Polynomials and Quadratic Equations
Polynomials and Quadratic Equations
A quadratic ax² + bx + c = 0 is fully described by two numbers: the sum of its roots, −b/a, and their product, c/a. NTPC asks expressions in the roots (α² + β², α⁴ + β⁴), the nature of roots from the discriminant, equations that share their roots, and the remainder and factor theorems for polynomials.
1. Formula Box
Roots of ax² + bx + c = 0: x = (−b ± √(b² − 4ac))/(2a)
Sum α + β = −b/a Product αβ = c/a
Equation with roots α, β: x² − (α + β)x + αβ = 0
Remainder theorem: p(x) divided by (x − a) leaves p(a); (x − a) is a factor when p(a) = 0
| Discriminant D = b² − 4ac | Nature of roots |
|---|---|
| D > 0 and a perfect square (a, b, c rational) | Real, distinct, rational |
| D > 0, not a perfect square | Real, distinct, irrational (pairs p ± √q when a, b, c are rational) |
| D = 0 | Real and equal, each −b/(2a) |
| D < 0 | No real roots |
2. Expressions in α and β
| Expression | In terms of S = α + β and P = αβ |
|---|---|
| α² + β² | S² − 2P |
| (α − β)² | S² − 4P |
| α³ + β³ | S³ − 3PS |
| α⁴ + β⁴ | (S² − 2P)² − 2P² |
| 1/α + 1/β | S/P |
| α/β + β/α | (S² − 2P)/P |
Two quadratics have both roots common when a₁/a₂ = b₁/b₂ = c₁/c₂
✗ Roots of 2x² − 5x + 1 = 0 have sum −5/2 | ✓ Sum = −b/a = −(−5)/2 = 5/2
✗ x² − kx + 9 = 0 has equal roots only for k = 6 | ✓ k² = 36 gives k = 6 or k = −6
हिंदी नोट: द्विघात समीकरण ax² + bx + c = 0 के मूलों का योग −b/a और गुणनफल c/a होता है। विविक्तकर b² − 4ac शून्य हो तो मूल बराबर, ऋणात्मक हो तो वास्तविक मूल नहीं। शेषफल प्रमेय: p(x) को (x − a) से भाग देने पर शेष p(a)।
Exam Pointer: Verified NTPC patterns: α⁴ + β⁴ type expressions from the sum and product of roots (June 2025 Graduate CBT-1) and two quadratics sharing both roots, with the coefficients in proportion (June 2025 Graduate CBT-1). Discriminant, factor-theorem and quadratic word problems had no verified NTPC shift and are tagged syllabus-based.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Sum and product of roots: symmetric expressions
[PYQ: NTPC Graduate CBT-1 21-Jun-2025 Shift-2]
EXAM LEVEL
Q. If α and β are the roots of x² − 7x + 10 = 0, find α² + β² and 1/α + 1/β.
S = 7 and P = 10. α² + β² = 49 − 20 = 29. 1/α + 1/β = S/P = 7/10. (The roots are 5 and 2.)
Answer: 29 and 7/10
EXAMATLAS LEVEL
Q. If α and β are the roots of 2x² − 5x + 1 = 0, find α/β + β/α and α³ + β³.
S = 5/2 and P = 1/2. α² + β² = 25/4 − 1 = 21/4, so α/β + β/α = (21/4) ÷ (1/2) = 21/2. α³ + β³ = S³ − 3PS = 125/8 − 15/4 = 95/8.
Answer: 21/2 and 95/8
Pattern 2: Nature of roots from the discriminant
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. For what values of k does x² − kx + 9 = 0 have equal roots?
D = k² − 36 = 0, so k = 6 or k = −6.
Answer: k = ±6
EXAMATLAS LEVEL
Q. Find the values of k for which kx² + 6x + 3 = 0 has real and distinct roots, and the greatest integer such k.
The equation must stay quadratic, so k ≠ 0, and D = 36 − 12k > 0 gives k < 3. So k < 3, k ≠ 0, and the greatest integer is 2 (check: 2x² + 6x + 3 has D = 12 > 0).
Answer: k < 3, k ≠ 0; greatest integer 2
Pattern 3: Forming equations and equations with the same roots
[PYQ: NTPC Graduate CBT-1 5-Jun-2025 Shift-1]
EXAM LEVEL
Q. Form the quadratic equation whose roots are 3 and −5.
Sum = −2 and product = −15, so x² − (−2)x + (−15) = 0, that is x² + 2x − 15 = 0.
Answer: x² + 2x − 15 = 0
EXAMATLAS LEVEL
Q. The equations px² + 10x + q = 0 and 2x² + 5x − 3 = 0 have the same roots. Find p and q, and the roots.
Both roots common means the coefficients are in proportion: p/2 = 10/5 = q/(−3), so p = 4 and q = −6. The roots of 2x² + 5x − 3 = (2x − 1)(x + 3) are 1/2 and −3.
Answer: p = 4, q = −6; roots 1/2 and −3
Pattern 4: Remainder and factor theorems
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the remainder when x³ − 2x² + 4x − 5 is divided by x − 2.
p(2) = 8 − 8 + 8 − 5 = 3.
Answer: 3
EXAMATLAS LEVEL
Q. Both x − 1 and x + 2 are factors of x³ + ax² + bx − 6. Find a and b and the third factor.
p(1) = 0 gives 1 + a + b − 6 = 0, so a + b = 5. p(−2) = 0 gives −8 + 4a − 2b − 6 = 0, so 2a − b = 7. Adding, 3a = 12, so a = 4 and b = 1. The product of the three roots is 6 (from −d/a), and 1 × (−2) × r = 6 gives r = −3, so the third factor is x + 3.
Answer: a = 4, b = 1; x + 3
Pattern 5: Solving quadratics and quadratic word problems
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Solve x² − 5x − 24 = 0.
Two numbers with product −24 and sum −5 are −8 and 3, so (x − 8)(x + 3) = 0 and x = 8 or −3.
Answer: 8 or −3
EXAMATLAS LEVEL
Q. The product of two consecutive positive odd numbers is 323. Find them and the sum of their squares.
n(n + 2) = 323 gives n² + 2n − 323 = 0 = (n − 17)(n + 19), so n = 17 and the numbers are 17 and 19. Sum of squares = 289 + 361 = 650.
Answer: 17 and 19; 650
60-Second Revision
- Sum −b/a, product c/a; build every expression from S and P.
- D = b² − 4ac decides equal, real or no roots; keep a ≠ 0.
- Same roots: all three coefficient ratios equal.
- p(a) is the remainder on division by (x − a); p(a) = 0 means a factor.