Chapter 4: Quadratic Equations

Mathematics · CBSE Class 10 · NCERT Solutions

📖 Chapter 4 — Quadratic Equations (CBSE Class 10 Maths): Detailed notes — standard form, solving by factorisation & the quadratic formula, and the nature of roots.

Chapter 4 — Quadratic Equations

🔢 1. Standard Form

A quadratic equation is ax² + bx + c = 0, where a, b, c are real and a ≠ 0. A value of x that satisfies it is a root (or solution/zero).

✖️ 2. Solving by Factorisation

Split the middle term so that ax² + bx + c = 0 becomes a product of two factors = 0; then set each factor to 0. e.g. x² − 3x − 10 = (x−5)(x+2) = 0 → x = 5 or −2.

📐 3. Quadratic Formula

x = [ −b ± √(b² − 4ac) ] / 2a

🎯 4. Discriminant & Nature of Roots

Discriminant D = b² − 4ac:

Discriminant	Nature of roots
D > 0	Two distinct real roots
D = 0	Two equal real roots (−b/2a)
D < 0	No real roots

💡 5. Word Problems

Assign a variable → form a quadratic equation from the condition → solve → reject any answer that is not physically valid (e.g. a negative length).

🧠 Remember: a ≠ 0. Factorise by splitting the middle term (product = ac, sum = b). Formula x = (−b ± √D)/2a. D = b²−4ac decides the nature: >0 distinct, =0 equal, <0 none. For equal roots put D = 0.

⭐ Exam Tips: ① Check if an equation is quadratic (simplify first). ② Roots by factorisation. ③ Roots by the quadratic formula. ④ Nature of roots using D. ⑤ Find k for equal roots (D=0). ⑥ Word problems (area, age, speed, numbers).

📘 EXERCISE 4.1

1Check whether (x+1)² = 2(x−3) is a quadratic equation.

Expand both sides: LHS (x+1)² = x² + 2x + 1 ; RHS 2(x−3) = 2x − 6. Bring all terms to one side: x² + 2x + 1 − 2x + 6 = 0 → x² + 7 = 0. Highest power of x is 2 (and coefficient ≠ 0) → it IS a quadratic equation. ∴ Quadratic. [ (iii) & (vi) reduce to linear → NOT quadratic ; (vii) reduces to cubic → NOT quadratic ; rest are quadratic]

2Represent as a quadratic equation: plot area 528 m², length = one more than twice the breadth.

Let breadth = x m, then length = (2x + 1) m. Area = length × breadth = x(2x + 1) = 528. Expand: 2x² + x = 528 → 2x² + x − 528 = 0. ∴ 2x² + x − 528 = 0. [consecutive integers: x²+x−306=0 ; Rohan & mother: x²+32x−273=0 ; train: x²−8x−1280=0]

📗 EXERCISE 4.2 (Factorisation)

1Find the roots of 2x² + x − 6 = 0 by factorisation.

a = 2, b = 1, c = −6 Split the middle term: product = a×c = 2×(−6) = −12; two numbers that multiply to −12 and add to +1 are +4 and −3. 2x² + 4x − 3x − 6 = 2x(x+2) − 3(x+2) = (x + 2)(2x − 3). Set each factor = 0: x + 2 = 0 → x = −2 ; 2x − 3 = 0 → x = 3/2. ∴ Roots are 3/2 and −2. [others: x²−3x−10 → 5,−2 ; 100x²−20x+1 → 1/10,1/10]

3Find two numbers whose sum is 27 and product is 182.

Let one number be x, then the other is (27 − x). Product: x(27 − x) = 182 → 27x − x² = 182 → x² − 27x + 182 = 0. Factorise: (x − 13)(x − 14) = 0 → x = 13 or 14. ∴ The numbers are 13 and 14.

5The altitude of a right triangle is 7 cm less than its base; the hypotenuse is 13 cm. Find the base and altitude.

Let base = b cm, then altitude = (b − 7) cm. By Pythagoras: base² + altitude² = hypotenuse² → b² + (b−7)² = 13². b² + b² − 14b + 49 = 169 → 2b² − 14b − 120 = 0 → b² − 7b − 60 = 0. (b − 12)(b + 5) = 0 → b = 12 (reject −5, a length can't be negative). ∴ Base = 12 cm, altitude = 5 cm.

📙 EXERCISE 4.3 (Formula & Nature of Roots)

1Find the nature of the roots of 2x² − 6x + 3 = 0; if real, find them.

a = 2, b = −6, c = 3 Discriminant D = b² − 4ac = (−6)² − 4(2)(3) = 36 − 24 = 12. D = 12 > 0 → two distinct real roots. x = (−b ± √D)/2a = (6 ± √12)/4 = (6 ± 2√3)/4 = (3 ± √3)/2. ∴ Distinct real roots: (3 + √3)/2 and (3 − √3)/2. [2x²−3x+5: D=−31 → no real roots ; 3x²−4√3x+4: D=0 → equal roots 2/√3]

2Find k so that 2x² + kx + 3 = 0 has two equal roots.

Equal roots ⇒ discriminant D = 0. D = k² − 4(2)(3) = k² − 24 = 0 → k² = 24. k = ±√24 = ±2√6. ∴ k = ±2√6. [kx(x−2)+6=0 → k=6]

3Can a rectangular mango grove have length twice its breadth and area 800 m²? Find its dimensions.

Let breadth = b, then length = 2b. Area = length × breadth = 2b × b = 2b² = 800 → b² = 400. b = 20 (reject −20). Then length = 2b = 40. (Since real roots exist, it IS possible.) ∴ Yes; breadth = 20 m, length = 40 m.

Chapter 4: Quadratic Equations