1. Chapter at a Glance
A quadratic equation is an equation of the form ax² + bx + c = 0 with a ≠ 0. Almost every chapter of JEE mathematics — coordinate geometry, calculus, probability — eventually leads to a quadratic. JEE Main questions focus on relations between roots and coefficients, nature of roots, location of roots and equations with modulus.
- Values of symmetric expressions such as α² + β², α³ + β³ or αn + βn.
- Range of a parameter for which roots are real, equal, of opposite signs, or lie in an interval.
- Common roots of two quadratics.
- Number of real solutions of equations like x² − 5|x| + 6 = 0.
2. Roots and the Discriminant
The roots of ax² + bx + c = 0 are
The discriminant D decides the nature of the roots (when a, b, c are real):
| Discriminant | Nature of roots | Graph of y = ax² + bx + c |
|---|---|---|
| D > 0 | Real and distinct | Cuts the x-axis at two points |
| D = 0 | Real and equal (x = −b/2a) | Touches the x-axis |
| D < 0 | Non-real, complex conjugates | Does not meet the x-axis |
| D a perfect square (a, b, c rational) | Rational roots | — |
3. Relation between Roots and Coefficients
If α and β are the roots of ax² + bx + c = 0, then
A quadratic with roots α, β is x² − (α + β)x + αβ = 0. Symmetric expressions can be written in terms of the sum S and product P:
- α² + β² = S² − 2P
- α³ + β³ = S³ − 3PS
- (α − β)² = S² − 4P, so |α − β| = √D/|a|
- 1/α + 1/β = S/P
- α⁴ + β⁴ = (α² + β²)² − 2P²
If α, β are roots of x² − Sx + P = 0 and Tn = αn + βn, then
with T0 = 2 and T1 = S. This is the fastest way to find α⁵ + β⁵ or to simplify expressions like (a10 − 2a8)/a9.
Cubic equations
For ax³ + bx² + cx + d = 0 with roots α, β, γ: Σα = −b/a, Σαβ = c/a, αβγ = −d/a.
4. Location of Roots
Let f(x) = ax² + bx + c with a > 0 and real roots α ≤ β. Use three tools: the sign of D, the sign of f at a given point, and the position of the vertex −b/2a.
| Condition on the roots | Conditions (a > 0) |
|---|---|
| Both roots greater than k | D ≥ 0, f(k) > 0, −b/2a > k |
| Both roots less than k | D ≥ 0, f(k) > 0, −b/2a < k |
| k lies between the roots | f(k) < 0 (D > 0 is then automatic) |
| Exactly one root in (k1, k2) | f(k1)·f(k2) < 0 |
| Both roots in (k1, k2) | D ≥ 0, f(k1) > 0, f(k2) > 0, k1 < −b/2a < k2 |
| Roots of opposite signs | c/a < 0 |
| Both roots positive | D ≥ 0, −b/a > 0, c/a > 0 |
Sign of a quadratic
- ax² + bx + c > 0 for all real x ⇔ a > 0 and D < 0.
- ax² + bx + c < 0 for all real x ⇔ a < 0 and D < 0.
- Minimum value (a > 0) is −D/4a at x = −b/2a; maximum value (a < 0) is −D/4a.
5. Common Roots and Special Equations
One common root of a1x² + b1x + c1 = 0 and a2x² + b2x + c2 = 0 requires
In practice it is quicker to subtract the equations to find the common root first. Both roots common means a1/a2 = b1/b2 = c1/c2.
Equations with modulus
Since x² = |x|², an equation like x² − 5|x| + 6 = 0 is a quadratic in t = |x| with t ≥ 0. Each positive value of t gives two values of x (±t), and t = 0 gives one.
Reciprocal-type substitutions
Equations such as x⁴ + ax³ + bx² + ax + 1 = 0 are solved by dividing by x² and putting t = x + 1/x.
6. Formula Master Table
| Result | Formula |
|---|---|
| Roots | (−b ± √(b² − 4ac))/2a |
| Sum and product | α + β = −b/a, αβ = c/a |
| Difference of roots | |α − β| = √D/|a| |
| Sum of squares | α² + β² = (α + β)² − 2αβ |
| Sum of cubes | α³ + β³ = (α + β)³ − 3αβ(α + β) |
| Equation from roots | x² − (α + β)x + αβ = 0 |
| Vertex of parabola | (−b/2a, −D/4a) |
| Power sums | Tn = S·Tn−1 − P·Tn−2 |
| Roots of reciprocal equation | cx² + bx + a = 0 has roots 1/α, 1/β |
| Roots increased by k | replace x by (x − k) |
| Roots multiplied by k | replace x by x/k |
7. Mind Map
- Formula (−b ± √D)/2a
- D decides nature
- Complex roots in pairs
- α + β = −b/a, αβ = c/a
- Symmetric expressions
- Newton's recurrence
- Sign of f(k)
- Vertex −b/2a
- Both roots in an interval
- Common roots
- Modulus equations
- Max/min value −D/4a
8. Tips and Tricks
- If a + b + c = 0, then x = 1 is a root and the other root is c/a.
- If a − b + c = 0, then x = −1 is a root and the other root is −c/a.
- To find a parameter for a common root, subtract the two equations — the x² terms cancel and you get the common root directly.
- For "for all real x" conditions, remember the pair (sign of a, D < 0). No need to find the roots.
- Draw a rough parabola for every location-of-roots question; the conditions become obvious from the picture.
- Forgetting the condition a ≠ 0. If the coefficient of x² contains the parameter, check separately what happens when it is zero.
- Using D ≥ 0 when the question asks for "distinct" real roots (need D > 0).
- In modulus equations, forgetting that a negative value of |x| must be rejected.
9. Important One-liners
- A quadratic equation cannot have more than two roots; if it has three, it is an identity (all coefficients zero).
- If the coefficients are real and one root is p + iq, the other is p − iq.
- If the coefficients are rational and one root is p + √q, the other is p − √q.
- If c and a have opposite signs, the roots are real and of opposite signs.
- If both roots are equal, the quadratic is a perfect square: a(x + b/2a)².
- The graph of y = ax² + bx + c is symmetric about the line x = −b/2a.
- Roots are reciprocals of each other when c = a.
- Roots are equal in magnitude and opposite in sign when b = 0.
10. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
If α and β are the roots of x² − 5x + 3 = 0, find α³ + β³.
Show solution
S = α + β = 5, P = αβ = 3.
α³ + β³ = S³ − 3PS = 125 − 45 = 80.
Answer: 80
The values of k for which x² + kx + 4 = 0 has real and equal roots are:
- k = ±2
- k = ±4
- k = 4 only
- k = ±8
Show solution
Equal roots need D = 0: k² − 16 = 0, so k = ±4.
Answer: (B)
The set of all real values of k for which both roots of x² − 2kx + k² + k − 5 = 0 are less than 5 is:
- (−∞, 4)
- (4, 5]
- (−∞, 5]
- (5, ∞)
Show solution
Let f(x) = x² − 2kx + k² + k − 5 (a = 1 > 0). Conditions:
(i) D ≥ 0: 4k² − 4(k² + k − 5) ≥ 0 ⇒ −4k + 20 ≥ 0 ⇒ k ≤ 5.
(ii) f(5) > 0: 25 − 10k + k² + k − 5 = k² − 9k + 20 = (k − 4)(k − 5) > 0 ⇒ k < 4 or k > 5.
(iii) Vertex −b/2a = k < 5.
Common part: k < 4.
Answer: (A) (−∞, 4)
If the equations x² + ax + 1 = 0 and x² + x + a = 0 have exactly one common root, then a equals:
- 1
- −2
- 2
- 0
Show solution
Subtracting: (a − 1)x + (1 − a) = 0 ⇒ (a − 1)(x − 1) = 0. If a = 1 the two equations are identical (both roots common), so a ≠ 1 and the common root is x = 1.
Putting x = 1 in x² + x + a = 0: 1 + 1 + a = 0 ⇒ a = −2.
Answer: (B) −2
Find the number of real solutions of x² − 5|x| + 6 = 0.
Show solution
Put t = |x| ≥ 0: t² − 5t + 6 = 0 ⇒ t = 2 or t = 3. Both are positive, so x = ±2 and x = ±3.
Answer: 4
If α and β are the roots of x² − x − 1 = 0, find α⁵ + β⁵.
Show solution
Here S = 1 and P = −1, so Tn = Tn−1 + Tn−2 with T0 = 2, T1 = 1.
T2 = 3, T3 = 4, T4 = 7, T5 = 11.
Answer: 11
The set of values of m for which x² − mx + 4 > 0 for all real x is:
- (−4, 4)
- [−4, 4]
- (−∞, −4) ∪ (4, ∞)
- (0, 4)
Show solution
The leading coefficient is 1 > 0, so we need D < 0: m² − 16 < 0 ⇒ −4 < m < 4.
At m = ±4 the expression becomes a perfect square and equals 0 at one point, so the endpoints are excluded.
Answer: (A)
If α and β are the roots of x² − 4x − 3 = 0 and an = αn − βn, then (a10 − 3a8)/(2a9) equals:
- 4
- −2
- 2
- 1
Show solution
Since α² = 4α + 3, multiply by α8: α10 = 4α9 + 3α8. Similarly β10 = 4β9 + 3β8.
Subtracting: a10 = 4a9 + 3a8 ⇒ a10 − 3a8 = 4a9.
So (a10 − 3a8)/(2a9) = 4a9/(2a9) = 2.
Answer: (C) 2
11. Quick Revision Summary
- D = b² − 4ac: > 0 distinct real, = 0 equal, < 0 complex conjugate roots.
- α + β = −b/a, αβ = c/a; build every symmetric expression from these two.
- Tn = S Tn−1 − P Tn−2 for power sums.
- Location of roots: sign of D, sign of f(k), position of the vertex.
- Always positive ⇔ a > 0 and D < 0; min value −D/4a at x = −b/2a.
- Common root: subtract the equations; modulus equations: substitute t = |x| ≥ 0.