≠ 0 always changes sign, so it is never positive for all x.
Worked problem 1: a parameter in the middle
For what k is x² − 2kx + k + 6 > 0 for all x?
Here a = 1 > 0, so only D matters. D = 4k² − 4(k + 6) = 4(k² − k − 6) = 4(k − 3)(k + 2). We need D < 0, so −2 < k < 3.
Check the edges. At k = 3 the expression is x² − 6x + 9 = (x − 3)², which equals 0 at x = 3, so k = 3 fails the strict inequality. At k = 0 it is x² + 6, positive everywhere.
Worked problem 2: a parameter in the leading coefficient
For what m is (m − 2)x² + 8x + m + 4 > 0 for all x?
m = 2: the expression is 8x + 6, which is negative for x < −3/4. So m = 2 fails.
Otherwise we need m − 2 > 0 and D = 64 − 4(m − 2)(m + 4) < 0. Simplify: D = −4(m − 4)(m + 6), so D < 0 when m > 4 or m < −6.
Combine with m > 2: the answer is m > 4.
Check m = 5: 3x² + 8x + 9 has D = 64 − 108 = −44, and a = 3 > 0. The part m < −6 answers the opposite question: at m = −7 the expression is −9x² + 8x − 3, with a < 0 and D = −44, so it is negative for all x.
Worked problem 3: the range of a rational function
Find the range of y = (x² − x + 1)/(x² + x + 1).
The denominator has a = 1 and D = 1 − 4 = −3, so it is positive for all x, and you can multiply through safely:
(y − 1)x² + (y + 1)x + (y − 1) = 0.
For a real x to exist, this quadratic in x needs D ≥ 0: (y + 1)² − 4(y − 1)² ≥ 0, which factorises as (3 − y)(3y − 1) ≥ 0. So 1/3 ≤ y ≤ 3. (At y = 1 the equation is 2x = 0, so x = 0 works.) Check: x = 1 gives 1/3 and x = −1 gives 3.
Which mistakes cost marks?
Forgetting a > 0 and keeping m < −6 in problem 2.
Skipping the a = 0 case.
Using D ≤ 0 when the question says strictly positive.
Multiplying an inequality by a denominator without first checking its sign.
A practice routine
Mixed practice helps here, because the hard part is picking the case. In a classroom study of 126 seventh-grade students, Rohrer, Dedrick and Stershic (2015) found that interleaved maths practice beat blocked practice on tests 1 and 30 days later. Make a set of eight: two "positive for all x", two "negative for all x", two with a parameter in a, and two range questions, shuffled. For questions on where the roots lie relative to a number, see our guide to the discriminant and location of roots.
When a step does not make sense, look it up one idea at a time. The Fermi Concept Library covers every concept in mathematics with its own lesson, so you can revise completing the square before returning to the set.
Today, solve problem 2 again with the sign flipped: find m so that the expression is negative for all x.
