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Complete NCERT Solutions for CBSE Class 10 Maths Chapter 4 — Quadratic Equations: solving by factorisation, the quadratic formula, the nature of roots (discriminant), and word problems.
Key Concepts (Quick Recap)
- Standard form: ax² + bx + c = 0, a ≠ 0.
- Quadratic formula: x = (−b ± √(b² − 4ac)) / 2a.
- Discriminant D = b² − 4ac: D > 0 → two distinct real roots; D = 0 → equal roots; D < 0 → no real roots.
Exercise 4.2 — Factorisation Method
Find the roots of the following by factorisation.
(i) x² − 3x − 10 = 0 (x − 5)(x + 2) = 0 → x = 5 or x = −2.
(ii) 2x² + x − 6 = 0 (2x − 3)(x + 2) = 0 → x = 3/2 or x = −2.
(iii) √2 x² + 7x + 5√2 = 0 (√2 x + 5)(x + √2) = 0 → x = −5/√2 or x = −√2.
(iv) 2x² − x + 1/8 = 0 Multiply by 8: 16x² − 8x + 1 = 0 → (4x − 1)² = 0 → x = 1/4, 1/4.
(v) 100x² − 20x + 1 = 0 (10x − 1)² = 0 → x = 1/10, 1/10.
Exercise 4.3 — Quadratic Formula & Nature of Roots
A. Find the roots using the quadratic formula (where they exist).
(i) 2x² − 7x + 3 = 0 D = 49 − 24 = 25; x = (7 ± 5)/4 → x = 3 or x = 1/2.
(ii) 2x² + x − 4 = 0 D = 1 + 32 = 33; x = (−1 ± √33)/4.
(iii) 4x² + 4√3 x + 3 = 0 D = 48 − 48 = 0; x = −4√3 / 8 = −√3/2 (equal roots).
(iv) 2x² + x + 4 = 0 D = 1 − 32 = −31 < 0 → no real roots.
B. Find the nature of the roots; if real, find them.
(i) 2x² − 3x + 5 = 0 D = 9 − 40 = −31 < 0 → no real roots.
(ii) 3x² − 4√3 x + 4 = 0 D = 48 − 48 = 0 → equal real roots; x = 4√3 / 6 = 2/√3, 2/√3.
(iii) 2x² − 6x + 3 = 0 D = 36 − 24 = 12 > 0 → two distinct real roots; x = (6 ± √12)/4 = (3 ± √3)/2.
Word Problems (Application)
Q1. The product of two consecutive positive integers is 306. Find the integers. Let them be n and n + 1: n(n + 1) = 306 → n² + n − 306 = 0 → (n + 18)(n − 17) = 0 → n = 17. Integers: 17 and 18.
Q2. The sum of the squares of two consecutive positive integers is 365. Find them. n² + (n + 1)² = 365 → 2n² + 2n − 364 = 0 → n² + n − 182 = 0 → (n + 14)(n − 13) = 0 → n = 13. Integers: 13 and 14.
Q3. The sum of a number and its reciprocal is 10/3. Find the number. x + 1/x = 10/3 → 3x² − 10x + 3 = 0 → (3x − 1)(x − 3) = 0 → x = 3 or x = 1/3.
Q4. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides. Let base = x, altitude = x − 7. Then x² + (x − 7)² = 13² → 2x² − 14x + 49 = 169 → x² − 7x − 60 = 0 → (x − 12)(x + 5) = 0 → x = 12. Base = 12 cm, altitude = 5 cm.
Common Mistakes
- Using the formula without writing the equation in standard form first.
- Sign error in the discriminant (b² − 4ac, with b² always positive).
- Rejecting a valid root, or keeping a negative/invalid root in a word problem (lengths and counts must be positive).
Exam Tips
- Always compute D first — it tells you immediately whether real roots exist.
- For "find the value of k for equal roots," set D = 0 and solve for k.
- In word problems, discard the root that does not fit the context (e.g., negative length).