Quadratic Equations and Inequalities — JEE Main & Advanced Notes

Discriminant, root nature, location of roots, parameter ranges and sign charts are core JEE algebra tools.

Material signal: Mapped from files marked QE and sequence-series revision assignments.

Priority: Must Do. Treat this as a foundation chapter inside the JEE Mathematics ladder.

• Discriminant and root nature: The discriminant D=b²−4ac is the single quantity that determines how many real roots the quadratic has. For real coefficients: D>0 gives two distinct real roots (rational when D is a perfect square); D=0 gives a repeated root at −b/2a; D<0 gives conjugate complex roots. When a problem asks for 'real and distinct roots', translate this to the condition D>0.
• Vieta's formulas — computing without finding roots: For ax²+bx+c=0 with roots α,β: the sum α+β=−b/a and product αβ=c/a. From these two quantities you can compute any symmetric function: α²+β²=(α+β)²−2αβ, α²β+αβ²=αβ(α+β), |α−β|=√D/|a|. JEE frequently asks for expressions like α⁴+β⁴ or 1/α+1/β — always express in terms of sum and product, never find α and β individually.
• Sign chart for inequalities: To solve f(x)>0 where f is a quadratic with roots r₁0: f(x)>0 when xr₂; f(x)<0 when r₁
• Location of roots — the five standard JEE conditions: (1) Both roots > k: need D≥0, vertex right of k (−b/2a>k), and f(k)>0 for a>0. (2) Both roots < k: D≥0, −b/2a0 for a>0. (3) One root on each side of k: a·f(k)<0 (no need for D condition separately since this guarantees two real roots). (4) Both roots in (k₁,k₂): D≥0, f(k₁)>0, f(k₂)>0, k₁<−b/2a
• Parameter problems — translating into discriminant conditions: Questions like 'find values of k for which the equation has real roots' or 'the roots are positive' translate into inequalities on D and on Vieta's formulas. Set up all conditions simultaneously and find their intersection. A common trap: after finding D≥0 condition, forget to also check α+β>0 and αβ>0 for both roots positive.
• Quadratic inequalities with rational expressions: For f(x)/g(x)>0, find roots of numerator and denominator separately. These points divide the number line into intervals. Test one point in each interval by checking the sign. Points where g(x)=0 must be excluded from the solution even if f(x)=0 there as well.
• Range of a rational function y=f(x)/g(x): Cross-multiply to get a quadratic in x: f(x)−y·g(x)=0. This has real solutions in x iff D≥0 in terms of y. The condition D(y)≥0 gives the range of y. This technique is used in problems like 'find the range of (x²+x+1)/(x²+1)'.
• Transformation of equations: If roots of f(x)=ax²+bx+c=0 are α,β, then the equation with roots α+k,β+k is f(x−k)=0; with roots kα,kβ is f(x/k)=0; with roots 1/α,1/β is x²·f(1/x)=0, i.e., cx²+bx+a=0. These are used to avoid solving for α,β directly.

• For ax²+bx+c=0: roots α,β satisfy α+β = −b/a and αβ = c/a
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Discriminant D = b²−4ac: D>0 two distinct real roots; D=0 repeated real root; D<0 no real roots
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Quadratic formula: x = (−b ± √D) / 2a
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Axis of symmetry: x = −b/2a (vertex x-coordinate); Vertex: f(−b/2a) = −D/4a
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Symmetric functions of roots: α²+β² = (α+β)²−2αβ; α³+β³ = (α+β)³−3αβ(α+β); |α−β| = √D/|a|
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• New equation with roots transformed: if roots are α,β, new equation with roots (α+k),(β+k) is a(x−k)²+b(x−k)+c=0
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Location of roots — both roots > k: D≥0, f(k)>0 (when a>0), −b/2a > k
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Location of roots — both roots < k: D≥0, f(k)>0 (when a>0), −b/2a < k
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Location of roots — k lies between roots: a·f(k) < 0
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Location of roots — exactly one root in (k₁,k₂): f(k₁)·f(k₂) < 0
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Range of y = f(x)/g(x) where f,g are quadratics: set y = f(x)/g(x), rearrange to quadratic in x, require D≥0
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.

• Flipping inequality sign only sometimes: whenever you multiply or divide both sides of an inequality by a negative number (or a negative expression), the direction flips. When dividing by (x−1) in an inequality, you cannot treat it as positive unless x>1. Instead, multiply both sides by (x−1)² (always positive) and rearrange.
• Finding roots but not the solution interval: solving x²−5x+6>0 correctly begins by finding roots x=2 and x=3, but the answer is x<2 or x>3 (not 2
• Ignoring a=0 in parameter questions: if a problem says 'the quadratic ax²+bx+c=0 has real roots' and then a takes a specific value that makes a=0, the equation becomes linear, not quadratic. Treat a=0 separately.
• Using Vieta's formulas before confirming real roots: Vieta's formulas hold for all roots (real or complex) of a quadratic, but if a JEE problem assumes real roots, check D≥0 as an additional constraint before applying sum/product formulas.
• Forgetting repeated root in product rule for inequalities: if f(x) = a(x−r)²·(x−s), the factor (x−r)² is always non-negative. The sign of f(x) is determined only by the sign of a and (x−s). A double root does not split the number line into alternating signs; it just makes the function touch zero without crossing.

For JEE Main, aim for fast recognition and clean substitution. Finish the first pass of Quadratic Equations and Inequalities questions in 60–90 seconds each. Prioritise standard formulas, short sign/domain checks and option elimination only after the setup is correct.

For JEE Advanced, expect combined conditions, hidden domains, multi-correct traps and integer-style answers. Build the solution from definitions, not memorised tricks. When a parameter appears, solve the general case and then filter using restrictions.

• D>0: two distinct real roots. D=0: one repeated root = −b/2a. D<0: no real roots, roots are complex conjugates.
• Vieta's: sum = −b/a, product = c/a. Build any symmetric expression from these two alone.
• Sign chart: find roots, place on number line, check sign in one interval, alternate (unless a root is repeated).
• Location of roots: draw the parabola mentally. Use f(boundary)>0 (same side as the parabola's opening) combined with the axis of symmetry condition.
• For 'one root in (k₁,k₂)', the quickest test is f(k₁)·f(k₂)<0 — no D condition needed.