🎯 Key Points
- Quadratic formula: x = (-b ± √(b²-4ac)) / 2a — works for ANY quadratic, even when factorization fails
- Discriminant D = b²-4ac: D>0 → 2 distinct real roots; D=0 → 1 repeated root; D<0 → 2 complex conjugate roots
- Vieta's formulas: sum of roots α+β = -b/a, product αβ = c/a — lets you build equations from given roots without solving
- Vertex (minimum if a>0, maximum if a<0) occurs at x = -b/2a; irrational roots always come in conjugate pairs (p+√q and p-√q)
Quadratic Equations
Parabola y = x^2 - 4x + 3 with its two roots, vertex, and axis of symmetry marked.
A quadratic equation is of the form ax² + bx + c = 0 (where a ≠ 0). The solutions are called roots.
Methods of Solving
- Factorization: Write as (x - p)(x - q) = 0 and read off p, q as roots. Works when roots are rational.
- Quadratic formula: x = (-b ± √(b²-4ac)) / 2a: works always.
- Completing the square: Rearrange to (x + b/2a)² = (b²-4ac)/4a².
Discriminant (D = b² - 4ac)
- D > 0 : Two distinct real roots
- D = 0 : Two equal real roots (repeated root)
- D < 0 : No real roots (complex roots)
Vieta's Formulas
For roots α and β: α + β = -b/a and αβ = c/a. Use these to form equations given roots.
Nature of Roots
For rational roots, D must be a perfect square. If one root is irrational (like 2+√3), the other is its conjugate (2-√3).
Key Facts
- Maximum/minimum of ax² + bx + c is at x = -b/2a
- Sum of squares of roots: α² + β² = (α+β)² - 2αβ
- Every quadratic has exactly 2 roots (may be complex)
Forming a Quadratic Equation from Its Roots
If a quadratic has roots α and β, it can be reconstructed as x² − (α + β)x + αβ = 0, i.e. x² − (sum of roots)x + (product of roots) = 0. Any non-zero multiple k[x² − (α+β)x + αβ] = 0 is the same equation.
- Roots 2 and 5 → sum = 7, product = 10 → x² − 7x + 10 = 0.
- Roots 3 + √2 and 3 − √2 → sum = 6, product = 9 − 2 = 7 → x² − 6x + 7 = 0.
- To build an equation whose roots are related to α, β (say 1/α, 1/β or α², β²), first compute the new sum and product from Vieta's formulas, then plug into x² − (sum)x + product = 0.
Transforming Roots
- Reciprocal roots (1/α, 1/β): new sum = (α+β)/αβ, new product = 1/αβ. Shortcut: reverse the coefficients, so ax²+bx+c → cx²+bx+a.
- Roots increased by h (α+h, β+h): replace x by (x−h) in the original equation.
- Roots scaled by k (kα, kβ): replace x by x/k.
Symmetric Functions of the Roots
Expressions unchanged when α and β are swapped can be written using only the sum s = α+β = −b/a and product p = αβ = c/a, so the roots need never be found individually:
- α² + β² = s² − 2p
- (α − β)² = s² − 4p = D/a² → |α − β| = √D / |a|
- α³ + β³ = s³ − 3ps = s(s² − 3p)
- α² β + α β² = αβ(α + β) = ps
- 1/α + 1/β = s/p, 1/α² + 1/β² = (s² − 2p)/p²
Sign of a Quadratic Expression
The sign of f(x) = ax² + bx + c across the real line depends on the discriminant D and the leading coefficient a:
- D < 0: f(x) never changes sign — it keeps the sign of a for every real x (always positive if a>0, always negative if a<0).
- D = 0: f(x) keeps the sign of a everywhere except at the single repeated root x = −b/2a, where it is 0.
- D > 0: with real roots α < β, f(x) has the sign of a outside [α, β] and the opposite sign of a between the roots (α < x < β).
This is the basis of the "same sign as a outside the roots, opposite between the roots" rule used to solve quadratic inequalities like ax² + bx + c > 0.
Maximum and Minimum Value of a Quadratic
Since f(x) = ax² + bx + c = a(x + b/2a)² + (4ac − b²)/4a, the extreme value occurs at the vertex x = −b/2a and equals (4ac − b²)/4a = −D/4a.
- If a > 0 the parabola opens upward, so this is the minimum value and the range is [−D/4a, ∞).
- If a < 0 the parabola opens downward, so this is the maximum value and the range is (−∞, −D/4a].
- Example: for f(x) = 2x² − 8x + 3, x = −(−8)/(2·2) = 2 and minimum value = f(2) = 8 − 16 + 3 = −5.
🚀 JEE Advanced Edge
Common roots condition: Two quadratics a₁x²+b₁x+c₁=0 and a₂x²+b₂x+c₂=0 share BOTH roots only if their coefficients are proportional (a₁/a₂ = b₁/b₂ = c₁/c₂). If they share exactly ONE common root α, eliminating x² between the two equations gives α = (b₁c₂-b₂c₁)/(a₂b₁-a₁b₂) — a frequently tested elimination technique that avoids solving either quadratic directly.
Location of roots relative to a number: For f(x) = ax²+bx+c with a>0, both roots exceed a value k exactly when THREE conditions hold simultaneously: D≥0, f(k)>0, and the vertex x-coordinate -b/2a > k. Checking only the discriminant (a common mistake) is insufficient — all three conditions are needed together.
Worked problem: Find the value of k for which x²-(k+1)x+k²+k-8=0 has roots that are reciprocals of each other (i.e. product of roots = 1). Approach: Product of roots = c/a = k²+k-8. Setting this equal to 1: k²+k-8=1, so k²+k-9=0, giving k = (-1±√37)/2.
Worked Example: Sum and Product of Roots
If α and β are the roots of 2x² − 5x + 3 = 0, find α² + β² and α³ + β³.
By Vieta's formulas: α + β = 5/2, αβ = 3/2.
α² + β² = (α + β)² − 2αβ = 25/4 − 3 = 13/4.
α³ + β³ = (α + β)(α² − αβ + β²) = (5/2)(13/4 − 3/2) = (5/2)(7/4) = 35/8. Vieta's formulas eliminate the need to find α and β individually — use them whenever asked for symmetric expressions.
Worked Example: Equation Reducible to Quadratic
Solve x^(2/3) − 9x^(1/3) + 8 = 0.
Substitute t = x^(1/3): t² − 9t + 8 = 0 → (t − 1)(t − 8) = 0 → t = 1 or t = 8.
Reverse: x^(1/3) = 1 → x = 1; x^(1/3) = 8 → x = 512. Recognising a quadratic substitution is the key skill for this question type.