CLASS 10 · MATHEMATICS · CHAPTER 4
Recognise quadratic equations, form them from situations, solve by factorisation and use the discriminant to test whether real solutions exist.
13 worked answers · short steps first, explanations when you need them
Textbook pages 41–42
Question 1
(i) x² + 7 = 0: quadratic
(ii) x² − 4x + 6 = 0: quadratic
(iii) −3x + 1 = 0: not quadratic
(iv) x² − 10x − 3 = 0: quadratic
(v) x² − 11x + 8 = 0: quadratic
(vi) 7x − 3 = 0: not quadratic
(vii) x³ − 6x² − 14x − 8 = 0: not quadratic
(viii) 2x² − 13x + 9 = 0: quadratic
Quadratic:
(iv), (v), (viii). Not quadratic: (iii), (vi), (vii).
Question 2
(i) Breadth x: x(2x + 1) = 528 ⇒ 2x² + x − 528 = 0
(ii) First integer x: x(x + 1) = 306 ⇒ x² + x − 306 = 0
(iii) Rohan's age x: (x + 3)(x + 29) = 360 ⇒ x² + 32x − 273 = 0
(iv) Speed x: 480/(x − 8) − 480/x = 3 ⇒ x² − 8x − 1280 = 0
The equations are 2x²+x−528=0; x²+x−306=0; x²+32x−273=0; x²−8x−1280=0.
Textbook pages 44
Question 1
(i) (x − 5)(x + 2) = 0 ⇒ x = 5, −2
(ii) (2x − 3)(x + 2) = 0 ⇒ x = 3/2, −2
(iii) (2x + 5)(x + 1) = 0 ⇒ x = −5/2, −1
(iv) (4x − 1)²/8 = 0 ⇒ x = 1/4, 1/4
(v) (10x − 1)² = 0 ⇒ x = 1/10, 1/10
Question 2
Marbles: x² − 45x + 324 = (x − 9)(x − 36) = 0
They began with 9 and 36 marbles, in either order.
Toys: x² − 55x + 750 = (x − 25)(x − 30) = 0
25 toys at ₹30 each or 30 toys at ₹25 each.
Marbles: 9 and 36. Toys: either 25 at ₹30 each or 30 at ₹25 each.
Question 3
x(27 − x) = 182 ⇒ x² − 27x + 182 = 0
(x − 13)(x − 14) = 0
13 and 14.
Question 4
x² + (x + 1)² = 365 ⇒ x² + x − 182 = 0
(x − 13)(x + 14) = 0
13 and 14.
Question 5
Let base = x; altitude = x − 7
x² + (x − 7)² = 13² ⇒ x² − 7x − 60 = 0
(x − 12)(x + 5) = 0
Base 12 cm; altitude 5 cm.
Question 6
n(2n + 3) = 90 ⇒ 2n² + 3n − 90 = 0
(2n + 15)(n − 6) = 0
6 articles at ₹15 each.
Textbook pages 47
Question 1
(i) D = (−3)² − 4·2·5 = −31: no real roots
(ii) D = (−4√3)² − 4·3·4 = 0: equal roots 2√3/3
(iii) D = (−6)² − 4·2·3 = 12: roots (3 ± √3)/2
Question 2
(i) k² − 24 = 0 ⇒ k = ±2√6
(ii) D = 4k² − 24k = 4k(k − 6) = 0
k = 0 would not be quadratic, so k = 6.
Question 3
2b² = 800 ⇒ b² = 400 ⇒ b = 20
Yes. Breadth 20 m; length 40 m.
Question 4
Let ages be x and 20 − x.
(x − 4)(16 − x) = 48 ⇒ x² − 20x + 112 = 0
D = 400 − 448 = −48 < 0
No; there are no real ages satisfying both conditions.
Question 5
l + b = 40 and lb = 400
b(40 − b) = 400 ⇒ (b − 20)² = 0
Yes. It is 20 m by 20 m.