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Quadratic equations look like Class 10 material, and that is the trap. JEE questions rarely ask you to solve one. They ask for which values of a parameter the roots are real, equal, both positive, or on opposite sides of a number. These need a picture of the parabola, not just the formula. This post builds that picture step by step and checks every answer a second way.
What does the syllabus cover?
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, the formation of quadratic equations with given roots". The JEE (Advanced) 2026 syllabus adds symmetric functions of roots. "Location of roots" is not a separate syllabus line; it is an application of these ideas that appears often in practice papers. NCERT's Class 10 chapter on quadratic equations introduces the discriminant and the nature of roots. Check the documents for your own exam year.
How does the discriminant decide the nature of the roots?
For ax² + bx + c = 0 with real coefficients and a ≠ 0, let D = b² − 4ac.
D > 0: two distinct real roots.
D = 0: two equal real roots, x = −b/2a.
D < 0: no real roots; the roots are a complex conjugate pair.
If a, b, c are rational and D is a perfect square, the roots are rational.
Example. For which k does kx² + 4x + 1 = 0 have equal roots? D = 16 − 4k = 0, so k = 4. Note that k = 0 is not allowed, because the equation would then be linear.
How do you use the sum and product of roots?
If α and β are the roots, α + β = −b/a and αβ = c/a. Many questions ask for an expression in α and β without solving.
Example. For 2x² − 6x + 3 = 0, α + β = 3 and αβ = 3/2.
α² + β² = (α + β)² − 2αβ = 9 − 3 = 6.
1/α + 1/β = (α + β)/αβ = 3 ÷ 3/2 = 2.
Check by solving: x = (6 ± √12)/4 = (3 ± √3)/2. Their squares are (12 ± 6√3)/4, and adding them gives 24/4 = 6.
What decides where the roots lie?
Picture the graph of f(x) = ax² + bx + c with a > 0: an upward parabola with its vertex at x = −b/2a. Three facts tell you where the roots are relative to a number k:
Real roots need D ≥ 0.
The sign of f(k) tells you whether k is outside the roots (f(k) > 0) or between them (f(k) < 0).
The vertex −b/2a tells you which side of k the roots are on when both are on the same side.
For both roots greater than k you need all three: D ≥ 0, f(k) > 0 and −b/2a > k. For k to lie between the roots you need only a·f(k) < 0. Multiplying by a handles parabolas that open downward.
Worked example: both roots in an interval
Find all m for which both roots of x² − 2mx + m² − 1 = 0 lie between −2 and 4.
Discriminant: D = 4m² − 4(m² − 1) = 4, which is always positive.
f(−2) > 0: 4 + 4m + m² − 1 = m² + 4m + 3 = (m + 1)(m + 3) > 0, so m < −3 or m > −1.
f(4) > 0: 16 − 8m + m² − 1 = m² − 8m + 15 = (m − 3)(m − 5) > 0, so m < 3 or m > 5.
Vertex in the interval: the vertex is at x = m, so −2 < m < 4.
All four together give −1 < m < 3.
Check another way. This quadratic factorises as (x − m)² = 1, so the roots are m − 1 and m + 1. We need m − 1 > −2 and m + 1 < 4, which gives −1 < m < 3. The same answer.
Worked example: a number between the roots
For which a does x² + (a − 3)x + a = 0 have one root less than 1 and one greater than 1?
Here the leading coefficient is 1, so we need f(1) < 0: 1 + (a − 3) + a = 2a − 2 < 0, so a < 1. No discriminant condition is needed, because f(1) < 0 already forces the parabola to cross the axis twice. Check with a = 0: x² − 3x = 0 has roots 0 and 3, on either side of 1.
Which mistakes cost the most marks?
Forgetting a ≠ 0 when the x² coefficient contains the parameter.
D > 0 when the question says "real roots", which allows D = 0 as well.
Leaving out the vertex condition, which lets in values like m > 5 above, where both roots sit to the right of 4.
Using f(k) < 0 for a downward parabola. Use a·f(k) < 0 to cover both cases.
Strict versus non-strict inequalities when a root may equal k. Check the boundary values by substitution.
A practice routine
Mixing problem types matters here, because the hard part is choosing which conditions apply. In a study of 126 seventh-grade students over three months, Rohrer, Dedrick and Stershic (2015) found that interleaved practice, where assignments mixed different kinds of problems, produced higher test scores than blocked practice, both one day and 30 days later.
Sketch drill (three times a week): five location questions; draw the parabola before writing any inequality.
Check by a second route: where the quadratic factorises, compare your answer with the actual roots, as in the first worked example.
Mixed sets (weekly): combine roots with complex numbers and sequences and series, since JEE often joins them.
Graph to confirm: a free graphing tool shows instantly whether your range of m is right.
If you cannot see why a condition is needed, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson, so you can revisit the idea behind it.
Today, solve one location-of-roots problem with conditions, then check it by sketching or factorising.
