Quadratic Equations & Theory of Equations

Beyond the discriminant, Advanced tests the location of roots relative to numbers, common-root conditions, and polynomial theory (Vieta's relations, transformation of equations). Reasoning about a quadratic's sign and graph placement is the recurring high-order skill.

Key results

• Roots: x = (−b ± √D)/2a, D = b²−4ac. Real distinct D>0, equal D=0, complex D<0.
• Vieta (quadratic): sum = −b/a, product = c/a. Cubic ax³+bx²+cx+d: Σα = −b/a, Σαβ = c/a, αβγ = −d/a.
• Sign of quadratic: same sign as a except between the roots; f(x)>0 ∀x iff a>0 and D<0.
• Root location: for a root in (k, ∞) or an interval, use signs of f(k), the vertex −b/2a, and D together.
• Common root: two quadratics share a root when the cross-multiplication (elimination) condition holds; identical roots when coefficients are proportional.

Condition	Requirement
Both roots > k	af(k)>0, D≥0, −b/2a>k
Roots opposite signs	c/a < 0
f(x)>0 for all x	a>0 and D<0
One root in (k1,k2)	f(k1)·f(k2) < 0

Exam Tricks & Tips

• 🎯 f(k1)·f(k2) < 0 guarantees exactly one root between k1 and k2 — the fastest interval-root test.
• 🎯 For both roots on one side of k, check three things together: sign of a·f(k), the vertex position −b/2a, and D ≥ 0.
• 🎯 A quadratic keeps the sign of a everywhere except strictly between its real roots — use for inequality (sign-scheme) problems.
• 🎯 Product of roots negative ⇒ roots straddle zero (opposite signs) — read straight from c/a without solving.
• 🎯 Use Vieta to build a new equation whose roots are transformed (reciprocals, squares, shifted) without finding the roots explicitly.
• ❌ Common mistake: forgetting the D ≥ 0 requirement in root-location problems — the sign conditions alone can be satisfied by complex roots.

Expected exam pattern

1–2 questions: a root-location or common-root problem and a Vieta/transformation-of-equations conceptual (often multiple-correct or integer).

Quick recap

D sets root nature; Vieta gives sum/product. Quadratic sign = sign of a except between roots. Root-location combines f(k) signs, the vertex and D. f(k1)f(k2)<0 isolates one root in (k1,k2). Use Vieta to transform equations.

Cover the answer, recall, then check. 11 cards on quadratics for JEE Advanced.

Q1. Discriminant and root nature.
A1. D = b² − 4ac: D>0 real distinct, D=0 equal, D<0 complex conjugates.

Q2. Sum and product of roots of ax² + bx + c = 0.
A2. Sum = −b/a, product = c/a.

Q3. Condition for a quadratic to be positive for all real x.
A3. a > 0 and D < 0.

Q4. Sign of a quadratic between and outside its real roots.
A4. Opposite to a between the roots, same as a outside.

Q5. Test that exactly one root lies in (k1, k2).
A5. f(k1)·f(k2) < 0.

Q6. Conditions for both roots greater than a number k.
A6. a·f(k) > 0, vertex −b/2a > k, and D ≥ 0.

Q7. What does a negative product of roots tell you?
A7. The roots have opposite signs (one positive, one negative).

Q8. Vieta's relations for a cubic ax³+bx²+cx+d.
A8. Σα = −b/a, Σαβ = c/a, αβγ = −d/a.

Q9. Condition for two quadratics to have both roots common.
A9. Their coefficients are proportional (a1/a2 = b1/b2 = c1/c2).

Q10. How to form an equation with roots that are reciprocals of a given one?
A10. Replace x by 1/x and clear denominators (reverses the coefficient order).

Q11. If a, b, c are real, what can be said about complex roots?
A11. They occur in conjugate pairs.

Quadratics are the first place JEE Advanced tests conditional reasoning — location of roots, sign of a quadratic over an interval, common roots — rather than plug-and-chug. The link between roots and coefficients (Vieta) plus the discriminant unlocks the whole topic.

Core concept: for ax² + bx + c = 0, the roots relate to coefficients (sum −b/a, product c/a) and the discriminant D = b² − 4ac determines their nature, while the graph (a parabola) governs sign and root-location conditions.

Core ideas

Beginner — nature of roots and Vieta

Roots α, β = (−b ± √D)/2a. Discriminant D = b² − 4ac: D > 0 real distinct, D = 0 real equal, D < 0 complex conjugate. Vieta: α + β = −b/a, αβ = c/a. Build symmetric functions from these (α² + β² = (α+β)² − 2αβ, etc.) without solving. For rational coefficients, irrational/complex roots occur in conjugate pairs.

Intermediate — sign, range and common roots

The sign of ax² + bx + c: same sign as a for all x if D < 0 (never crosses zero); between the roots it has the sign of −a. This gives "for what values of the parameter is the quadratic always positive?" (need a > 0 and D < 0). Common root of two quadratics: subtract to get a linear relation, or use the cross-multiplication (resultant) condition. For a common root with two equations, eliminate x² and x.

Advanced — location of roots

Using the parabola f(x) = ax² + bx + c (assume a > 0), root-location conditions relative to a number k or interval (p, q):

• Both roots > k: D ≥ 0, f(k) > 0, and vertex −b/2a > k.
• k between the roots: f(k) < 0 (a > 0) — the single cleanest condition.
• Both roots in (p, q): D ≥ 0, af(p) > 0, af(q) > 0, and p < −b/2a < q.
  These four ingredients (sign of D, sign of f at the point, position of the vertex) solve every root-location problem — a JEE-Advanced signature. For higher-degree "theory of equations," use Vieta's general relations (Σα = −a_{n−1}/a_n, etc.), the rational-root/transformation of roots, and Descartes' rule of signs.

Worked example

Find all values of k for which both roots of x² − 2kx + (k² − 1) = 0 lie between −2 and 4. Here a = 1 > 0, D = 4k² − 4(k²−1) = 4 > 0 (always two distinct real roots). Apply the three location conditions: f(−2) > 0: 4 + 4k + k² − 1 > 0 ⇒ k² + 4k + 3 > 0 ⇒ (k+1)(k+3) > 0 ⇒ k < −3 or k > −1. f(4) > 0: 16 − 8k + k² − 1 > 0 ⇒ k² − 8k + 15 > 0 ⇒ (k−3)(k−5) > 0 ⇒ k < 3 or k > 5. Vertex −b/2a = k in (−2, 4). Intersecting: −1 < k < 3. The three-condition recipe (f at both ends positive, vertex inside) is exactly the graded method.

How JEE Advanced tests this

Parameter conditions for roots real/equal/of a given sign; root-location relative to a number or interval (the f-value/vertex/discriminant trio); common-root problems; range of a rational function via "quadratic in x has real roots"; sign analysis of quadratics; and symmetric-function evaluation without solving.

Exam tricks & shortcuts

• k lies between the roots ⇔ a·f(k) < 0 (the single fastest test).
• Always positive ⇔ a > 0 and D < 0.
• Build α²+β², α³+β³ from Vieta, never by solving.
• For range of y = (quadratic)/(quadratic), cross-multiply and demand D ≥ 0 in x.

Key formulas (ax² + bx + c = 0)

• Roots: x = [−b ± √(b²−4ac)]/2a; discriminant D = b²−4ac.
• D > 0 real distinct, D = 0 equal, D < 0 complex conjugate.
• Sum α+β = −b/a; product αβ = c/a.
• Cubic (roots α,β,γ): Σα = −b/a, Σαβ = c/a, αβγ = −d/a.
• Nature: for real coefficients, complex/irrational roots occur in conjugate pairs.

Vieta's relations connect roots to coefficients; the discriminant classifies the nature of the roots.

Worked Example

Problem: The equations x² + ax + b = 0 and x² + bx + a = 0 (with a ≠ b) have a common root. Show that a + b + 1 = 0, and find the common root.

Solution:
Let α be the common root. Then it satisfies both equations:
α² + aα + b = 0 … (i)
α² + bα + a = 0 … (ii)

Subtract (ii) from (i):
(aα + b) − (bα + a) = 0
(a − b)α − (a − b) = 0
(a − b)(α − 1) = 0.

Since a ≠ b, we must have α − 1 = 0, so the common root is α = 1.

Substitute α = 1 into equation (i):
1 + a + b = 0, i.e. a + b + 1 = 0.

Answer: The common root is x = 1, and the coefficients satisfy a + b + 1 = 0.