Quadratic Equations: Roots, Graph & JEE Main Tricks

Quadratic Equations is the algebraic backbone of JEE Main — it directly contributes one to two questions per session and indirectly underlies Coordinate Geometry (tangent conditions), Calculus (finding extrema), and inequalities. The topic is deceptively deep: the discriminant, Vieta's formulae, and the location-of-roots theory together produce a rich question variety that tests algebraic fluency as much as it tests quadratic-specific knowledge. This guide covers all of it.

Nature of Roots and the Discriminant

For ax²+bx+c = 0 (a ≠ 0): roots α, β = [−b ± √(b²−4ac)]/(2a). Discriminant D = b²−4ac: D > 0 → two distinct real roots; D = 0 → one repeated real root (α = β = −b/2a); D < 0 → two complex conjugate roots. For real coefficients, complex roots always come in conjugate pairs. If coefficients are rational and D is not a perfect square, roots are irrational conjugates (p+√q and p−√q). If D > 0 and a perfect square, roots are rational. JEE tests: given constraints on a, b, c find the nature of roots; or given the nature of roots, find the range of a parameter.

Condition for a quadratic with one root as the square of the other: let roots be α and α². Then sum = α+α² = −b/a and product = α³ = c/a. From product: α = (c/a)^(1/3). Substitute into sum condition to get the required relation between a, b, c. JEE has used this exact setup (with slightly modified conditions) multiple times. Test your quadratic problem-solving speed with a free mock. For related algebra, see our quadratic equations guide.

Vieta's Formulae and Symmetric Functions of Roots

Vieta's formulas for ax²+bx+c=0: α+β = −b/a, αβ = c/a. These two equations let you compute any symmetric function of the roots without finding α and β individually. Common JEE derivatives: α²+β² = (α+β)²−2αβ = b²/a²−2c/a; α³+β³ = (α+β)³−3αβ(α+β) = (−b/a)³−3(c/a)(−b/a); 1/α+1/β = (α+β)/(αβ) = −b/c; 1/α²+1/β² = (α+β)²/((αβ)²) − 2/(αβ) = b²/(c²) − 2a/c; (α−β)² = (α+β)²−4αβ = b²/a²−4c/a = D/a².

Forming a new quadratic with transformed roots: if new roots are α+k and β+k, replace x by x−k in the original equation. If new roots are kα and kβ, replace x by x/k. If new roots are 1/α and 1/β, replace x by 1/x (and multiply through to clear fractions). If new roots are α² and β², find the new sum and product from the above Vieta-derived expressions and write x²−(α²+β²)x+(αβ)² = 0. These transformation methods appear in JEE as "form the equation whose roots are..." questions. For the complete Algebra chapter strategy, see our Math 2026 strategy guide.

Location of Roots: The Most Tested Advanced Topic

Location of roots of ax²+bx+c=0 (a>0) with respect to a fixed value k: (1) Both roots > k: D ≥ 0, vertex x-coordinate > k (i.e., −b/2a > k), and f(k) > 0. (2) Both roots < k: D ≥ 0, −b/2a < k, and f(k) > 0. (3) Exactly one root > k: f(k) < 0 (since the parabola opens upward, if f(k) < 0, one root is below k and one above k, regardless of discriminant). (4) Both roots in (k₁,k₂): D ≥ 0, k₁ < −b/2a < k₂, f(k₁) > 0, f(k₂) > 0. (5) At least one root in (k₁,k₂): not directly — use the complementary conditions.

The key insight: for a > 0, the parabola opens upward. f(k) > 0 means k is outside the roots; f(k) < 0 means k is between the roots. This graph-based interpretation makes all location-of-roots conditions visually intuitive. JEE uses "find the range of parameter p such that both roots of ax²+bx+p=0 are positive" — a direct application of the Both roots > 0 conditions: D ≥ 0, sum = −b/a > 0, product = p/a > 0. For inequalities involving quadratics, see our mathematical inequalities guide.

Maximum and Minimum of Quadratic and Bounded Range Problems

For f(x) = ax²+bx+c: minimum at x = −b/(2a) with value c − b²/(4a) if a > 0; maximum if a < 0. The expression c − b²/(4a) = (4ac−b²)/(4a) = −D/(4a). So: maximum of f (when a < 0) = −D/(4a). JEE uses this when finding the range of a quadratic on all of ℝ: range = [−D/4a, ∞) for a > 0, (−∞, −D/4a] for a < 0. For a quadratic defined on [p,q], the range requires checking both endpoints and the vertex. A common JEE question: "Find values of k such that x²+kx+1 > 0 for all real x" — condition is D < 0, i.e., k²−4 < 0, i.e., k ∈ (−2,2). For a complete look at inequalities and quadratics working together, see our quadratic equations guide.