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Chapter 4: Quadratic Equations
Exercise & In-Text Questions & Answers

1Check whether the following are quadratic equations:
Answer & Reference:
Based on definition (Page 39): A quadratic equation is of the form ax² + bx + c = 0, where a ≠ 0.

(i) (x + 1)² = 2(x – 3)
⇒ x² + 2x + 1 = 2x – 6 ⇒ x² + 7 = 0 → Quadratic (Page 40, Example 2 style)
(ii) x² – 2x = (–2)(3 – x)
⇒ x² – 2x = –6 + 2x ⇒ x² – 4x + 6 = 0 → Quadratic
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
⇒ x² – x – 2 = x² + 2x – 3 ⇒ –3x + 1 = 0 → Not quadratic
(iv) (x – 3)(2x + 1) = x(x + 5)
⇒ 2x² – 5x – 3 = x² + 5x ⇒ x² – 10x – 3 = 0 → Quadratic
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
⇒ 2x² – 7x + 3 = x² + 4x – 5 ⇒ x² – 11x + 8 = 0 → Quadratic
(vi) x² + 3x + 1 = (x – 2)²
⇒ x² + 3x + 1 = x² – 4x + 4 ⇒ 7x – 3 = 0 → Not quadratic
(vii) (x + 2)³ = 2x(x² – 1)
⇒ x³ + 6x² + 12x + 8 = 2x³ – 2x ⇒ x³ – 6x² – 14x – 8 = 0 → Not quadratic
(viii) x³ – 4x² – x + 1 = (x – 2)³
⇒ RHS = x³ – 6x² + 12x – 8
⇒ 2x² – 13x + 9 = 0 → Quadratic (Page 41, Example 2(iv) style)
2Represent the following situations in the form of quadratic equations:
Answer & Reference:
(i) Let breadth = x m. Then length = 2x + 1 m.
Area = x(2x + 1) = 528 ⇒ 2x² + x – 528 = 0 (Page 38, similar to prayer hall)
(ii) Let integers be x and x + 1.
x(x + 1) = 306 ⇒ x² + x – 306 = 0 (Page 42)
(iii) Let Rohan’s age = x. Mother’s age = x + 26.
After 3 years: (x + 3)(x + 29) = 360 ⇒ x² + 32x – 273 = 0 (Page 42)
(iv) Let speed = x km/h. Time = 480/x.
New time: 480/(x – 8) = 480/x + 3
⇒ 480x = (x – 8)(480 + 3x) ⇒ 3x² – 24x – 3840 = 0 ⇒ x² – 8x – 1280 = 0 (Page 42)
3Find the roots of the following quadratic equations by factorisation:
Answer & Reference:
(i) x² – 3x – 10 = 0 ⇒ (x – 5)(x + 2) = 0 ⇒ Roots: 5, –2 (Page 43, Example 3 style)
(ii) 2x² + x – 6 = 0 ⇒ (2x – 3)(x + 2) = 0 ⇒ Roots: 3/2, –2 (Page 43, Example 4)
(iii) √2 x² + 7x + 5√2 = 0 ⇒ (√2 x + 5)(x + √2) = 0 ⇒ Roots: –5/√2, –√2 (Page 43, Example 5)
(iv) 2x² – x + 1/8 = 0 ⇒ Multiply by 8: 16x² – 8x + 1 = 0 ⇒ (4x – 1)² = 0 ⇒ Root: 1/4 (repeated)
(v) 100x² – 20x + 1 = 0 ⇒ (10x – 1)² = 0 ⇒ Root: 1/10 (repeated) (Page 44)
4Solve the problems given in Example 1.
Answer & Reference:
(i) Equation: x² – 45x + 324 = 0 ⇒ (x – 36)(x – 9) = 0 ⇒ John had 36 or 9 marbles (Page 40)
(ii) Equation: x² – 55x + 750 = 0 ⇒ (x – 25)(x – 30) = 0 ⇒ Toys produced: 25 or 30 (Page 40)
5Find two numbers whose sum is 27 and product is 182.
Answer & Reference:
Let numbers be x and 27 – x.
x(27 – x) = 182 ⇒ x² – 27x + 182 = 0 ⇒ (x – 13)(x – 14) = 0
Numbers: 13 and 14 (Page 44, Exercise 4.2 Q3 style)
6Find two consecutive positive integers, sum of whose squares is 365.
Answer & Reference:
Let integers be x and x + 1.
x² + (x + 1)² = 365 ⇒ 2x² + 2x – 364 = 0 ⇒ x² + x – 182 = 0
⇒ (x + 14)(x – 13) = 0 ⇒ x = 13 (positive)
Integers: 13 and 14 (Page 44)
7The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Answer & Reference:
Let base = x cm, altitude = x – 7 cm.
By Pythagoras: x² + (x – 7)² = 13² ⇒ 2x² – 14x – 120 = 0 ⇒ x² – 7x – 60 = 0
⇒ (x – 12)(x + 5) = 0 ⇒ x = 12
Sides: 12 cm (base), 5 cm (altitude) (Page 44, similar to Example 8)
8A cottage industry produces pottery articles. Cost per article = ₹(2x + 3), total cost = ₹90. Find number of articles and cost per article.
Answer & Reference:
x(2x + 3) = 90 ⇒ 2x² + 3x – 90 = 0 ⇒ (2x + 15)(x – 6) = 0 ⇒ x = 6
Articles: 6, Cost per article: ₹15 (Page 44, Exercise 4.2 Q6)
9Find the nature of the roots of the following quadratic equations. If real roots exist, find them:
Answer & Reference:
Discriminant D = b² – 4ac (Page 44–45)
(i) 2x² – 3x + 5 = 0 ⇒ D = 9 – 40 = –31 < 0 → No real roots (Page 47, Ex 4.3 Q1(i))
(ii) 3x² – 4√3 x + 4 = 0 ⇒ D = 48 – 48 = 0 → Equal roots: x = (4√3)/(2×3) = 2/√3 (Page 47, Ex 4.3 Q1(ii))
(iii) 2x² – 6x + 3 = 0 ⇒ D = 36 – 24 = 12 > 0 → Real roots: x = (6 ± √12)/4 = (3 ± √3)/2 (Page 47)
10Find the values of k for which the following equations have two equal roots:
Answer & Reference:
For equal roots, D = 0 (Page 45)
(i) 2x² + kx + 3 = 0 ⇒ k² – 24 = 0 ⇒ k = ±2√6 (Page 47, Ex 4.3 Q2(i))
(ii) kx(x – 2) + 6 = 0 ⇒ kx² – 2kx + 6 = 0 ⇒ 4k² – 24k = 0 ⇒ k(k – 6) = 0
But k ≠ 0 (else not quadratic), so k = 6 (Page 47)
11Is it possible to design a rectangular mango grove whose length is twice its breadth, and area is 800 m²? If so, find its dimensions.
Answer & Reference:
Let breadth = x, length = 2x.
2x² = 800 ⇒ x² = 400 ⇒ x = 20 (positive)
Yes. Breadth = 20 m, Length = 40 m (Page 47, Ex 4.3 Q3)
12Sum of ages of two friends is 20 years. Four years ago, product of their ages was 48. Is this possible?
Answer & Reference:
Let ages be x and 20 – x.
Four years ago: (x – 4)(16 – x) = 48 ⇒ –x² + 20x – 64 = 48 ⇒ x² – 20x + 112 = 0
D = 400 – 448 = –48 < 0 → No real roots → Not possible (Page 47, Ex 4.3 Q4)
13Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find length and breadth.
Answer & Reference:
Perimeter = 2(l + b) = 80 ⇒ l + b = 40. Let b = x, then l = 40 – x.
Area: x(40 – x) = 400 ⇒ x² – 40x + 400 = 0 ⇒ (x – 20)² = 0
So, x = 20. Yes. Length = Breadth = 20 m (square park) (Page 47, Ex 4.3 Q5)
Fig. 4.1: Prayer Hall Breadth = x m Length = (2x+1) m
Fig. 4.1: Rectangular prayer hall with length = (2x + 1) m and breadth = x m (Page 38)
A B P x m (x+7) m Circular Park (Diameter = 13 m)
Fig. 4.2: Pole P on boundary of circular park; AB is diameter, ∠APB = 90° (Page 45–46)
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